I believe you're not paying attention to what I'm saying. I'm not saying a set can't be a member of more than one set. I am saying:

A) When a set is not a member of itself, it is a member of another set
B) When a set is a member of itself, it is not a member of another set — Philosopher19

See again my replies to you:

Evidently, in A, A is a member of itself.
Evidently, in B, A is not a member of itself because A is a member of B. — Philosopher19

You are not a set. You can never be a member of yourself. But a set can either be a member of itself or a member of other than itself. And it is logically the case that when a set is a member of itself, it is a member of itself. And when it is a member of other than itself (as was the case with A in B), it is not a member of itself. — Philosopher19

A is a member of both A and B.

I'll explain it to you in non-math terms:

I am a member of the football team and a member of the tennis team.

These are two different claims:

1. I am not a member of the football team
2. I am a member of a non-football team

(1) is false and (2) is true. — Michael

So at least you're engaging me in clear meaningful language. I will respond in kind.

You are not a set. You can never be a member of yourself. But a set can either be a member of itself or a member of other than itself. And it is logically the case that when a set is a member of itself, it is a member of itself. And when it is a member of other than itself (as was the case with A in B), it is not a member of itself.

So your non-math example doesn't apply because you are talking about something that by definition can't be a member of itself, whereas in your A and B example, you were talking about something that by definition was a member of itself in its own set whilst a member of other than itself in another set.

I repeat:

A) When a set is not a member of itself, it is a member of another set
B) When a set is a member of itself, it is not a member of another set — Philosopher19

Regarding Russell's paradox, it is simply this:

1. x is a member of R if and only if x is not a member of x.

Is R a member of R?

Either answer entails a contradiction, and so (1) is a contradiction. Given that naive set theory entails (1), naive set theory is shown to be inconsistent. — Michael

I understand Russell's paradox. Here is what I say in response:

z = any set that is not the set of all sets
v = any set
The v of all vs = the set of all sets
The z of all zs = The not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets — Philosopher19

Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different standards/contrexts as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?) — Philosopher19

These are two different claims:

1. A is not a member of itself
2. A is a member of some other set

Given this:

A={A}B={A,0}

(1) is false and (2) is true. — Michael

Evidently, in A, A is a member of itself.
Evidently, in B, A is not a member of itself because A is a member of B.

So you have not shown that 1 and 2 are two different claims.

I repeat:

A) When a set is not a member of itself, it is a member of another set
B) When a set is a member of itself, it is not a member of another set

THAT is dogma. You have no proof of it, and I gave an exact disproof of it. — TonesInDeepFreeze

My proof was here:

z = any set that is not the set of all sets
v = any set
The v of all vs = the set of all sets
The z of all zs = The not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets — Philosopher19

Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different standards/contrexts as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?) — Philosopher19

As if "the rejection of the set of all sets is by definition contradictory" is not proof enough.
Or as if "a set is not a member of itself when it is a member of another set" is not proof enough.

I think you can understand this if, for a few moments, you clear your mind of the voice in it that keeps saying "I am right. I know I am right. I must be right. All the logicians and mathematicians are wrong and I am right", then very carefully, very slowly, consider: — TonesInDeepFreeze

I tried to look at your reply, but it is unclear to me as to what it's doing. I believe it deliberately strays from what is clear simple language to try and force something that cannot be forced (perhaps due to dogma).

A) When a set is not a member of itself, it is a member of a set other than itself.
B) When a set is a member of itself, it is not a member of another set.

Rejection of either A or B is blatantly contradictory, yet, you seem to be arguing that rejection of B is not contradictory. Until you acknowledge that the rejection of B is contradictory, I don't see how we can progress.

Is there any way you can provide an example of a set that IS a member of itself, other than the set of all sets, in plain language? I can't think of one. — Fire Ologist

That's because speaking in absolute terms, only the set of all sets is a member of itself.

In short, if we were to focus on absolutely all sets, then only the set of all sets is a member of itself. However, if we were to only focus on all sets other than the set of all sets, then another set is a member of itself (but then we are not speaking in absolute/complete terms). I will try and show this to you:

Call any set a v.
Call the set of all sets the v of all vs.
If we were to focus on all sets, we would be focused on all vs.

