How does axiom of regularity ensures that Russell's Paradox doesn't occurs. Given the axiom of regularity states: If A is a non-empty set, then there is at least one element x of A which is either not a set, or is disjoint from A.
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In ZF set theory the axiom of regularity (also called axiom of foundation) says that:     In all nonempty sets x there is an element y such that x∩y=∅   As I been told that  of the axiom is to prohibit constructions as $x\in x$, but is this a possible strategy?  Given a binary predicate $R$, written in infix notation, there are unitary predicates $p$ such that:  $\qquad \exists y \forall x: x R y \leftrightarrow p(x)$.  But there also are some convincing counterexamples:   $  p(x) \leftrightarrow \neg(x R x)$ $  p(x) \leftrightarrow \neg(x R f(x))$, where $f$ is any surjection $  p(x) \leftrightarrow$ all chains $x_1 R x, \: x_2 R x_1, \: x_3 R x_2,  \ldots \:$ are finite.   I think there is a major mistake to try to avoid certain paradoxical loops. First, there are no guarantees that every possibility is prohibited. And second, it seems pathological to make constructions that avoid certain proof-techniques (it's like the attitude of a shyster).  So, what protection comes from this axiom? Historically, has it anything to do with the Russell-predicate $x\notin x$?  See also
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Is there anything wrong with the following proof?  Theorem. Every non-empty subset $B \subset\mathbb{N}$ has a least member.  Proof. Assume not. Then, of necessity, we'd have to have $B=\varnothing$, for if $B$ contained even one $b\in\mathbb{N}$, then that $b$ would satisfy $b=min(B). $ Contradiction, end proof.  I get the sense there is something wrong here, but I can't seem to define exactly what. Also, is there a context where one can simply take this principle as an axiom, and not have to prove it? After all, it is extremely intuitive. Thanks.
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Is it possible to prove the following fact without axiom of choice ?  " Every infinite set has a countably infinite subset".  Can it be proved that axiom of choice is necessary here ?
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I'm having trouble understanding the necessity of the Axiom of Choice. Given a set of non-empty subsets, what is the necessity of a function that picks out one element from each of those subsets? For instance, take a look at following paragraph from Wikipedia:     A sketch of the proof of Zorn's lemma follows, assuming the axiom of   choice. Suppose the lemma is false. Then there exists a partially   ordered set, or poset, P such that every totally ordered subset has an   upper bound, and every element has a bigger one. For every totally   ordered subset T we may then define a bigger element b(T), because T   has an upper bound, and that upper bound has a bigger element. To   actually define the function b, we need to employ the axiom of choice.   Using the function b, we are going to define elements a0 &lt; a1 &lt; a2 &lt;   a3 &lt; ... in P. [...]   To me it is clear that the ai's exist so why do we need singling them out? Sure enough, the binary relation representing b may not exist as a set, but what do we need it for?  Finally... if we actually need the Axiom of Choice to single out the ai's in the above paragraph, since the axiom only postulates the existence of choice functions, how do we single out the choice function itself so that we can select the ai's with it? Certainly not with another choice function, since we would need to choose this function by a third choice function.  So when and why do we really need the Axiom of Choice?
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There are things which are illogical/logically impossible (like saying that 2+2=4 and 2+2=5. Without changing anything in the axioms of mathematics or logic, this would be a contradiction and would be inconsistent and illogical/logically impossible.   There are other types of logic systems apart from classical logic, like paraconsistent logics or even trivialism, that allow these contradictions to occur, prove them as right and work with them.   We can make a paraconsistent or trivialist system and work with it. For example, with trivialism, in theory, we would be able to derive and state everything we would want (since literally everything, even including illogical/logically impossible inconsistencies and contradictions), but we as humans (or as brains), are limited and can't conceive everything we want (at least to what I know). Therefore, no matter how many trivialist models we create and how much time we would work with them, we would never find or conceive many illogical/logically things because they are just that: impossible. There are impossible things to describe and conceive. For example, Russel's set is the set of all sets that do not contain themselves. If Russel's set contains itself, then it cannot contain itself, since it only contains sets that don't contain themselves. But if Russel's set does not contain itself, then it must contain itself, since it contains all sets that don't contain themselves.  There are quite a few logic bombs like this. You cannot ever compute the contents of Russel's set, and there are more formal, mathematical ways to present it. All of them have in common that you can't actually compute what the set is, whether you do it by hand, in your head, or on a computer. It's just a statement that cannot be fully logically processed.  If you take every possible state the human brain can be in, none of them include the computation of Russel's Set's contents. That is, not only can the contents not be computed, they cannot even be represented. No stimulus can cause us to comprehend Russel's Set, since such comprehension is not possible to begin with. It does not have a solution. Even if we try to solve it using trivialism, we would just be able to write a solution that does not make sense and prove it has sense and it is the real solution, but we could not be able to have a solution that would make sense "outside" the realm of trivialism (for example in classical logic), even though, using trivialism, we could prove that such solution would have sense in whatever context and logic system.   But what about hypercomputational machines (for example oracle-like machines)? I've read about some models of hypercomputation which are compatible with paraconsistent or trivialism logics. I've also read there are some models of hypercomputation (particularly those oracle-like models which use a black box) where, essentially, the hypercomputer is an algorithm that cannot exist. It might be because such an algorithm is fundamentally forbidden by logic itself (which is hided in a black box). Would any of these be capable of computing/"conceiving" these impossible things I wrote before? Do you know of anything that would help?
