• fdrake
    6.5k
    If yes, can you please try to explain this in a better way? I don't even have much time for this, sorry. I have even to go to the hospital for a couple of days next week.Mephist

    The Heyting algebras are for intuitionist logic (of some sort). I wrote that previously.

    Hope you feel better soon.
  • jgill
    3.8k
    I apologize for interrupting a productive flow of thought. But I was curious what you guys were talking about. Seems pretty esoteric. :chin:
  • fdrake
    6.5k
    I apologize for interrupting a productive flow of thought. But I was curious what you guys were talking about. Seems pretty esoteric. :chin:jgill

    What do logic and topology have to say about each other?

    Specifically; if a logic has a model is there a correspondence between a topological space on the set which models it and how proof works in the logic?
  • fishfry
    3.4k
    I apologize for interrupting a productive flow of thought. But I was curious what you guys were talking about. Seems pretty esoteric.jgill

    I can summarize. Short answer is that these days you can do logic via category theory; and when you do that, you get intuitionist logic (denial of the law of the excluded middle (LEM) and all that) in a natural way.

    This relates to topology via the idea of fiber bundles from differential geometry if you know what those are. If not just ask). The examples presented so far aren't clear to me and I haven't worked through @fdrake's promising-looking examples. @Mephist may or may not have presented coherent examples but there's a gap between his expositions and my understanding that only gets worse over time. The fault may all be mine.

    There's a very nice illustration of how this works if you consider any topological space and restrict your attention to the open sets. The "complement" of an open set in a topology, if you only care about open sets, is the interior of the complement of the set. (In other words in general the complement is not an open set, but if we only look at the open sets, it makes sense to define funny complements this way).

    So it is in general NOT true that the complement of the (true) complement of an open set is going to give you back the original open set. This corresponds to a failure of the law of the excluded middle.

    For a very nice overview of the history and meaning of all this I recommend the prologue of Mac Lane's Sheaves in Geometry and Logic. One need not understand the details to get the big picture from this very clearly written book.

    Intuitionism was developed in the 1930's but didn't get any mindshare in mainstream math. Now with the advent of computers (where the complement of a noncomputable set of natural numbers may also be noncomputable), denial of LEM is back in fashion, especially in computer science and categorical logic. I call this Brouwer's revenge. Brouwer invented intuitionism in the old days but it didn't catch on. His form of intuitionism had a touch of mysticism to it, but the modern versions are mathematically solid. Fifty years from now (or sooner) they'll be teaching this to undergrads and set theory will be a relic of the past like Euclidean geometry. Set theory of course won't become wrong, just out of fashion.

    Another thread of development is that mathematicians want to use computers to check their proofs for accuracy. It turns out that intuitionist type theory (which I know nothing about) is the key. In homotopy type theory one uses the idea of homotopy from topology (continuously deforming one path into another) to do intuitionist logic in such a way that you can build working computerized proof assistance for professional mathematicians. Also see intuitionistic type theory.

    These are the broad outlines I've picked up, but I haven't spent much if any time on the details. One name you'll hear a lot is Vladimir Voevodsky, a Fields medal winning mathematician who became frustrated at longstanding errors in published proofs and devoted himself to the project of computerized proof assistants. He died tragically young just recently, in 2017.

    Voevodsky's contribution is the Univalent foundations of mathematics. The idea here is that mathematicians routinely conflate equality and equivalence whenever it's convenient. For example there's only one cyclic group of order four, even though there are lots of isomorphic copies of it that are not equal as sets. For example the integers mod 4 and the powers of the imaginary unit are the "same" group.

    I should mention that this kind of thing bothers @Metaphysician Undercover greatly, and he's right that mathematical equality can sometimes be stretched past what he would consider true equality. The answer to this is that if it quacks like a duck it's a duck, and if it's isomorphic to the cyclic group of order four, it doesn't matter what representation we choose. They're all the "same" in the appropriate technical sense.