Now call any set that is not the set of all sets a z.
If we were to focus only on all sets other than the set of all sets, we would be focused on all zs as opposed to all vs.

Since z = any set that is not-the-set-of-all-sets, the z of all zs means "the not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets. The z of all zs, is a member of itself as a z, but it is not a member of itself as a v. As a v, it is a member of the v of all vs.

So if we were to talk about all vs, then only the set of all sets is a member of itself.
If we were to only talk about all zs, then only the z of all zs is a member of itself.

I'm not sure I get what I'm saying. — Fire Ologist

I think your instincts/intuition is in the right place (or at least trying to get to the right place)

Help me out. Besides the set of all sets, what is an example of a set that is a member of itself? — Fire Ologist

I recommend the following: http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/

So revised: Let x not equal b. Let b not be a member of x. Let x be a member of x. But x is a member of {x b}. So x is a member of x and x is a member of a set different from x, viz {x b}.

That refutes your claim. — TonesInDeepFreeze

How does this show there is a meaningful/semantical difference between 1 and 2? Therefore, how has this refuted my claim? It is clear that "not itself" and "other than itself" mean the same thing.

Wrong. You cannot produce a valid demonstration that — TonesInDeepFreeze

I did. I think it was very clear. Here it is again in a numbered format. Tell me which number doesn't follow from (or is irrelevant to) which number or tell me which number is wrong if you are sincere in this discussion.

1) There is no meaningful/logical difference between "not a member of itself" and "a member of other than itself".

2) A set cannot be a member of itself and not a member of itself

3) If a set is a member of itself, it is not a member of other than itself (precisely because it is a member of itself)

4) If a set is a member of itself, it is not a member of another set (precisely because it is a member of itself. If it is a member of another set, it is not a member of itself precisely because it is a member of another set).

The 1-4 point I'm making is clear. If I get a reasonable/meaningful response to it, I believe I will respond to that response.

What you skipped is my refutation (posted twice) of your claim that a set can't be both a member of itself and a member of another set. — TonesInDeepFreeze

You agreed with me that a set cannot be both a member of itself and not a member of itself. You said it was an important point in your refutation to me. The above logically implies you are rejecting it. First look at the part I underlined in the quote above. Then look at how there is no difference between:

1) A member of other than itself (which is the same as saying a member of another set that is not itself)
2) Not a member of itself

If it's not a member of itself, it's a member of other than itself. If it's a member of itself, it's not a member of other than itself. You need to show a meaningful difference between 1 and 2 since you're the one claiming that on the one hand a set can't be both a member of itself and not a member of itself, and on the other hand, a set can be both a member of itself and a member of another set (to be a member of another set is to be a member of other than itself)

"is a member of itself, as a member of itself" has no apparent meaning to me. — TonesInDeepFreeze

This is perhaps why you didn't get my point/refutation. Perhaps if you try to reply to my above point, you will start to get my point.

You said that it is perhaps fair to say that such locutions have no apparent meaning, but then you proceed to post them again — TonesInDeepFreeze

No, z and v were clearly defined. What I did not say (which is why I said "perhaps fair to say") was that the v of all vs = the set of all sets and that the z of all zs = the not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets. I thought they could be inferred from v and z, but still, in my next post to you I clearly defined them, yet you did not address that main refutation of mine.

The set of all sets is really the set of all members, which is also how we use the word "all" in the first place. — Fire Ologist

Isn't the set of all sets equivalent to the set of all members? — Fire Ologist

And by definition all those members are sets.

There aren't actually any sets within the set of all sets. There are only members. — Fire Ologist

I get where you're coming from. I believe the issue lies in correctly determining what it is for something to be a member of itself and what it is for something to be not a member of itself. This is something I address in my last reply to TonesInDeepFreeze (which is literally the post that is above/before this post) in case you're interested.

A set of all sets has as members all the sets — TonesInDeepFreeze

Yes, this is true by definition.

