• Banno
    22.9k
    Nice try, .

    I see no point in continuing this discussion.Philosopher19

    Then have a look here:

    Theorem 1.29 (Russell’s Paradox). There is no set R = {x : x ∉ x}.
    Proof. If R = {x : x ∉ x} exists, then R ∈ R iff R ∉ R, which is a contradiction.
    Open Logic: Complete build

    Banno's Law says that it is easier to critique something if you begin by misunderstanding it. That is what your OP does.

    You, too,

    , you may find the Open Logic text useful.
  • Philosopher19
    273


    Again, I have said this multiple times. I recognise and acknowledge the following:
    Proof. If R = {x : x ∉ x} exists, then R ∈ R iff R ∉ R, which is a contradiction.Open Logic: Complete build

    What has not been shown to me is how this logically obliges us to view the set of all sets as contradictory.

    Again, it is clearly contradictory for any set to contain all sets that are not members of themselves and no other set. But this does not mean that it is contradictory for the set of all sets to contain all sets that are not members of themselves. The set of all sets contains absolutely all sets that are not members of themselves and it is a member of itself.
    Just because one set is contradictory (In the above case R), doesn't mean another set is also contradictory when they are not the same sets.

    The set of all sets and R are not the same sets.
  • Banno
    22.9k
    What has not been shown to me is how this logically obliges us to view the set of all sets as contradictory.Philosopher19

    And this shows that you have not understood R = {x : x ∉ x}:

    How do we set up a set theory which avoids falling into Russell’s Paradox, i.e., which avoids making the inconsistent claim that R = {x : x ∈/ x} exists? Well, we would need to lay down axioms which give us very precise conditions for stating when sets exist (and when they don’t). — On the next page...

    :wink:
  • Philosopher19
    273
    And this shows that you have not understood R = {x : x ∉ x}:Banno

    To me it looks like no meaningful answer has been given to me and that what I have posted has not been paid attention to.
  • Banno
    22.9k
    Then I'm sorry for you. All I can suggest is that you read the section of the Open Logic project on Russell's paradox again, very carefully.

    Because the choice is between the whole of the remainder of that project being founded on an error unnoticed by more than a century of study by logicians world wide; and your being mistaken.

    Which is most likely?
  • TonesInDeepFreeze
    2.3k
    There are too many confusions in this thread. This post can be used for correct and definite formulations. (For more, see https://plato.stanford.edu/entries/russell-paradox)

    Below are three theorems, using only intuitionistic (thus classical too) rules of inference, applied to certain axioms:

    One may wish to conceive mathematics in some other way, but if the discussion pertains to the rules of inference and axioms of ordinary mathematics, then below are English renditions of formalizations that have formalized proofs that are are machine-checkable from the stated axioms and rules of inference; that cannot be rationally disputed.

    (1) There is no y such that for all x, x is a member of y if and only if x is not a member of itself.

    Proof is by pure logic alone, no set theory need be invoked, as it is merely a case of:

    There is no relation M such that there is a y such that for all x, x bears M to y if and only if x does not bear M to x.

    (2) There exists a unique y such that for all x, x is member of y if and only if x is a member of x. That y is 0 (the empty set).

    Proof uses the axiom of regularity, the schema of separation for existence, and the axiom of extensionality for uniqueness.

    (3) There is no z such that for all x, x is a member of z.

    Proof uses (1) and the axiom schema of separation.

    EXPLANATIONS:

    'The set of all sets that are not members of themselves' refers to a set, call it 'R', such that for any set x, x is a member of R if and only if x is not a member of x.

    Russell's paradox is this contradiction:

    The set of all sets that are not members of themselves is a member of itself, and the set of all sets that are not members of themselves is not a member of itself.

    A proof* is obvious:

    Suppose there is a set R such that for all sets x, x is a member of R if and only if x is not a member of x.

    So substituting R for x, we have:

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

    Suppose R is a member of R. So R is not a member of R. So both R is a member of R and R is not a member of R.

    Suppose R is not a member of R. So R is a member of R. So both R is member or R and R is not a member of R.

    But either R is member of R or R is not a member of R. And, in either case, as above, R is a member or R and R is not a member of R. So R is a member of R and R is not a member of R. QED.