    In category theory this conflating of equality and equivalence becomes semi-formalized via universal properties. Any two objects that satisfy the same universal property are isomorphic and are regarded as the same thing.

    Univalent foundations takes this one step farther by formalizing an axiom that says that equivalent things are equal. My high-level understanding is that the univalence axiom makes mathematically precise the informal practice mathematicians have been accustomed to for decades.

    https://en.wikipedia.org/wiki/Univalent_foundations

    All of what I've written is of course hopelessly vague and should not be relied on as gospel. But I've hit most of the buzzwords and major concepts in their broad outlines.

    ps -- The immediate subject of the thread recently is to see how we can view intuitionist logic as an example of a fiber bundle. Maybe that was the short answer to the question.
  • Mephist
    352
    What do logic and topology have to say about each other?

    Specifically; if a logic has a model is there a correspondence between a topological space on the set which models it and how proof works in the logic?
    fdrake

    I see that there is a misunderstanding between us on what it means "a logic has a model".

    A logic is a bunch of rules that describe how you can build sentences that speak about "something".
    What I call model is that "something". For example, the real numbers can be the model. The model is the thing that we are speaking about. The rules of logic have nothing to do with it! If we speak about the waves of the sea, then "the set of all waves of the sea" is the model. It is "the real thing" that we are speaking about.

    Now, the essential change in the point of view that allows you to see the correspondence between topology and logic it this one: consider sets to be more "fundamental" than their elements.
    So, if our model are the real numbers, the sets of real numbers are more "fundamental" than the single real numbers. If you think about it, that's what boolean algebra does: boolean algebra speaks about sets and operations between sets (union, intersection, complement): you build sets starting from other sets, without mentioning their elements.

    From this point of view, we have an "universe" set (in our example the set of all real numbers), and the set of all sets of real numbers (the powerset of the "universe"), and a boolean algebra defined on the powerset of the "universe".

    Now, we can generalize logic by substituting the powerset of the "universe" with a topological space. A topological space is in general defined as the powerset of the "universe", plus a choice of which elements of the powerset are "open" (this choice of the open sets is what's usually called the "topology").
    A topological space, then is a generalization of the powerset of the "universe": instead of considering as the fundamental elements of your algebra the full powerset of the "universe", you consider as your fundamental elements the open sets of the universe (in our example, the open sets of real numbers). The algebra built on the open sets is a Heyting algebra. The algebra built on the full power set of the universe is a Boolean algebra. Of course, you can consider the Boolean algebra as a particular case of Heyting algebra by choosing as open sets all the subsets of the "universe" (the full powerset of the universe). This is what is called a "discrete" topology.
    So, logic built as an algebra based on a topological space is a generalization of the logic built as an algebra on the powerset of the universe, at the same way as a topological space is a generalization of the powerset of the "universe".

    ( I'll describe the part related to proofs the next time )
  • Metaphysician Undercover
    13.1k
    Now, the essential change in the point of view that allows you to see the correspondence between topology and logic it this one: consider sets to be more "fundamental" than their elements.
    So, if our model are the real numbers, the sets of real numbers are more "fundamental" than the single real numbers. If you think about it, that's what boolean algebra does: boolean algebra speaks about sets and operations between sets (union, intersection, complement): you build sets starting from other sets, without mentioning their elements.
    Mephist

    This is the mistaken procedure called Platonism. A set is a human creation, produced to categorize. If we produce an empty set, which may or may not be filled, and this is implied if a set is more fundamental than its elements, then the type, or universal Form, is prior to the particular, or individual. However, as Plato demonstrated, then the type, or universal, what you call "the set", must itself have some existence, and this would be as a particular, individual object. So the empty set has been created with the purpose of being a universal Form, a type, but upon creation, it actual exists as a particular object. To uphold the premise, that a set, or universal Form is more fundamental than an element, this created object, the empty set must already exist as an element of an existing larger set. Aristotle demonstrated this premise, that the Form, as a universal type, (what you call "the set") is more fundamental than its elements, leads to an infinite regress and is actually impossible, therefore false. This is because the set itself can only be represented and understood as a particular object, and understanding a particular requires relating it to a more universal, categorizing it.