I've refuted that claim. You skip the refutation. — TonesInDeepFreeze

I am not skipping. You say:

So whether a set x is a member of itself or not a member of itself, if there is a set of all sets then x is a member of that set of all sets. — TonesInDeepFreeze

This point implies that a set v can have more than one set that is a member of itself, as a member of itself/v.
I refuted this point with my z example of which you perhaps fairly said "the z of all zs has no apparent meaning to me". Here is another form of the refutation:

L = the list of all lists
LL = the list of all lists that list themselves

Is L a member of itself in L? Yes. Is L a member of itself in LL? No, because in LL, it is a member of LL as opposed to a member of itself/L. Does this not prove that L is not a member of itself in LL?. Just because a set is a member of itself in its own respective set, doesn't mean it is a member of itself in another set.

My original refutation with the z example was, I believe, more complete. So:

z = any set that is not the set of all sets
v = any set
The v of all vs = the set of all sets
The z of all zs = The not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets

Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different standards/contrexts as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?) — Philosopher19

This post is partly because of the discussion in the other discussion (the one titled "proof that infinity does not come in different sizes").

- We need a meaningful distinction between "member of self" and "not member of self"
- We need a set of all sets (math/logic would be incomplete without it, if not contradictory)

So:

The set of all sets encompasses all sets that are not members of themselves (precisely because they are members of it and not themselves) as well as itself (precisely because it is a set).

A subset of "all sets that are members of themselves" and "all sets that are not members of themselves" is contradictory because whether something is a member of itself or not, is purely dependent on the single set/context/reference that it is in. Here is the proof for this:

Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different standards/contrexts as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?)

I take you to be saying that a set cannot be a member of another set and also a member of itself. — TonesInDeepFreeze

Yes. The list of lists that list themselves is a member of itself in that list alone. Even though it is also a member of the list of all lists, it is not a member of itself in the list of all lists precisely because it is a member of the list of all lists and not the list of all lists that list themselves.

Note that the above shows the need to distinguish between "member of self" and "not member of self". To say no such distinction exists or is possible is to have an incomplete/contradictory theory in my opinion (contradictory because it argues the semantics of "member of self" and "not member of self" do not exist in Existence).

Do you mean: If S is a subset of some set T and x is member of S, then x cannot be a member of T ? — TonesInDeepFreeze

That's incorrect. By the definition of 'subset', if S is a subset of T, and x is a member of S, then x is a member T. — TonesInDeepFreeze

I get where you're coming from and I believe I completely get what you're saying. I don't deny the set of all natural numbers encompasses the set of all even numbers. Confusion occurs when one views sets that are not members of themselves (precisely because they are members of the set of all sets and not themselves) as being members of themselves.

To meaningfully talk about "member of self" and "not member of self", a set/context/reference is needed and adherence to it is necessary for the sake of consistency. The best way for me to convey to you what I'm saying in response to what you're saying, is the following:

Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different references as one (as in are we focused on the context of vs or the context of zs?) — Philosopher19

So that this point is not misunderstood:

An item in a subset cannot be both a member of the subset and the set. If it is a member of the subset, it is a member of the subset. If it is a member of the set, it is a member of the set. — Philosopher19

The following must be considered properly:

Something cannot be both a member of itself and a member of other than itself at the same time. For example, take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different references as one (as in are we focused on the context of vs or the context of zs?) — Philosopher19

Correct! Indeed that is a crucial point that is used in an important proof I gave you. — TonesInDeepFreeze

'There exists a z such that for all y, y is a member of z' contradicts this instance of the axiom schema of separation: For all z, there is a x such that for all y, y is a member of x iff (y is a member of z and yis not a member of y). — TonesInDeepFreeze

No, I believe it is a crucial point that is used in an important proof I gave you, which to put it in as short a manner as possible is: An item in a subset cannot be both a member of the subset and the set. If it is a member of the subset, it is a member of the subset. If it is a member of the set, it is a member of the set.

That is not "once again". Previously you said that "a set cannot be both a member of itself and a member of other than itself". That is different from "a set cannot be both a member of itself and not a member of itself". — TonesInDeepFreeze

If it's a member of other than itself, this means that it's not a member of itself. So my question to you is, what is the difference?

. I am more concerned with what issues you solve with your beliefs. — Lionino

I believe the solution to Russell's paradox is in here:
http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/

There are also other things on the website. I think they are appropriately titled with regards to what they try to do or highlight or solve or discuss.

It seems to be a collection of semantic games — Lionino

A triangle is triangular (or the angles in a triangle add up to 180 degrees) is not a semantic game. It is use of semantics in a non-contradictory manner. Counting to infinity, or there being no set of all sets is use of semantics in a contradictory manner.

