I would cut you more slack if you demonstrated more understanding. You've been thoughtful, but the rigour's lacking. This shows whenever someone spells out a mathematical idea with more precision and it turns out that the concepts (as they use them) contradict established theorems or the intuitions (conceptual understanding/imaginative background) that support them.

Specifically, your understanding of convergence needs refining. I also suspect you would discover more difficulties with your argument if you tried to define your function formally (what's its domain, what's its image, is it injective, surjective...) - one presentation of it is a way of associating reals with infinite sequences of rationals (if they do converge in the usual sense, you will successfully associate with only rationals, making the image countable, if they don't converge in the usual sense your argument doesn't work).

probably should have just kept my mouth shut. — Isaac

Nah. Well thought criticism is welcome.

if I do that infinetley it should end up on a rational numbe — Umonsarmon

Try and prove it! This discussion might be helpful.

It would converge on a rational number just as the sequence 1/2+1/4+1/8 converges on 1 after an infinite number of steps. — Umonsarmon

I don't think this is right. Infinite convergent sequences of rationals typically converge on non-rational reals, even though all the finite sums and elements are rational. The incompleteness of the rationals demonstrates this. (Well, more precisely, convergence doesn't make much sense for infinite sequences of rationals...)

I bring this up because the incompleteness property of the rationals shows that infinite convergent sequences of rationals do not have to converge to rationals. (or more precisely that convergence breaks when you don't have ensured existence of suprema and infima)

Hello fdrake. My next post is going to talk more about the structure, and what the structure is. It's not unique, in fact, it's quite familiar to us — Sam26

Very interested in your continued exegesis, then!

Each binary number will terminate on its own unique rational a/b as will each irrational number — Umonsarmon

The sequences don't terminate for any real number which has an infinite (non-repeating) binary expansion. EG pi/10 would never terminate. The consequence of allowing infinite sequences there means the function is simply from binary expansions to real numbers - essentially a way of encoding binary expansions. The representation of a set doesn't change any of its cardinality properties though.

Well as far as I can tell any number fed into this procedure should result in a terminating rational length which will produce a set of rationals which map to the natural numbers regardless of whether it is an irrational number or not. — Umonsarmon

Yes! You can quite happily map a countable set of irrationals to any other countable set.

Now I understand the point that you could argue that the set your feeding in is uncountable but this leaves us in a strange position because the two proofs directly contradict. — Umonsarmon

1)First convert all numbers into binary strings. — Umonsarmon

Convert all real numbers to binary strings!

2)Draw a square and a line down the middle — Umonsarmon

The line is an uncountable set.

3) Starting at the middle line do the following .If the digit in your string is a 1 move half the distance to the next line to the right. If the digit is a 0 move half the distance to the next line to the left. — Umonsarmon

You have to apply this to every real number. You're mapping an interval of reals to an interval of reals with (what looks to me like) an injection.

The structure rests on bedrock. What is the structure? — Sam26

Can you guarantee uniqueness of the structure? It seems to me "Here is one hand" is a bedrock for philosophy. But not for examining the self reports of a delusional phantom limb patient waving their hand about.

I know it's not my place to say, it's not my forum, but if there's a desire to attract involvement from serious academics then somehow (and I understand it's a lot of work) there's going to need to be more control over post quality. — Isaac

As far as I know, we'll continue to vet content which will be submitted to guest speakers. And we'll have a think about what to do to make guest speaker engagement more long lasting in the future.

Increasing the standard for content isn't particularly worthwhile in my view as: (1) users can selectively respond and read, like the under used "following" posters option in profiles (2) people's interest in philosophy usually starts long before researching much of it, and it's a valuable space for learning for that user type (3) less restricted posting stimulates discussion (4) increasing content standard to make the place more attractive to seasoned academics would simultaneously reduce our attractiveness for having a large and relatively high standard (for the internet) of discussion.

All the questions had good points and bad points, but some were somehow imbalanced and / or unanswerable; some had the questions too deeply buried, under too many layers of pre-explanations, is the feeling I got from them. — god must be atheist

Mismatch of expectations I think. I'm guessing we need a word limit.

If you're gonna ask a long question, you need to set up context. It's probably more to do with question length / complexity / number and how much time Pigluicci wanted to spend.