    The Neo-Platonists get beyond this problem by producing an Ideal particular as the most fundamental. The Ideal particular is the most fundamental, as a type, a universal Form, or a set, which is also an individual, or particular. It is both. However as the most fundamental, it cannot be an element of a further set, or part of a more fundamental or universal Form. As a particular, and also the most fundamental universal, it is identified as the "One".

    Making the "One" the most fundamental resolves the inherent contradiction of having the empty set as fundamental. The empty set is inherently contradictory because it is something, an object, which at the same time must be nothing.
  • tim wood
    9.2k
    For a very nice overview of the history and meaning of all this I recommend the prologue of Mac Lane's Sheaves in Geometry and Logic. One need not understand the details to get the big picture from this very clearly written book.fishfry

    From page 4 of the the text referenced:
    1) "Since the cardinality of the set R of reals is the same as that of the powerset P(N) of the set of natural numbers."

    Please help me out?
    2) Is there an error in thinking of a representation of a powerset as all the permutations of the elements of the original set?

    3) if 1 and 2 are correct (and if 2 is correct, then I'm thinking 1 obviously follows), then the question of the cardinality of the continuum, c, becomes the question of the existence of point on the line to which no real number can be applied - for some reason: is this a correct way to think of it?

    4) But if 3, and there is no such point on the line, then (it appears to me) that c = P(N).

    5) And it cannot be that simple. which implies there are points on the line that cannot be numbered.

    6) By "number on the line," I am assuming that to each point on the line is assignable some unique number representable as, say, some numeral in binary form, all of which points/binary numerals represented in the set of permutations of all the zeros and ones.

    Is 5 the true statement, that there are points on the line to which no real number can be applied?
  • Mephist
    352
    Aristotle demonstrated this premise, that the Form, as a universal type, (what you call "the set") is more fundamental than its elements, leads to an infinite regress and is actually impossible, therefore false.Metaphysician Undercover

    What I wrote is only an idea, that (in my opinion) is important to understand the "meaning" of a theory, but from the point of view of mathematics all explanations that you can give by words are worth nothing: at the end, the only thing that counts in a mathematical argument are proofs. If what you say cannot be proved, it's not mathematics. I know, neither of us presented any proof of what we said here, but we are on a philosophy forum here, right? :wink:
    What I want to say is that your argument "... leads to an infinite regress and is actually impossible, therefore false" (Aristotle's argument) would not be accepted as a valid prove in today's mathematics.
    In mathematics you are free to "invent new worlds" (I believe this is what Grothendieck was saying about his work), but you have to do it using proofs that are rigorous enough to be accepted by peer reviewers.
    Then, you can discuss if what you invented is "important", and what's it's "meaning".

    Making the "One" the most fundamental resolves the inherent contradiction of having the empty set as fundamental. The empty set is inherently contradictory because it is something, an object, which at the same time must be nothing.Metaphysician Undercover

    The empty set is only an abstract construction defined by a set of axioms that has nothing to do with the "One" of ancient Greek philosophers.
  • jgill
    3.8k
    I can summarize. Short answer is that these days you can do logic via category theory; and when you do that, you get intuitionist logic (denial of the law of the excluded middle (LEM) and all that) in a natural way.fishfry

    I really appreciate your explanation. Thank you. I looked briefly at your first link to get an idea of the univalent approach. I was completely unaware of this, being happily non-constructive at times! :smile:
  • aletheist
    1.5k
    2) Is there an error in thinking of a representation of a powerset as all the permutations of the elements of the original set?tim wood
    Combinations, not permutations; i.e., the different proper subsets, and the order of the members does not matter. For a set with n members, its power set has 2^n members.