The relative consistency of those theories indicates that it is not contradictory that a set is a member of itself and also a member of other sets. — TonesInDeepFreeze

It is blatantly contradictory for x to be both x and not x. It is blatantly contradictory for a set to be both a member of itself and not a member of itself. Yet you want to persist by saying things like the above. Again, I asked:

Is it logically possible for a set to be both a member of itself and a member of other than itself? If it is a member of other than itself, then it is not a member of itself, is it? And if it is a member of itself, it is not a member of other than itself is it? — Philosopher19

and I added:

And don't say to me something like "some set theories allow for this or that". I'm asking a basic logical question that has a basic and straight forward answer. There is no need to dance around anything. Just deal with the main issue at hand. — Philosopher19

It seems that what I added was ignored and what I asked was not answered. Until I see a good enough response, I'm done putting any more time into this. Once again:

It is blatantly contradictory for x to be both x and not x. It is blatantly contradictory for a set to be both a member of itself and not a member of itself.

Who would reject this but the contradictory/unreasonable/irrational/absurd/insincere?

I don't know anyone who has said that all others are ignorant. You are ignorant on the subject. That doesn't entail that others are ignorant on it. Indeed, there are people who critique classical set theory who are extremely knowledgeable about it. Critiques of set theory are quite fair game and bring profound insights into the subject. But those are knowledgeable, responsible and thoughtful critiques. And better yet, they are critiques that are followed up with actual mathematical alternatives to classical set theory. — TonesInDeepFreeze

I didn't say all others are ignorant. I just said there are people who are like this. I did not specify who.

Did you read anything from the link I gave you?
I believe my beliefs are not foundationally incomplete or contradictory in any way from a rational/semantical point of view.

I responded to you, you responded me with a refutation, I responded to your refutation with the following:

Is it logically possible for a set to be both a member of itself and a member of other than itself? If it is a member of other than itself, then it is not a member of itself, is it? And if it is a member of itself, it is not a member of other than itself is it? — Philosopher19

Where is my response? Is it me who ignores you or you who ignores me?

All of them are here:

godisallthatmatters.com

Regarding Russell's paradox, sets and infinity:
http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/

He did it again! He completely skipped recognizing the refutation given him. — TonesInDeepFreeze

Ok. Let me put it this way. I gave you a refutation with the z example. You started with insults, then you eventually said something like this:

By the way, we don't need to use temporal phrases such as "at the same time". Set theory does not mention temporality.

Then the rest of your z's and v's is irrelevant if it is supposed to refute the proofs I gave. — TonesInDeepFreeze

I decided discussing something with someone who seems to be emotional or biased is a waste of my time so I said I will stop, but I felt the need to add the following to the discussion:

I will just say this. That a set cannot be both a member of itself and a member of other than itself is the equivalent of saying that a shape cannot be both a square and a triangle (I have taken out the "at the same time" and the effect is still the same). — Philosopher19

This dealt with your temporal phrases response.

You have not yet answered:

Is it logically possible for a set to be both a member of itself and a member of other than itself? If it is a member of other than itself, then it is not a member of itself, is it? And if it is a member of itself, it is not a member of other than itself is it?

And don't say to me something like "some set theories allow for this or that". I'm asking a basic logical question that has a basic and straight forward answer. There is no need to dance around anything. Just deal with the main issue at hand.

If you look at the posts, I don't think I'm the one that has been showing the disrespect (if I have, it has been in response to disrespect). I wanted a discussion because I felt I had something to offer in response to something that I saw as contradictory. I don't think I entered the discussion closed-minded or dogmatic. And I think I tried to understand the other's point of view.

If someone is an "expert" in the field of something, but that something is evidently paradoxical or foundationally incomplete, it's absurd to treat them like an expert of anything useful. Some people are unreasonable/absurd. They want to hold on to their paradoxical or contradictory theory or belief at the cost of sincerity to Truth/Goodness/Existence/God

If people here witness that their beliefs or theories or axioms lead to no paradoxes or contradictions or foundational incompleteness, then I can't say to them they're misguided or lacking in knowledge.