3) Starting at the middle line do the following .If the digit in your string is a 1 move half the distance to the next line to the right. If the digit is a 0 move half the distance to the next line to the left. — Umonsarmon

I think your procedure does produce an injection between the sets, but the initial set you're feeding into the injection is actually uncountable. You're mapping the real numbers to the real numbers rather than the real numbers to the rationals.

If you wanna see this, there's an uncountable infinity of real numbers whose first digit right of the decimal point is 1 in the binary expansion. The same goes for any binary digit. I think you're not registering the distinction between "the set of sets of real numbers with x in a given digit in their decimal expansion" and "this real number has x with a given digit in their decimal expansion".



Sophisticat is right though. The diagonal argument does establish that no injection from the reals to the rationals exists. If your claim is correct, set theory is inconsistent (as it proves a contradiction), but it is provably consistent (within larger theories).

I'm not sure if the following is a proof that cantor is wrong about there being more than one type of infinity — Umonsarmon

Even if you grant your whole argument, it doesn't stop Cantor's theorem about powersets and cardinality from going through. And the broader claim about consistency rears its head again.

Seems Prof. Pigliucci is more busy than he thought and doesn't have much time to reply to anything, unfortunately.

What you think you know is knowledge. — ovdtogt

I think I know that what anyone thinks they know is not knowledge.

Ultimately, the stakes of the Liar sentence are the consequences it has: if you can formulate it in a language, it does weird shit; has unfavourable inferential consequences for; to the semantics and logic of that language.

Perhaps because the first has some determinant truth-condition (even if arbitrary, e.g. fewer than 10 letters) whereas the second doesn’t. — Michael

Let's press on the "determinant" thing there. One way to look at the conditions under which a statement is true or false is to submit it to a T-sentence and see what happens. At face value, you can T-sentence the Liar:

"This sentence is false" is true if and only if this sentence is false.

The T-sentence (in a deflationary manner) sets out the truth conditions for the statement. Whether it provides a full account of what it means for a sentence to be true doesn't seem too relevant to me here, it's about whether arbitrary sentence interpretation requires the universal applicability of the T-sentence.

Tarski intuited this, and tried to dissolve the paradox by appeal to the idea that the truth predicate "... is true" lives in a higher order meta-language. More precisely, that there are lots of truth predicates we equivocate over in natural language with "...is true", and the contradiction from the Liar arises by mistaking one truth predicate for another; object language and meta language truth predicates. This approach attempts to preserve the universal applicability of the T-sentence (and all statements having truth conditions and being true or false) at the expense of multiplying truth concepts.

In this view, it does not seem to matter whether the sentence is "evaluable" or not as there's always another truth predicate and meta language which can come in to save the day.

So the first approach I detailed here is pretty much Prior's - the liar isn't a paradox, it's just a disguised contradiction, and thus false. It keeps the underlying logic to have 2 truth values (true, false), it seems consistent with the universal applicability of the T-sentence (it just evaluates the statement as false), and there's one truth predicate operative within it.

I’m partial to Kripke’s take on this. It doesn’t seem to mean anything for the liar sentence to be either true or false. There’s no evaluable fact. — Michael

How does the "evaluability" idea block either of the above accounts? At what points does it intervene? And why is it a better response to the Liar?

The truth is always somewhere in the middle — ovdtogt

And so can be made indistinguishably close to either side of the middle?

Did you have only a fried egg?

Did you have a fried egg and did you have beans?

Sorted. Ultimately not a problem with conjunction, a problem with the awkwardness of rendering sentences conformably with propositional logic.

It has truth conditions. — creativesoul

The two claims you made mean the same thing. Why doesn't "This sentence is false" have truth conditions when "This sentence is short." Does?

I asked you to explain why the truth table says that. — frank

You have only a fried egg for breakfast.

Did you have a fried egg and beans?

No.

Ultimately, definition of conjunction.

Row three.

This sentence is short.
This sentence is false.

Why is the first truth apt but not the second?

Why? — frank

That's how conjunction works. Look at its truth table.

X = This sentence is false.

Assume X is true, then X is false.
Assume X is false, then X is true.

Y = This sentence is false and this sentence is true.

Assume Y is true, then Y is false (and true, but we knew that already)
Assume Y is false, then Y is true (and false, but we knew that already)

X and Y are equivalent.

But Y evaluates as:

Y is True and False.

True and False evaluates as False.