    /

    So, the supposition 'There is a set R such that for all sets x, x is a member of R if and only if x is not a member of x' yields a contradiction. So we infer the negation of that supposition, viz. 'There is no set R such that for any set x, x is a member of R is and only if x is not a member of x'.

    But the earliest version of set theory had an axiom schema we all 'unrestricted comprehension', which is (read 'P(x)' as 'x has property P'):

    For any property P that we can formulate, we have the axiom:

    There is a set y such that for all sets x, x is a member of y if and only if x has property P.

    Then let P(x) be 'x is not a member of x'.

    Then we have:

    There is a set y such that for all sets x, x is a member of y if and only if x is not a member of x.

    But above (using 'R' instead of 'y') we proved that there is no such set y.

    So the axiom schema of unrestricted comprehension yields a contradiction.

    /

    So set theory was reformulated to not have the unrestricted axiom axiom schema of comprehension but instead a restricted axiom schema (called the 'separation schema' or the 'specification schema' or the 'subset schema'):

    For any property P that we can forumulate, we have the axiom:

    For all sets z, there is a set y such that for all sets x, x is a member of y if and only if (x is a member of z and P(x)).

    And to that schema we add the axioms: Extensionality, Pairing, Union, Power Set, Infinity and Regularity. (This is known as 'Z set theory').

    In Z set theory (and in the extended theories, ZF, ZFC, etc.), there is no known proof of 'There is a set R such that for all sets x, x is a member of R if and only if x is not a member of x', and no known proof of any contradiction.

    Moreover, in Z set theory there is a proof of 'There is not a set R such that for all sets x, x is a member of Y if and only if x is not a member of x'. It's just Russell's argument again:

    Suppose there is a set R such that for all sets x, x is a member of R if and only if x is not a member of x.

    So substituting R for x, we have:

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

    Suppose R is a member of R, so R is not a member of R. So both R is a member of R and R is not a member of R.

    Suppose R is not a member of R, so R is a member of R. So both R is member or R and R is not a member of R.

    But either R is member of R or R is not a member of R. And, in either case, as above, R is a member or R and R is not a member of R. So R is a member of R and R is not a member of R.

    NOTICE that that argument didn't even need any axioms of set theory anyway. Moreover, we don't even have to mention the notion of 'set' nor 'member of'. We have (read 'M(x y) as 'x bears property M to y'; with one instance being 'x bears the property of being a member of y'):

    There is no 2-place relation T such that there is a y such that for all x, T(x y) if and only if it is not the case that T(x x). I.e., 'There is no 2-place relation T such that there is a y such that for all x, x bears T to y if and only if x does not bear T to x.'

    It's purely logical, not needing to mention sets nor membership:

    Suppose there is a T such that there is a y such that for all x, T(x y) if and only if it is not the case that T(x x).

    So, substituting y for x, we have T(y y) if and only if it is not the case that T(y y). And, as seen above, we then derive a contradiction.

    Indeed, this may be seen as the lesson of the Barber Paradox in its purest form (don't even have to mention 'village' or 'barber'). There, instead of the 2-place property 'is a member of' we use the property 'shaves':

    There is no y such that for all x, y shaves x if and only if x does not shave x.

    /

    Moreover, even in set theory, we don't have to mention the term 'set'. Rather, the only primitive is 'member of'. Indeed, though most people don't bother to do state these trivial steps, we can actually define 'set' ('iff' stands for 'if and only if'):

    First with the axiom of extensionality, we define a unique y such that for all x, x is not a member of y. We dub that unique y as '0':

    Axiom of extensionality:

    For all x and y, if for all z, z is a member of x iff z is a member of y, then x equals y.

    Also, from the logic of identity, we already have:

    For all x and y, if x equals y, then for all z, z is a member of x iff z is a member of y.

    So putting that together with the axiom of extensionality we have:

    For all x and y, x equals y iff for all z, z is a member of x iff z is a member of y.

    Then:

    From the schema of separation we have:

    For all z, there is a y, such that for all x, x is a member of y iff (x is a member of z and (x is a member of x and x is not a member of x)).

    Let z be any object. (There is at least one object, as provided by the logic itself).

    So there is a y, such that for all x, x is a member of y iff (x is a member of z and (x is a member of x and x is not a member of x)).