You have those who recognise/witness that their theories are incomplete and act as such (there is honesty to them), and then you have those who recognise/witness this, but act as though they are the knowledgeable ones whilst all others are ignorant (which to me is the very definition of a "bad guy"). I believe spending hours or years or decades on something that is foundationally corrupt, does not make you an expert in anything other than something that is useless. What good is an expert in multishapism geometry that deals with the study of shapes such as round triangles and circular pentagons?

I don't feel like I have any contradictory or paradoxical theories or beliefs that I need to reconcile. I was trying to address what I saw as contradictory. If it's not contradictory, then it's not contradictory. But if it is contradictory/paradoxical and some are hardcore with regards to holding on to this, what can I say?

I will just say this. That a set cannot be both a member of itself and a member of other than itself is the equivalent of saying that a shape cannot be both a square and a triangle (I have taken out the "at the same time" and the effect is still the same).

The above point I felt was worth adding to this discussion, but I will probably stop posting here as I don't think there's anything left to add to this discussion.

In certain alternative set theories, there are sets that both members of themselves and of other sets. — TonesInDeepFreeze

We go in a circles, as it is with cranks. The crank makes false claims and terrible misunderstandings. Then the crank is corrected and their error is explained. Then the crank ignores all the corrections and just posts the false claims and misunderstanding again as if the corrections and explanations never existed. — TonesInDeepFreeze

Evidently, there's no point in continuing this discussion. If you believe your mathematics is free from contradictions or paradoxes, then in my opinion, you are not blameworthy for upholding them or sticking to them (unless of course someone presents a better or more complete thing to you and you reject greater for lesser), but if you see paradoxes and contradictions or incompleteness and you treat them as other than paradoxes/contradictions/incompletions...
I see no paradoxes or contradictions or foundational incompleteness in the beliefs that I uphold (mathematical or otherwise).

Peace

There is no object called 'Infinity' in the sense you have been using it.

Here is a way to say what you want to say:

In mathematics, there are sets that are infinite but that have different cardinality from one another.

Better yet:

If x is infinite then there is a y that is infinite and y has greater cardinality than x. — TonesInDeepFreeze

But all of the above is exactly what I'm saying is contradictory. And my use of infinity which (if I've understood you correctly) you say is not the one that they use in maths, is the reason that I say all of the above is contradictory.

There is no object called 'Infinity' in the sense you have been using it. — TonesInDeepFreeze

So what semantic are mathematicians using when they use the world/label "infinite"?

There is no x such that for all y, y is a member of x iff y is not a member of y. Proof: — TonesInDeepFreeze

the axiom schema of separation — TonesInDeepFreeze

Something cannot be both a member of itself and a member of other than itself at the same time. For example, take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different references as one (as in are we focused on the context of vs or the context of zs?)

Note that the above shows the impossibility of a set that contains all sets that are members of themselves where all equals more than one.

For the fully fleshed out version of this, see my post on Russell's paradox which I posted the link to in this discussion and in the other one.

'There exists a z such that for all y, y is a member of z' contradicts this instance of the axiom schema of separation: For all z, there is a x such that for all y, y is a member of x iff (y is a member of z and yis not a member of y). — TonesInDeepFreeze

If some theory suggests that you can view the z of all zs as both a member of the z of all zs and a member of the v of all vs at the same time, then that theory is contradictory. The z of all zs is either to be treated like a z or a v. If it is to be treated like a z, it is a member of itself. If it is to be treated like a v, it is not a member of itself (precisely because it is a member of the v of all vs)

We do NOT claim that from "after each natural number there is a next number" and "there is no greatest natural number" that we can infer that there is a set of all the natural numbers. Indeed such an inference IS a non sequitur. And every mathematician and logician knows it is a non sequitur. So, we recognize that to have a set with all the natural numbers we need an AXIOM for that, which is NOT an inference. — TonesInDeepFreeze

I'm not sure what you mean by "So, we recognize that to have a set with all the natural numbers we need an AXIOM for that, which is NOT an inference."