Y is false.

Y and X are equivalent.

X is false.

X is just false. No contradiction here.

Problems?

Something I don't get, though I don't know enough to decide if this is actually an error in the proof or a limitation of understanding, is that the proof doesn't seem to require that $f$ is computable, just that it has a boolean image. The conditional "if f is computable and maps to T F then ~ f is computable and maps to T F" only uses that f has a boolean image. Where have you used the assumption that f is computable?

Posters:

Types of posters who are welcome here:

Those with a genuine interest in/curiosity about philosophy and the ability to express this in an intelligent way, and those who are willing to give their interlocutors a fair reading and not make unwarranted assumptions about their intentions (i.e. intelligent, interested and charitable posters).

Types of posters who are not welcome here:

Evangelists: Those who must convince everyone that their religion, ideology, political persuasion, or philosophical theory is the only one worth having.

Racists, homophobes, sexists, Nazi sympathisers, etc.: We don't consider your views worthy of debate, and you'll be banned for espousing them.

Advertisers, spammers: Instant deletion of post followed by ban.

Trolls: You know who you are. You won't last long

Sockpuppets

This is a whole field/. Not just confined to humans.

It was to deal with some stuff that came up. I think there will be separate forum wide threads for each discussion with the guest.

Deleted the personal remarks from your question. Also PM'd them to you in case you wanted to keep them anyway.

Done.

Title's too long. Not just editorially. It hits the character limit for a thread title.

I'm a bit hesitant about that being OK by itself, I think it would be OK so long as the quoted bit is contextualised explicitly by you in your discussion. Want to hear what the other mods and questioners think, hence putting this in feedback.

I mean by well-ordered a set (of numbers) ordered in such a way that given a starting value, say, on the left, one could move to the right step-by-step and enumerate/list/count all the elements without missing any. The natural numbers, ordered on ≤, would be such an ordering. — tim wood

There's really only one order like this.

You have a set like the naturals: {0,1,2,3,...} which are presented in their standard order with the successor function S (@SophistiCat "counting") always giving the "next" element in that order. The successor function for the naturals will be called S.

You define a bijection f from the naturals onto some countable set, giving {f(0),f(1), f(2), f(3), ...}. This set also needs a similar successor function T. But you additionally require that f(S(x) ) = T f( x ); the mapping of a successor is the successor of the mapping. (successor functions are how you move right step by step)

If you have that x<y in the naturals, then for some n, y=S^n ( x ), then f( y )=f(S^n (x) ) so f( y ) = T(f(S^(n-1) x) ) > f( x ), that is, f is an order preserving bijection. This is called an order isomorphism. IE - the only possible mathematical order you could be talking about (up to order isomorphism) is the standard order on the naturals.

It's not much of a definition of well ordered. You're artificially constraining math with a poor definition (if you think it's necessary that we accept your definition).

I'll have to appeal to selective black-band-blindness the refusal to accept that rabbits can wear black bands, but an enthusiastic belief that ducks always do. — Isaac

I've made a joke before about Dirac delta priors, don't get to wheel it out very often.

Alice: "What do you learn from the evidence?"
Bob: "I always learn I was right before"
Alice: "That's not how learning works"
Bob: "It is if you have the right prior"
Alice: "You have a Dirac delta prior, you can't check if it's right"
Bob: "Give me any evidence that contradicts it and I'll perform a Bayesian update"
Alice: "You know that won't do anything"
Bob: "This is because I have incorporated all available evidence and found it consistent with my prior"
Alice: "Your decision procedures are inadmissable"
Bob: "With respect to my measure and loss function they're Bayes optimal"
Alice: "Just look at the data, the performance is terrible"
Bob: "The sample is unrepresentative"
Alice: "You have the whole population"
Bob: "This is a model specification problem not a prior elicitation problem; if we correctly modelled the selection biases in the data generating mechanism we'd do better"
...
It can go on forever.

It's constantly in your field of vision, your brain just refuses to see it. In some unfortunate cases of brain injury, this effect gets shifted an the patient can't see any noses at all! — Isaac

That's amazing.

Not if you're 'black-band-blind'. — Isaac

The equivalence is ironically still true because both are false; the duck is seen without a black-band and so is the rabbit.

The rabbit wears a black band on its neck if and only if the duck wears a black band on its neck.