    By logic, we can reduce that to:

    There is a y, such that for all x, x is a member of y iff (x is a member of x and x is not a member of x)).

    By the axiom of extensionality, such a y is unique, since there is no x such that x is a member of x and x is not a member of x.

    Then:

    df. 0 is the unique y such that for all x, x is a member of y iff (x is a member of x and x is not a member of x).

    df. x is a class iff there is a y such that y is a member of x or x is 0.

    df. x is a set iff x is a class and there is a z such that x is a member of z.

    So, since the context here is set theory, we can drop mentioning 'set' throughout. Though, we use the nickname for 0 as "the empty set".

    /

    Next, using only the schema of separation we prove that there is no z such that for all x, x is a member of z:

    Suppose there is a z such that for all x, x is a member of z:

    So, by the schema of separation:

    There is a y such that for all sets x, x is a member of y iff (x is a member of z and x is not a member of x).

    With some routine logic steps, we get:

    There is a y such that for x, x is a member of y iff x is not a member of x.

    Thus, there is no z such that for all x, x is a member of z.

    NOTICE, we need no invoke extensionality nor regularity, only an instance of the axiom schema of separation.


    /

    * That proof uses excluded middle, but we can prove it intutionistically too:

    Instead of proving that both R is a member of R and R is not a member of R, we prove a different contradiction, viz. that both R is not a member of R and it is not the case that R is not a member of R:

    Suppose there is a set R such that for any set x, x is a member of R if and only if x is not a member of x.

    So substituting R for x, we have:

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

    If R is a member of R then both R is a member of R and R is not a member of R. But it is not the case that both R is a member of R and R is not a member of R. So, by modus tollens, R is not a member of R.

    If R is not a member of R then both R is a member of R and R is not a member of R. But it is not the case that both R is a member of R and R is not a member of R. So, by modus tollens, it is not the case that R is not a member of R.

    But both 'If R is a member of R then R is not a member of R' and 'If R is not a member of R then R is a member of R'. So both R is not a member of R and it is not the case that R is not a member of R. QED.
  • Philosopher19
    273


    Because the choice is between the whole of the remainder of that project being founded on an error unnoticed by more than a century of study by logicians world wide; and your being mistaken.Banno

    That sounds like bias and dogma to me as opposed to actual discussing of the argument with attention to detail.
  • TonesInDeepFreeze
    2.3k
    Regarding a notion that ZF is "inadequate/incomplete".

    If ZF is consistent then ZF is incomplete, in the sense that there are sentences in the language of ZF such that neither they nor their negations are theorems of ZF. But any consistent, recursively axiomatized theory that expresses a certain amount of arithmetic is incomplete.

    'inadequate' is not defined here.

    /

    Regarding the claim that set theory is contradictory.

    A contradiction is a sentence P along with not-P.

    There is no known proof of a contradiction in set theory.

    One may find that the theorems of set theory go against one's own intuitions or conceptions about mathematics, but that does not establish a claim that the theory is contradictory, since 'contradictory' doesn't mean "Goes against the way I conceive things" but rather means "Proves both a sentence P and not-P".)
  • Banno
    22.9k
    We can't help that you can not see what is going on here.
  • Michael
    13.9k
    What has not been shown to me is how this logically obliges us to view the set of all sets as contradictory.Philosopher19

    It doesn't. This is your misunderstanding. Russell's paradox only shows that the axiom schema of unrestricted comprehension leads to a contradiction, and so that naive set theory is inconsistent.
  • TonesInDeepFreeze
    2.3k
    Just to be clear, neither Regularity nor Pairing are needed to prove:

    There is no y such that for all x, x is a member of y iff x is not a member of x

    nor

    There is no z such that for all x, x is a member of z.
  • TonesInDeepFreeze
    2.3k
    To properly discuss 'lists' in the context of set theory, we need to have a formalization of the notion of lists:

    At least personally I would take 'list' to mean a function whose domain is either a natural number or the set of natural numbers.

    A finite list has a natural number as its domain (recall that every natural number is the set of its predecessor natural numbers).

    A denumerable list has the set of natural number as its domain.

    I guess one might say that any ordinal could be the domain of a list, including countable ordinals other than the set of natural numbers or even uncountable ordinals.