    4) But if 3, and there is no such point on the line, then (it appears to me) that c = P(N).tim wood
    This was Cantor's view, which is fairly standard among mathematicians today. However, there is a power set for the real numbers, and a power set for that power set, and so on ad infinitum. That being the case, some argue that the real numbers are not truly continuous, despite comprising what is conventionally called the analytical continuum.

    Is 5 the true statement, that there are points on the line to which no real number can be applied?tim wood
    There are no points in a truly continuous line, period. As a one-dimensional continuum, its parts are all likewise one-dimensional, rather than dimensionless points. We could hypothetically mark points on a line of any multitude--including that of the real numbers and that of their power set--or even beyond all multitude.
  • Metaphysician Undercover
    13.1k
    What I wrote is only an idea, that (in my opinion) is important to understand the "meaning" of a theory, but from the point of view of mathematics all explanations that you can give by words are worth nothing: at the end, the only thing that counts in a mathematical argument are proofs. If what you say cannot be proved, it's not mathematics. I know, neither of us presented any proof of what we said here, but we are on a philosophy forum here, right?Mephist

    It's not true that words are worth nothing in mathematics, because the axioms are written in words. My demonstration was a proof, a logical proof that a set cannot be more fundamental than its elements, because that creates an infinite regress. If you are satisfied with an infinite regress you have an epistemological problem. Such mathematics is not supported by sound epistemology.
  • tim wood
    9.2k
    Thank you for the reply. My post was not entirely thought out. My question really runs like this:

    The quote is
    "Since the cardinality of the set R of reals is the same as that of the powerset P(N) of the set of natural numbers."

    Now what can this mean? It implies that the elements of P(N) can be matched with the reals.

    Let's consider the numbers between zero and one. (I'm thinking) each has a representation in binary form (a concatenation of zeros and ones). Some of finite length, e.g., 1/2, 1/4, 1/10, & etc., and some of non-finite length, viz., all the others. It seems to me that those must all be of denumerably infinite length and not longer. That is, given an irrational number between zero and one, in binary form, its digits must be countable, meaning it must be possible to set the digits of such a number into a one-to-one correspondence with the counting numbers, which is my understanding of denumerable.

    So far so good? I'm sure I'm making a mistake somewhere....

    Let's start a list of them all.
    .1
    .01
    .11
    .001
    .011
    .101
    .111
    .0001
    .0011
    .0101
    .0111
    .1001
    .1011
    .1101
    ...
    You get the idea.

    This list will eventually take in all the numerals of denumerable length. Or at least can put into a one-to-one correspondence all the members of this list with the integers, N, meaning that this list is countable. In as much as N is denumerable, it seems to me that if this list is of denumerable length, then it's countable.

    On this list, however, is every sequence, every combination, of zeros and ones. I think it is accurate to say that "every combination" is equivalent to the power set. The power set of what? The power set of N zeros and ones. Which, according to the quote above, corresponds to the set of reals. In a private message, I've been advised that any listing like the above is subject to the diagonalization argument which shows that for any arbitrary listing, there is always an entry not on the list (and that can be added). I have no argument with this other than to point out two things: 1) this listing is designed to be a complete and exhaustive listing of all numerals of a denumerable number of digits. And 2) the image of the number line itself is a complete and ordered collection of the reals. Our difficulty lies in representing them.

    There is no attempt here to establish any limit on or value of N, but only to suggest that countability seems to coincide with being denumerable, in the same sense that N is denumerable and obviously countable.

    I'm not-so-much arguing, but rather presenting a picture of how P(N) can be listed, and on the basis of the quote above, it must then follow that the reals are countable. I'm educated to believe the reals are not countable, therefore I take it that card (P(N)) is not equal to card(R).
  • Mephist
    352
    What do logic and topology have to say about each other?

    Specifically; if a logic has a model is there a correspondence between a topological space on the set which models it and how proof works in the logic?
    fdrake

    (continuation: correspondence between a topological space and how proof works in the logic)

    The rules of logic should be valid for ANY topology and ANY subset of points. So, for example, they should work even for a discrete topology.