If I've understood him right, Cantor treats a number sequence that goes on forever as being infinite. But something going on forever does not make it infinite (if my counting to infinity goes on forever, that neither makes my counting infinite, nor does it mean I will eventually reach infinity). It also makes no sense to say something like "assume that your counting to infinity is completed such that you have counted the set of all natural numbers and have successfully proven that there are an infinite number of natural numbers" and then label this as {1,2,3,4,...}

If Cantor did not do this, he would not then be forced to conclude as he did with his diagonal argument. Now I feel the following is relevant:

But from what I've seen of mathematicians, they either have no part for infinity, or they're using infinity wrongly. I believe they're doing the latter which leads to the former (which I think is why I have heard it said before that "maths is incomplete") — Philosopher19

It makes a real difference. By saying 'infinity' as a noun and then that there are different sizes of infinity is to picture an object that has different sizes. There is no such object in mathematics. — TonesInDeepFreeze

I don't think I'm picturing an object. I think I'm just focused on the semantic of Infinity.

Good faith in posting a critique of mathematics would entail at least knowing something about it. — TonesInDeepFreeze

I think it is from all that I have seen and heard that I said the following:

But from what I've seen of mathematicians, they either have no part for infinity, or they're using infinity wrong. I believe they're doing the latter which leads to the former (which I think is why I have heard it said before that "maths is incomplete") — Philosopher19

Whether all that I have seen or heard is enough, is another matter. You don't think I have. I think I have.

He still persists in mischaracterizing mathematics as claiming that there is an "Infinity" [capitalized, no less] that has different sizes — TonesInDeepFreeze

To my understanding, mainstream maths claims:
There are infinites of various sizes (or at least infinite sets of various sizes, but that amounts to the same thing)
The set of all sets is contradictory

Is my understanding wrong?

That is egregious intolerance — TonesInDeepFreeze

If someone came to me and said I've seen a triangular square, I would say to them that that's nonsense to me and that it is impossible for them to have seen such a thing. It's in the semantic of square that it can't be triangular. I would not call this intolerance, but perhaps I could be more tolerant by trying to understand the person better. Perhaps what they really mean is that they saw some shape, that the best way that they could label it was "triangular square". Maybe they saw some kind of trapezium and did not know the label/word for the semantic of trapezium.

Similarly, if someone came to me and said they have demonstrated how infinity comes in various sizes, I would say to them that that's nonsense to me. It's in the semantic of Infinity that It does not come in various sizes.

I guess this "something" is "space", right? Like a balloon ...
But this seems impossible since space is part of the universe itself; it cannot be larger than it. E.g. like the space around a balloon that is inflated ... — Alkis Piskas

I would just say if the universe is expanding, then it is expanding in Existence (as opposed to 'space-like-the-space-in-our-universe')

If scientists have in fact measured the size of our universe, then our universe is finite (meaning that our universe is not all there is to Existence. I hear there's been more talk of parallel universes lately).

We know by way of pure reason that the universe cannot be expanding in non-Existence. Such a thing is not conceivable at all, therefore, it is not observable at all. And this is not an unknown like a 10th sense which some superior being may have that we can't comprehend. This is a clear case of something contradictory (like a round square) that no being would make sense of because it is a known contradiction as opposed to an unknown.

Doesn't infinity mean endless? i.e. unreachable eternal continuation in concept? — Corvus

I think Infinity is why something can go on forever. But if something goes on forever (or keeps going without end) it will not become infinite (just as if I keep counting without end, I will not reach Infinity)

To me, the only thing that is Infinite, is Existence. And Existence has always Existed and will always Exist (so It has no beginning and no end whilst all ends and beginning are within It. And if something goes on forever within It like a number sequence or a forever expanding universe, then that thing will never reach Infinity/Infiniteness

If it was reachable, then it wouldn't be infinity. Any set or size would be unknowable, if it were infinity. Therefore talking about different size, set or number of infinity, is it not a nonsense? — Corvus

I think it's nonsense to say Infinity comes in various sizes. But the semantic of Infinity itself is not nonsense because it is clearly meaningful. As for sets, the only thing that can be the set of all cardinalities or houses or any other meaningful/imaginable/understandable thing, is Existence/Infinity. Since Existence is Infinity, it allows for there to be no end to the number of numbers possible (because you can always add one and this can go on forever without Infinity being reached or exhausted).