    /

    Then, what would be involved here is not lists being members of lists, but rather lists being in the RANGE of lists. So it is not "a list being a member of itself" but rather a list being in the RANGE of itself. That is, to say "the list lists itself" doesn't mean "the list is as a member of itself" but rather "the list is in the RANGE of itself".

    For example, If I have a list L of three lists, then it looks like this

    L = {<1 List-a> <2 List-b> <3 List-c>}

    List-a, List-b, and List-C are not in L. Rather they are in the RANGE of L.

    /

    In any case, there is no known contradiction in set theory arising from questions of lists
  • TonesInDeepFreeze
    2.3k
    An argument was made that it is contradictory for a set to have as members all and only the sets that are not members of itself (correct) but that it is not contradictory for a set to have as members all the sets that are not members of themselves along with other set (incorrect in context of set theory).

    In set theory, from any set, we can take subsets defined by any property.

    So, let B be a set that has as members all the sets that are not members of themselves plus some other members. Now take the subset of B that has only the sets that are not members of themselves, and we are back to the contradiction.

    So it is a theorem of pure logic:

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

    And it is a theorem from the subset axiom schema:

    There is no y such that for all x, if x is not a member of x, then x is a member of y.

    /

    Moreover, given the subset axiom schema, it is a theorem:

    There is no z such that for all x, x is a member of z.

    /

    Again, one may wish to conceive of mathematics and sets in some other way than as usual in mathematics, but then we may ask, "What are your primitives, your rules of inference and your axioms to express your alternative conception and to prove things about it?"
  • TonesInDeepFreeze
    2.3k
    Again, what Russell's paradox shows is that by pure logic:

    There is no set y whose members are all and only the sets that are not members of themselves.

    Then, from Russell's paradox, along with the subset schema we show:

    There is no set z that has every set as a member.

    and

    There is no set such z such that for every x, if x is not a member of itself then x is a member of z.
  • TonesInDeepFreeze
    2.3k
    It was claimed that

    R = {x | x not-e x}

    has not been explicated.

    '{ x | }' is the abstraction operator. It is variable binding notation that takes any formula P and names a set {x | P(x)} that is the set of all and only the x such that P(x).

    However, proper use of the abstraction operator requires first proving that indeed there is such a set for a given property P.

    And we prove that for the property P, where P(x) iff x is not a member of x, there is no such set of all and only the x such that x is not a member of x.

    There are different ways of handling such situations (Russellian vs Fregan, e.g.).

    In ordinary discourse, mathematicians would say that

    R = {x | x not-e x} is an "illegal" definition, as there is no such object described by the right side of the equation.

    With the Russell treatment, names with abstraction operators are regarding not standalone but rather contextually, so that we evaluate the truth value or satisfiability of formulas without regarding the abstraction operator expression as naming but rather as a nested component indicating an assertion about existence of an object having certain properties.

    With the Fregean treatment, if we have at least one constant symbol, then we may set all non-referring abstraction operator expressions to that constant.
  • Banno
    22.9k
    A valiant effort.

    I suspect it is in vein, and that the issue here is not logical so much as pedagogic.
  • Philosopher19
    273
    I just want to say that I've had a look at the OP and I believe it to be very outdated. I believe that instead of reading the OP one is much better off reading the following (if one was to read my work):

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

    Peace
  • TonesInDeepFreeze
    2.3k
    NOTE: Earlier in this thread, with a different username, I used the word 'onto' not intending its mathematical meaning of a function onto a set, but rather to mean 'in and of itself'. I have later found out that it is not recognized English to use the word 'onto' that way, so I should have said 'in and of itself'. That caused confusion. I regret that.

    Where I said things like 'EyAx xey' is not inconsistent onto itself', I meant 'EyAx xey' is not inconsistent in and of itself' (though it is inconsistent with the axiom schema of separation).
  • TonesInDeepFreeze
    2.3k
    Posts are missing from this thread, including some of my own. What happened?