    Now, for intersection and union everything works fine, since the intersection of two open sets is again an open set, and the union of two open sets is again an open set. But the complement operation does not preserve the openness, so it cannot be a primitive operation in an algebra of open sets.

    Well, the "trick" to make it work is very simple: do NOT considered it as a primitive operation but define "complement of A" as "the largest OPEN set X such that the intersection of X and A is empty" (the largest open set with no points in common with A, but we don't need to speak about points)

    With this definition the two logics are exactly the same in the limit case when all subsets are open, but the second one (intuitionist logic) is weaker: every intuitionist derivation is even a standard derivation, but there are boolean logic derivations that are not intuitionistic derivations. The difference is due to the fact that intuitionistic logic does not "see" the points that are not part of open sets (the points on the border of an open set, or even isolated points). If you take a look at the example with fiber bundles that I posted a couple of days ago, you can see what I mean: even if you think of your model as a set of points, you can't speak about single points using this language: every eventual single point not included in an open set is simply "ignored".

    The effect on the logic rules, as @fishfry pointed out, is that double complement (double negation) does not give you back the original set: if you had a set A that was not open nor closed (let's say an open set plus some part of the border), the first complement operation gives you the complementary OPEN set; and then the second negation gives you back A WITHOUT BORDER (that now has become open. Taking again the complement a third time, now you obtain the same result as taking it once, and so on.

    The effect on the rules of logic (one of the effects) is that the excluded middle is not valid rule. But if you add the rule of excluded middle to intuitionistic logic as an additional axiom you don't obtain an unsound system: you simply get back boolean logic. This is equivalent to choosing as topology of the space the discrete topology. It's an additional assumption: intuitionistic logic works for any topology; boolean logic, instead, works only in the particular case of the discrete topology.
  • Mephist
    352
    It's not true that words are worth nothing in mathematics, because the axioms are written in words. My demonstration was a proof, a logical proof that a set cannot be more fundamental than its elements, because that creates an infinite regress. If you are satisfied with an infinite regress you have an epistemological problem. Such mathematics is not supported by sound epistemology.Metaphysician Undercover

    I meant words in plain english language (or in another natural language): you have to use a formal language to express mathematical theorems.
  • Metaphysician Undercover
    13.1k

    I see mathematical axioms expressed in plain English.
  • Mephist
    352
    I see mathematical axioms expressed in plain English.Metaphysician Undercover

    There is a way to translate any mathematical proposition (or axiom) into plain English, but there is no way to translate any English proposition into a mathematical proposition: formal languages are more limited than natural languages.
  • Metaphysician Undercover
    13.1k

    Sure, but how's that relevant? What is at issue is the postulate that a set is more fundamental than its elements. That's plain English.
  • Mephist
    352
    Yes, but that's not mathematics! The distinction of which concepts are more "fundamental" is very useful to "understand" a theory, but it cannot be expressed as part of the theory. Mathematical theorems don't make a distinction between more important and less important concepts: if a concept is not needed, you shouldn't use it. If it's needed, you can't prove the theorems without it.

    P.S. That's a very important point to understand: the words used in mathematical sentences are not chosen at random: they are carefully chosen to give some "intuition" of the things that we are speaking about. However, you cannot use that intuition in proofs. Proofs have to be completely "formal": they have to be valid even if you substitute the words with random strings of characters.
  • fishfry
    3.4k
    From page 4 of the the text referenced:
    1) "Since the cardinality of the set R of reals is the same as that of the powerset P(N) of the set of natural numbers."

    Please help me out?
    tim wood

    The powerset of a set is the set of all subsets of the set. So for example .

    It's easy to show that . For any subset of , create a bitstring that has 1 in the n-th position if n is in the subset, 0 otherwise. Put a binary point in front of the string and you have the binary expression of a real number in the unit interval, and vice versa.