Not 'infinity' as a noun, as if there is an object named 'infinity', but rather 'is infinite' as an adjective to name a property. — TonesInDeepFreeze

It makes no difference. Existence is Infinity (here it is a noun). Existence is Infinite (here it is an adjective). You cannot become Infinite (adjective) even if you expand forever. You are not Infinity (noun) if you are not Infinite (adjective).

You keep saying I ignore your points, but rightly or wrongly, I also think you have not read or considered what I've written with sufficient attention to detail.

What you've seen is what you've allowed yourself to see, which is virtually nothing about the actual mathematics you've not even bothered looked up. — TonesInDeepFreeze

I am not claiming to have seen everything. But (again, rightly or wrongly) I think I've seen enough to say:

But from what I've seen of mathematicians, they either have no part for infinity, or they're using infinity wrong. I believe they're doing the latter which leads to the former (which I think is why I have heard it said before that "maths is incomplete") — Philosopher19

In any case, in the event that you have made good points and I have failed to give them the right amount of attention, I apologise. I do think that I am being sincere and honest in this discussion (as well as not closed-minded).

When it comes to the empirical application of the concept of infinity, it is indeed reasonable to think that it is fundamentally unverifiable whether something is infinite — DanCoimbra

I get your point with regards to empirically verifying infinity, but I believe the a priori is superior to the a posteriori in that whatever observation we make (scientific or otherwise), has to be interpreted in line with the dictates of pure reason. It also has to be such that it does not contradict the semantics that we are aware of (for example, we must not have a theory that amounts to saying or logically implying that triangles don't have three sides because that contradicts the semantic of triangle) .

If a scientist says something like "I have observed something pop in and out of Existence" because it may have looked that way to him, we have to reject him because 'something popping in and out of Existence' is clearly contradictory. Non-Existence does not Exist for something to pop into or come out of. Things can be turned on and off but this is not the same as things popping in and out of Existence.

I see what you mean. Well, the words "exist" and "existence" can be used in different ways. And it can be used strictly (substantially, concretely) and loosely (insubstantially, abstactly). And I guess the second form applies to what you say above. — Alkis Piskas

I don't mean to use Existence loosely/abstractly. By "Existence" I mean that which encompasses all things physical or otherwise (if otherwise is possible). So dreams (which some may view as non-physical) are clearly a part of Existence. The term universe seems limited to me in terms of accounting for all that exists (I cannot comfortably say something like "the universe has a space for dream worlds" whereas I can comfortably say "Existence has a space for all worlds including the universe and dreams"). To me, Existence/Infinity clearly fits the bill of 'encompasses all things/existents' whilst universe does not.

But doesn't an expanding universe mean that this process is infinite and thus the universe itself is limitless? It is not much different than if we consider the universe as being static, in which case it can also be infinite. — Alkis Piskas

If the universe is expanding, it is expanding in something. Some thing has to be Infinite to allow for the possibility/potentiality for the universe to forever expand. But it is also the case that even if the universe expands forever, it will not become infinite (this is not unlike me saying even if I count 1, 2, 3 ad infinitum, I will never reach infinity).

So to me, Infinity/Existence is the reason that something can expand forever or go on forever. As for the thing that expands (like the universe), it is a part of the Infinite. It is not itself infinite.

As an analogy, consider looking at a dictionary and judging how truthfully the definitions represent how people are actually using the words which are defined there. — Metaphysician Undercover

Of course, it is possible, for example, for mathematicians to be using the label "infinity" to refer to a semantic that is different to the semantic of infinity. But from what I've seen of mathematicians, they either have no part for infinity, or they're using infinity wrong. I believe they're doing the latter which leads to the former (which I think is why I have heard it said before that "maths is incomplete")

It may well be the case that existence determines truth, like you say, but that's not relevant to the selection of mathematical axioms. — Metaphysician Undercover

I just think if mathematical axioms are to be selected, they have to be such that they do not lead to what is contradictory to Existence/Truth (or just semantics in general).

The axioms do not give mathematicians rules for how to do things, because the mathematicians get to create and choose their own axioms. So the axioms simply provide a representation of what mathematicians are doing. Since they are descriptions, "truth" is to be found in how well the axioms represent what the mathematicians are actually doing — Metaphysician Undercover

If a mathematician or a philosopher decides on an axiom or theory that requires belief in the following (or at least logically implies it or leads to it): Nothing can be the set of all things (which logically implies Existence is not the set of all existents), or one infinity is a different bigger than another (or is a different quantity than another), I believe that axiom or theory should be disregarded or at least viewed as contradictory to Existence/Truth (or at least contradictory to the semantic of infinity).