    And a post of mine was deleted in another thread. So I listed the reasons why it and the posts in this thread should not be deleted. Then that post was deleted and an admin made a post saying where discussions of deletions should be, but when I came back to read the admin's post and take note of where the admin had in mind for discussion of deletions, that post by the admin was also deleted. Thus, I didn't get to save my own post about the deletions to instead put it in the location suggested, and I didn't even get to read the admin's post about the suggested location. That doesn't even include all the posts by me and others in this thread that were wiped out, without notification or stated reason.
  • jgill
    3.5k
    Posts are missing from this thread, including some of my own. What happened?TonesInDeepFreeze

    Good question. This thread has been a particularly vexing one and has to a large extent been a conversation, sometimes a bit heated, between two of TPF's top people in math (set theory in particular). It seems to have resulted in one of these, @fishfry, leaving the forum. I question why such a technical thread on set theory didn't go to the Lounge. I suppose the obvious fact that the initiator was not conversant with the subject might have argued for keeping it on page one, seeing what interesting philosophical ideas might emerge.
  • TonesInDeepFreeze
    2.3k
    What evidence do you have that fishfry left because of this thread?

    The reason technical content should not be shunted elsewhere is that if the philosophical discussion is ABOUT the mathematics, especially a critique of the mathematics, then the mathematics should not be misrepresented, otherwise (1) Misinformation about mathematics is permitted to thrive and (3) The philosophical claims themselves are errant for not even being correctly about what they are supposed to be about.

    Anyway, the recent deletions were not about the earlier exchanges with fishfry.
  • jgill
    3.5k
    What evidence do you have that fishfry left because of this thread?TonesInDeepFreeze

    You are correct. Coincidence is not proof.

    The philosophical claims themselves are errant for not even being correctly about what they are supposed to be about.TonesInDeepFreeze

    So true. The OPs lay out belief systems in one form or another, and sometimes they don't budge. Which I find acceptable in @Metaphysician Undercover's pronouncements, for he dwells with the ancients as they ponder space, time, and points and curves - although he balks at 1+4=5 and has little patience with Weierstrass and his limit ideas: admittedly useful, but fundamentally flawed. But I see where he is coming from there. Others, like this thread, are more or less unmovable in their opinions, which clash with standard mathematics. How you deal with the frustration of offering knowledge to those unwilling to accept it is admirable.
  • Metaphysician Undercover
    12.3k
    So true. The OPs lay out belief systems in one form or another, and sometimes they don't budge. Which I find acceptable in Metaphysician Undercover's pronouncements, for he dwells with the ancients as they ponder space, time, and points and curves - although he balks at 1+4=5 and has little patience with Weierstrass and his limit ideas: admittedly useful, but fundamentally flawed. But I see where he is coming from there. Others, like this thread, are more or less unmovable in their opinions, which clash with standard mathematics. How you deal with the frustration of offering knowledge to those unwilling to accept it is admirable.jgill

    Now, now, let's have fair representation. I balk at the claim that "1+4=5" implies that "1+4" refers to the exact same thing as "5" does. And, I have no patience for people like fishfry who simply assert over and over again, that the axiom of extensionality proves that "1+4" and "5" must refer to the exact same thing in common applications. Further, although I am very interested in the fundamental incompatibility between the proposal of non-dimensional points, and the proposal of a continuous one-dimensional line, I believe no one until today has brought Weierstrass to my attention.
  • TonesInDeepFreeze
    2.3k
    In mathematics, with or without the axiom of extensionality, '=' means identity.
  • jgill
    3.5k
    I believe no one until today has brought Weierstrass to my attention.Metaphysician Undercover

    (I am one of his 35K math descendants) I think you have indicated that the limit concept is useful but doesn't tell the whole story. Now's the time to elaborate. :chin:
  • Philosopher19
    273
    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?)
  • Fire Ologist
    65
    Isn't the set of all sets equivalent to the set of all members?

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

    For example, if we take the set of all numerical values (0 through 9 - ten members) and the set of all alphabetical values (a to z - twenty-six members), we could make the set of these two sets and call it the Set of all Characters (for sake of argument). So does this set of all characters (including 1, 2, 3, etc., and a, b, c, etc.) now contain a total of 36 members? I say yes. Or does it contain 2 more members, total of 38, being the prior sets called "numerical" and "alphabetical" plus their members? If we call it the set of two sets, do we need to add in the set designated "numerical" and the set designated "alphabetical" and count the members separately from the sub-sets, when we join the sub-sets into a new super-set?