    For example the real number whose binary representation is .10101010101... corresponds to the set {1, 3, 5, 7, ...}. So we have a bijection between the real numbers (in the unit interval) and the subsets of the natural numbers. (For convenience I'm numbering positions to the right of the binary point starting at 1, and excluding 0 from the natural numbers).

    [We can ignore dual representations (.5 = .4999...) because there are only countably many of those and countable sets don't make any difference to the cardinality of an infinite set].


    2) Is there an error in thinking of a representation of a powerset as all the permutations of the elements of the original set?tim wood

    That doesn't work because for example the set of permutations of {1.2.3} is 123, 132, 321, 312, 213, and 231. It's not the same thing.

    3) if 1 and 2 are correct (and if 2 is correct, then I'm thinking 1 obviously follows), then the question of the cardinality of the continuum, c, becomes the question of the existence of point on the line to which no real number can be applied - for some reason: is this a correct way to think of it?tim wood

    No not really. We just proved above that the cardinality of the reals is , the set of functions from to the set {0,1} (which we can think of as the set of bitstrings). Now if the transfinite cardinals are , the question is which Aleph is ? The claim that it's is the Continuum hypothesis. For all we know it's some other Aleph, perhaps a very large one. The answer is independent of the usual axioms of set theory.

    4) But if 3, and there is no such point on the line, then (it appears to me) that c = P(N).tim wood

    I'm not sure I follow your idea of a particular point on the line.

    5) And it cannot be that simple. which implies there are points on the line that cannot be numbered.tim wood

    I don't follow your idea but that's not what CH is. CH is just the question of which Aleph is the cardinality of the reals.

    6) By "number on the line," I am assuming that to each point on the line is assignable some unique number representable as, say, some numeral in binary form, all of which points/binary numerals represented in the set of permutations of all the zeros and ones.tim wood

    The points on the real line are just the real numbers and vice versa.

    Is 5 the true statement, that there are points on the line to which no real number can be applied?tim wood

    I'm afraid I don't follow this idea. The "real line" is just a synonym for the set of real numbers.
  • fishfry
    3.4k
    I see that there is a misunderstanding between us on what it means "a logic has a model".Mephist

    I found a paper that indicated the the fibers are "L-structures." Not too sure what those are, or what the base set is. I'm not sure I entirely believe it's a discrete topological space. I'm thinking you've probably explained this point to me several times over but I still don't get it. My apologies for giving you a hard time out of frustration at my inability to understand how fiber bundles can be used to model logical structures.
  • fishfry
    3.4k
    consider sets to be more "fundamental" than their elements.Mephist

    I'm afraid I share @Metaphysician Undercover's misgivings about this remark. I understand the categorical viewpoint of sets, but I would not characterize that viewpoint via this particular way of phrasing it.
  • fishfry
    3.4k
    There are no points in a truly continuous line, period.aletheist

    That's a Peircean view and not a standard mathematical view; and I think it's important to make that distinction when explaining things. The standard mathematical view is that "the continuum," "the real line," and "the set of real numbers" are synonymous. Philosophical considerations do not alter the conventional mathematical meanings.
  • fishfry
    3.4k
    Let's start a list of them all.
    .1
    .01
    .11
    .001
    .011
    .101
    .111
    .0001
    .0011
    .0101
    .0111
    .1001
    .1011
    .1101
    ...
    You get the idea.

    This list will eventually take in all the numerals of denumerable length.
    tim wood


    No, your idea only lists all the bitstrings of FINITE length, of which there are only countably many. For example 1010101010101010... never appears on your list.
  • tim wood
    9.2k
    No, your idea only lists all the bitstrings of FINITE length, of which there are only countably many. For example 1010101010101010... never appears on your list.fishfry

    Well, you've arrived just where I have a problem, or, rather, where I'm unclear and confused, because I'm not arguing against any well-known fact, but rather I seem to be stuck in some misconception or misperception.

    Taking your .10101010..., how long is it? How many zeros and ones? As many as there are counting numbers? Or more? ℵo or ℵ1?