An analysis of the duck/rabbit experience reveals ideas and uninterpreted stuff (matter in the old sense of the word). — frank

It reveals that some environmental patterns can generate more than one perceptual feature. Notice that we don't disagree where the lines are drawn even as it shifts from duck to rabbit. The shifts are however unambiguously perceptual events; the lines aren't changing.

I'm with you so far, but it seems unwarranted to extend this to literally all cases, just on principle. And 'fleshing out the contexts' in which differences might be actualized, is a good aim, but again seems unwarranted to assume will be possible in all cases. That essentially back to where necessity matters in your N1.. — Isaac

I guess we don't want to be anti-realist and say that any system just is our representation of it. Otherwise there'd be no possibility of disaccord with model and modelled or representation and represented; since there's strict identity between modelled and model or representation and represented. In terms more related to the essay: that is, in order for there to be true statements, the propositional content of those statements would have to be identical to their associated perceptual features.

I also guess we both agree that you can weaken this identity to an equivalence; like counting the same things as cakes, or agreeing upon what my cat is and in what configurations it is in when it is on the mat.

What I would like to say is that the truth of a statement does not require the identity between perceptual features associated with statement propositional content and that propositional content, it only requires that the perceptual features are equivalent to the propositional content. That is, in order for a statement to be true, the propositional content of the statements would have to be equivalent to its associated perceptual features. We can demonstrate any such equivalence fallibly and contextually; whether it is true or not does not depend upon this demonstration (hence the fallibility, we can be in error).

Another way of saying this is that propositional content occurs in the same way as perceptual features; they are of the same ontological order/stratum/regional ontology. They're all events under some representation that tracks some generating conditions, so long as the conditions which generate the propositional content are tracking (strongly informationally constrain or are accurately modelled by) the conditions which generate the perceptual features; differences in one track differences in another, content in one track content in another, changes in hidden states in one track changes in hidden states in another, we're in a relative accord whereby we can state truths of what is modelled by counting it as a model output. "Sticks really do look like they bend in water, why?".

I can say that "my arms are on my body" because I have no reason to doubt that they are not; that is, I have no evidence that I'm in any context in which any doubt regarding that is warranted. What count as my arms are what count as on what count as my body. But what makes this true is that what count as my arms are what count as on what count as my body! Not whether the perceptual features are necessarily of the hidden states generating my arms, my body and their attachment.

Yes, I think I'd agree with that, but is it commensurable? Is it impossible for someone else to have a different set of hidden states combine to make a slightly different entity? If so, their entities (and relations) may be incommensurable with yours because, despite the fact that we're happy to accept whatever aggregate we perceive as real, we cannot refer to the simples constituting it (they're hidden). So if someone did have a different aggregate it would not be possible to translate it by reference to shared simples. — Isaac

Why would it need to be impossible? What's the reasoning behind (N1)?

I don't think the hidden states matter here. The perceptual features or internal states do. The perceptual features or internal states count as the entity (whose dynamics are given by the dynamics of hidden states). This is active perception and thought/predisposition. Then the words concerning the perceptual features or internal states count as the perceptual features or internal states (which count as the entity). This is speech acts. When we follow the chain backwards, we understand the speech act when we can output perceptual or internal states which count as the other's perceptual or internal states; that is, we can attribute "beliefs" and "propositional attitudes" to them which are in accord with their behaviour (and self reports).

The criterion for commensurability I don't think should be the possibility of different perceptual features being associated with entities, it should be which perceptual features count as that entity; and the behavioural/language components which count as those perceptual features give us fallible indicators that we count entities as entities identically; that is, for when we coincide in how we count what as what. When we coincide in how we count what as what, we are commensurable, when we have strong indicators that we coincide in how we count what as what, we have strong indicators of commensurability.

Trying to count the same stuff as the same stuff as much as possible is a form of the principle of charity; maximise agreement. The mere possibility of difference is largely irrelevant when we can agree upon that possibility and flesh out contexts which would actualise it.

(N1) In order for an output of a model to be real for certain, the connection between the model output (model results) and model input (what is modelled) must be necessary. — fdrake

Yes, I think so. I sense there's a commitment resulting from this that I'm not going to like, so I'm wary of the fact that it's not exactly how I would word it (laying out my escape route early on!), but yes,. It's related to the same answer I would give to Banno, so I've put them in the same post. — Isaac

The thing that bugs me about the argument is (N1). If I can replace "real for certain" with "true", the thing that bugs me about it is I have to assume (N1) is true in order for the syllogism to be valid. It's like we've found a necessary truth that there are no necessary truths.