    I say no. We are smart animals, so we can look at the single set of all Characters which has 36 members and simultaneously see that there are sub-sets of 10 numbers and 26 letters, but we are not counting those two distinctions as 2 additional members of the set of all Characters. The set of numbers has 10 members, and the set of letters has 26 members, and any set containing only these two sets would have 36 members.

    When you bring sets together, under a new super-set, the sub-set distinctions that were named for example "numerical set" and "alphabetical set" no longer exist - these distinctions are irrelevant or non-existent to the set of all characters.

    Apply this when joining all sets into the "set of all sets". We take the set of numbers, the set of letters, the set of atomic particles, the set of forces, etc., etc., until we take up all sets, list them on the blackboard, and fashion the idea of "the set of all sets". Aren't we just overcounting this new "set of all sets" if we count the sets within it, and not just the members of those sets (ignoring the prior set distinctions themselves)?

    If so, then we have no need to call it "the set of all sets" - the set of all sets (incoherent paradoxical term) becomes the set of all members, or just a misapplication of the term "all".

    This doesn't solve the paradox. It just shows that the paradox is where logical process, which occurs between multiple things (as in between sets and members), borders on the identity of single things (like what is a member, or what is a set). Logic lives between things, in their relations. Once we say the set of all sets, we have a logical problem if we lose site of the members of all those sub-sets and just look at those sub-sets themselves as if those sub-sets could be members without their own sub-members in the first place. As we gather up sets into bigger sets, the distinctions between this set and that set are no longer relevant and do not count in the membership, and only the new set exists.

    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.
  • TonesInDeepFreeze
    2.3k
    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).Philosopher19

    That is confused. A set of all sets has as members all the sets. Let that sink in. ALL sets are members of the set of all sets. 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.

    You claim that a set cannot be a member of itself and also be a member of another set.

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

    A subset of "all sets that are members of themselves" and "all sets that are not members of themselves" is contradictoryPhilosopher19

    Maybe you mean that no set x can be both a subset of a set of all sets that are members of themselves and a subset of a set of all sets that are not members themselves.

    That is true of all sets except the empty set.

    The z of all zs iPhilosopher19

    "The z of all zs" has no apparent meaning to me. But if you mean a set z such that every set is a member of z, then that is said as "the z such that for all x, x is a member of z". Let me know that you've fixed that, then maybe I'll go to your next sentence.

    We need a meaningful distinction between "member of self" and "not member of self"Philosopher19

    We have it.

    We need a set of all sets (math/logic would be incomplete without it, if not contradictory)Philosopher19

    Again, you skip my whole explanation to you about incompleteness.

    Again, you skip my refutation of the claim that without a set of all sets we have a contradiction.

    Again, discussion goes in a circle with you as you skip the counterpoints given you. And, again, you've been given a refutation that it is your interlocuters who are circular, so skipping that refutation and instead yet again claiming the fault is with your interlocuters would be yet more circularity from you.
  • TonesInDeepFreeze
    2.3k
    Isn't the set of all sets equivalent to the set of all members?Fire Ologist

    Members of what?

    Of course, every set is the set of all and only its members.

    And in ordinary set theory, every set is a member of another set.

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

    If by "within" you mean 'member of', then the above nonsense. If x is the set of all sets, then, by that description alone, every member of x is a set and every set is a member of x

    And in ordinary set theory (without urelements and without proper classes) every object is a set.

    .
    Or does it contain 2 more members, total of 38, being the prior sets called "numerical" and "alphabetical" plus their members?Fire Ologist

    The set {0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z} has 36 members.

    If we call it the set of two setsFire Ologist

    We call it 'the union' of two sets.

    {0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z}
    is the union of
    {0 1 2 3 4 5 6 7 8 9} and {a b c d e f g h i j k l m n o p q r s t u v w x y z}

    But let N = {0 1 2 3 4 5 6 7 8 9} and L = {a b c d e f g h i j k l m n o p q r s t u v w x y z}.

    Then we also have the set:

    {0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z N L}

    which has 38 members.

    Aren't we just overcounting this new "set of all sets" if we count the sets within it, and not just the members of those setsFire Ologist

    No.
  • Philosopher19
    273


    A set of all sets has as members all the setsTonesInDeepFreeze

    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
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