    I'm thinking the number of digits must be countable. And I'm thinking my listing, then, being ordered, is also countable. It's all countable. But clearly that's not correct. So the questions are 1) are the digits in the numeral, the binary expansion, countable, by which I mean denumerably infinite? (Assuming "countable" is what denumerably infinite means, as opposed to not-countable). And 2) is the listing countable (same assumption)? On its construction it seems it must be.

    And if the list is denumerable and complete (just as N is denumerable and complete), then the diagonal argument seems not to work, because any new number generated by the diagonal process will already be somewhere on the list. If the list is non-denumerable, then I suppose that there is no method for making a complete list, and that new diagonal numbers won't be on the list.
  • Mephist
    352
    I found a paper that indicated the the fibers are "L-structures." Not too sure what those are, or what the base set is. I'm not sure I entirely believe it's a discrete topological space. I'm thinking you've probably explained this point to me several times over but I still don't get it. My apologies for giving you a hard time out of frustration at my inability to understand how fiber bundles can be used to model logical structures.fishfry

    I don't know what are "L-structures", but I think I know what's the source of misunderstanding: the words "discrete" and "continuous" used to refer to finite structures. In my example the "space" of the model is made of 12 points, but it's NOT a discrete space: not all subsets of the set {1...12} are open sets BY DEFINITION. The definition of which sets are open is arbitrary: the only required conditions is that it has to include the empty set, the full set, and all possible unions and intersections.

    You should see the topology as a kind of "blurring glass" that is put over the set {1...12} and does not allow you to distinguish the individual points: you can see groups of points, but not individual points. Think of the set of real numbers when they are interpreted as results of physical experiments: you explained this to me very clearly: you can never get a real number as the result of an experiment (and you can't split a physical sphere in distinct points as in the Banach-Tarsky theorem).
    The same thing can be true for the set of 12 points in my example: you cannot distinguish the point 1 from the point 2, because there are no open set {1} and open set {2} in the topology.
    Of course, the most interesting cases of open sets are infinite sets (as real numbers), not finite ones as in my example. But I especially chose a finite set to make it crystal clear: topology is not about the cardinality of the "universe" set.
  • fishfry
    3.4k
    Taking your .10101010..., how long is it? How many zeros and ones? As many as there are counting numbers? Or more? ℵo or ℵ1?tim wood

    There's one decimal place for each natural number. A decimal expression .abcdef... means a/10 + b/100 + c/1000 + ... There's one place for each negative power of 10.

    I'm thinking the number of digits must be countable. And I'm thinking my listing, then, being ordered, is also countable. It's all countable. But clearly that's not correct.tim wood

    Your list is countable. You've listed all the FINITE bitstrings. Where is .10101010101010... on your list? It's not there.
  • fishfry
    3.4k
    But I especially chose a finite set to make it crystal clearMephist

    Not clear to me. I literally and honestly did not understand what you said in this post. Perhaps it's a lost cause.

    A fiber bundle is like the collection of tangent planes to a sphere. Somehow, one can replace the tangent planes with logical structures of some sort, and the points of the sphere with .... something, and intuitionist logic drops out. Perhaps it's not explainable in elementary terms. But I couldn't relate what you wrote with any attempt to clarify this point.
  • Mephist
    352
    I'm afraid I share Metaphysician Undercover's misgivings about this remark. I understand the categorical viewpoint of sets, but I would not characterize that viewpoint via this particular way of phrasing it.fishfry

    All right. Fair point.
  • Mephist
    352
    Not clear to me. I literally and honestly did not understand what you said in this post. Perhaps it's a lost cause.fishfry

    Well, OK, never mind. However, the book that I gave you the link is very clear and contains proofs and exact definitions. Surely that's easier to understand than my explanations...
  • fishfry
    3.4k
    Well, OK, never mind. However, the book that I gave you the link is very clear and contains proofs and exact definitions. Surely that's easier to understand than my explanations...Mephist

    Ok I will have a look. Many links have been posted recently. Can you repost the one you want me to look at please?
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