How? — Lionino

By Being. Existence just Is. It just is the case that triangles are triangular or that Existence is Infinite or that 1 plus 1 = 2. Or if you're interested in more on Existence, it just is the case that Existence indubitably exists and is Perfect. I won't go into detail with regards to how Existence is Perfect and indubitably exists. I'll just provide the link to the argument: http://godisallthatmatters.com/2021/05/03/the-image-of-god-the-true-cogito/

But Euclidian triangles don't exist in nature. — Lionino

To my knowledge they don't exist in our universe due to gravity. But I see our universe as just a part of Existence/Nature/Infinity. Something has to account for why we are aware that "the angles in a triangle add up to 180 degrees". To me, the nature of Existence/Infinity accounts for this awareness.

Existence can accommodate both perfect and imperfect triangles. We have experienced imperfect triangles (as in we have visually seen them), we have not experienced perfect triangles. But somehow, we have the knowledge that the angles in a triangle add up to 180 degrees. This is the awareness we have got in Existence and have gotten from Existence.

You cannot start counting 1,2,3,4,... ad infinitum and reach somewhere, anywhere. Infinity has neither a start or an end — Alkis Piskas

Then, counting (natural) numbers you can never reach infinity because that infility would be also a number, and infinity is not a real or natural number. — Alkis Piskas

Agreed (especially with "Infinity has neither a start or an end").

A set is a collection of objects (elements, members). I'm not sure if we can talk about an infinite set — Alkis Piskas

I don't believe we can talk about different infinite sets because it will lead to contradictions. But I do believe Infinity and Existence denote the same Entity. I see Infinity/Existence as the set of all existents. I see there being no end to the number of existents purely because the nature of Infinity/Existence allows for such possibilities.

All this raises questions about the infiniteness of the Universe, whether it started (created) from something or it always existed, etc. And, as I see it, since we don't have a proof that it is created from nothing, it must have always existed, even in the form of extremely high density and temperature, which at some point exploded (re: Big Bang), or in any other form. But I'm not the right person to talk about these things. — Alkis Piskas

It is clearly contradictory for something to come from nothing. And since I have heard some say that the universe is expanding, my view is that the universe is not infinite (if it's expanding, it's not infinite). But I do view Existence/Omnipresent as Infinite. I see Infinity as the reason for why an endless number of things can be imagined or thought about or experienced (dream or otherwise). Infinity has Infinite potential, therefore, an endless number of things can be imagined or thought about or experienced (we and our minds are wholly contingent on Existence. There is no non-Existence for us or our minds to draw anything from).

First, imagine you have achieved immortality and are presented with two options: to receive $1 every day forever or $1 every year. Intuitively, you would choose $1 every day because, over the same infinite duration, you would accumulate more money. This illustrates that while both options extend to infinity in time, the rate at which you receive money differs, leading to a larger "size" of wealth in one scenario over the other. — punos

Thank you for that clear and easy to understand example.

When you say "extend to infinity in time", I assume you mean go on forever. It follows that both will forever add to their money. It also follows that one will always have more than the other. But it also follows that neither will ever have an infinite amount of dollars precisely because their wealth will not amass to infinity dollars. To say that it would is to say that one can count to infinity.

Now, let's consider a spatial analogy. Imagine two pipes, both of infinite length, but one has a diameter of 1 inch and the other has a diameter of 10 inches. Despite their lengths being equally infinite, the pipe with the larger diameter has a greater volume. This demonstrates that even with one dimension being infinite, other finite dimensions can contribute to a difference in "size" or capacity. — punos

I believe there is no contradiction in saying that something can go on forever. So I believe a pipe can go on forever. But to me, Infinity is the reason something can go on forever or be endlessly added to. It is not the measurement of the thing that goes on forever.

I think Infinite and Infinity should be exclusively used to refer to Existence, and a part of Existence is not equal to the whole of Existence. Trying to divide Infinity into parts seems contradictory to me. Another reason for why I think a pipe that measures infinite in length is an impossibility.