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I need to prove that $A ⊂ B$ if and only if $A − B = ∅$. I have the following "proof":  $$ A \subset B \iff A - B = \emptyset$$ proof for $\implies:$  $$\forall x \in A, x \in B$$  Therefore,  $$A - B = \emptyset$$  proof for $\impliedby$:  If $$A - B = \emptyset$$ then $$\forall x \in B, x \in A$$ since $\forall x \in B, x \in A$,   $$ A \subset B $$  However the whole thing seems to be incredibly "fragile" and relies on circular logic (see how I just switched the sets in the for all statements)  Is this a valid proof? Is there a better way to write it?
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My understanding of logic is really basic, and I ask this question out of curiosity.  Is there an explicit example of a statement whose proof uses the continuum hypothesis and is unprovable in $ZFC + \lnot CH$.   ADDED: Preferably I'm looking for an example outside set theory.
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Is axiom of choice required to show the existence of non-measurable sets? Is there a Lebesgue non-measurable set that can be constructed without axiom of choice?  Related question on MO says it is consistent:
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I'm reading through How to Prove It by Velleman and I'm having trouble with this exercise in the section about Existence and Uniqueness proofs. Here is the exercise:     Suppose $A$ is a set and for every family of sets $\mathcal{F}$, if $\cup \mathcal{F}=A$ then $A \in \mathcal{F}$. Prove that $A$ has exactly one element.    He hints that for both the existence and uniqueness parts of the proof it would be a good idea to use contradiction. I've been playing around with this proof for a while but I can't seem to make any substantial progress.   I current idea is considering some cases  where for some family of sets $\mathcal{G}$, $A \in \mathcal{G}$ and $A \notin \mathcal{G}$. I thought if I could show that if $A= \varnothing$ lead to contradictions, I could at least say that there is something in $A$, and try to prove that it is unique from there. I haven't been able to make any progress with this though. Any help with this problem would be greatly appreciated!
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When I study Topology, I met with a problem. On my book, it says 'we cannot admit that there exists a set whose members are all the topological spaces. That will lead to a logical contradiction, that there will be a set who is a member of itself.' But why we cannot have a set who is a member of itself?
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This question is quite subjective, but here goes:  The axiom of choice notoriously leads to many extremely counterintuitive results, like the Hausdorff, Von Neumann, and (most famously) Banach-Tarski paradoxes. Unfortunately, the proof of these paradoxes is also notoriously complicated, and it's not at all obvious why an axiom as seemingly innocuous (and, in fact, obvious) as the axiom of choice has such strange consequences.  Do there exist any examples of a simple chain of reasoning that leads from the axiom of choice to a counterintuitive result? (Though not necessarily a result as spectacularly counterintuitive as the Banach-Tarski paradox.) One which is simple enough that someone without training in advanced mathematics could understand the general train of logic (with perhaps a few steps fudged over)? I've never see an example that lets me intuitively understand why naive intuition "goes wrong" exactly at the step where we use the axiom of choice.  (To be clear: I'm not looking for a simple counterintuitive result that follows from the axiom of choice - Banach-Tarksi would qualify for that. I'm instead looking for a counterintuitive result that follows from the axiom of choice with a (relatively) simple proof.)
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In the proof of 'Every open set in $\mathbb{R}^1$ is a countable union of disjoint open intervals', we need to pick one rational representative from each of the intervals hence establish the countability.  This seems to depend on the axiom of choice.  Thus I wonder does this property of $\mathbb{R}$ is equivalent to the axiom of choice, that is, is there any construction such that this property is not true once we abandon the axiom of choice?  Thanks!
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I would like to know if someone can explain in a somehow down to earth (almost logic free) way why is it true that in an axiom system where there is some statement $P$ such that $P$ and its negation $\lnot P$ are true, then every statement in the system is true?  I'm not sure if this can be done, but basically since I don't know any formal logic at all, I'm interested in seeing if at least the argument can be conveyed in an intuitive way, or if the idea can be explained without talking about first or second order logic and using symbols like $\top$, $\bot$, and $\vdash$.  This  is like the formal version (which I don't understand) so maybe my question can be thought of as a version for dummies of that question.   Thanks a lot in advance.
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Set theory (Jech) $\text{p.}\;27:$     It is an open problem whether one can prove without the axiom of choice that there exists a regular uncountable $\aleph_{\alpha}\;($the informed guess is that one cannot$)$.   My question is on the remark in brackets, in particular the use of the word informed: could someone briefly expand on the reasons for believing that this is likely to be impossible?
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I came to point where I suppose for case 1 that $A\subseteq B$ and conclusion is trivial. For case 2 I suppose that $A\not\subseteq B$ and try to prove $B\subseteq A$, but that gets me nowhere. Any pointers here are most welcome.
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This was  on Quora. I thought about it a little bit but didn't make much progress beyond some obvious upper and lower bounds. The answer probably depends on AC and perhaps also GCH or other axioms. A quick search also failed to provide answers.
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A partially ordered set $(X, \leq)$ is called a lattice if for every pair of elements $x,y \in X$ both the infimum and suprememum of the set $\{x,y\}$ exists. I'm trying to get an intuition for how a partially ordered set can fail to be a lattice. In $\mathbb{R}$, for example, once two elements are selected the completeness of the real numbers guarantees the existence of both the infimum and supremum. Now, if we restrict our attention to a nondegenerate interval $(a,b)$ it is clear that no two points in $(a,b)$ have either a suprememum or infimum in $(a,b)$.   Is this the right way to think of a poset that is not a lattice? Is there perhaps a more fundamental example that would yield further clarity?