Regardless, I'm quite happy with the idea that there are no necessary truths (for some account of necessity). What I'm really interested in is why this sense of necessity seems relevant at all to you, and what it is. If we can find defeater contexts for every model, we can clearly revise our knowledge.

For example, it isn't necessary that the cat is on the mat (the cat could be elsewhere). It isn't necessary that I believe the cat is on the mat if and only if the cat is on the mat in order for the cat to be on the mat (the cat could be on the mat and I could be out of the house and believing the cat is outside). That which my perceptual features aggregate into "my cat" counts as the cat, but the represented entity also counts as my cat. This "counting as" works both ways - it's relational and context sensitive.

What I'd replace the notion of necessity with is (fallible) accord of (fallible) perceptual features; then treat the perceptual features as real objects with regularities that (fallibly, contextually) ensure the (fallible, contextual) accord. Phantom limb patients don't have to be delusional ("my leg's still there") to have phantom limb sensations.

It looks to me that once you remove that need for necessity, a fallibly perceived reality is literally at your fingertips. No more see through veil, a seen through veil. Rather than "unmediated contact", say, mediation is a shared style of being in the world (the operation of embodiment as it works in human bodies). (Though I don't think I'm arguing exactly what Davidson is here, but we're talking about something (truth-belief relations) that weren't fleshed out in the paper anyway; an intuition I have is that Davidson's "principle of charity", if it works, works because we already do share so much)

In terms of your example of even onto the even integers, and evens onto all integers, the latter, because the odd integers aren't matched, isn't a bijection. — tim wood

A bijection between evens and odds.

The set of all even numbers is given by {2k} for k in {0,1,2,...,}
The set of all odd numbers is given by {2k+1} for k in {0,1,2,...}

The even numbers look like {0,2,4,6,...}
The odd numbers look like {1,3,5,7,...}

Define f from the odds to the evens by f(x) = x-1
If x is odd, then x-1 is even. So this works.

f is an injection: f(y) = f(x) => y-1 = x-1 => y=x
f is a surjection: any even number is of the form 2k for some k, then 2k+1 is odd, then f(2k+1)=2k

f is therefore a bijection.

2 sets have the same cardinality if they have a bijection between them. f is a bijection between the evens and the odds. Therefore the evens and the odds have the same cardinality.

(If you want to work through the case for the evens or odds to the naturals; f(k) = 2k is a bijection from the naturals to the evens, f(k) = 2k+1 is a bijection from the naturals to the odds.)

With regards to the order stuff, this is an order preserving function

assume x<y then f(x) = x-1 < y-1 = f(y), so x<y => f(x) < f(y) which was to be demonstrated. The order is a well order since the evens and odds are subsets of the naturals. The naturals are generated by the following function: f(x) = x+1, repeated application of f to 0 gives every natural. The order defined by x<y iff y = f^n (x) for some natural n>0 and x=y iff f(x) = f(y). This is a total order.

That order is also a well order. Take some nonempty subset of the naturals. Assume for reductio that it does not have a least element. Let some element x belong to the set, then this element cannot be 0 (all others are greater). Then this element cannot be 1 (all others are greater and it can't be 0). Then this element cannot be 2 (all others are greater and it can't be 0 or 1)... Then this element cannot be x (all others are greater and it cannot be in {0,1,2,...,x-1}). This holds for arbitrary x by induction. Then arbitrary x isn't in the subset. Then the subset is empty. Contradiction. Therefore the set is well ordered. Call an even (odd) less than another even (odd) when it is less than it in this order on the naturals. That is also a well order through the same argument and symbol substitution.

Edit: this well order does not require the well ordering principle to construct. Therefore there are sets that can be well ordered without the well ordering axiom. This says nothing about whether well orders are first order predicable/definable for arbitrary sets in general, but shows that there are first order predicable well orders (well, once you rewrite the f^n thing into a formula using x+n for some n to save quantifying over a function symbol by instead quantifying over the variable n).

But I still have a problem with bijection in uncountable sets: how do you do it? — tim wood

I gave you a worked example.