Exercise 3.2.2 Use the axiom of regularity and the singleton set axiom to show that if $A$ is a set then $A \notin A$.  Axiom of Regularity: 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.  Singleton Set Axiom: If $a$ is an object, then there exists a set $\{a\}$ whose only element is $a$, we refer to $\{a\}$ as the singleton set whose element is $a$ Furthermore, if $a$ and $b$ are objects, then there exists a set $\{a,b\}$ whose only elements are $a$ and $b$    I think I can give the proof when A is the non-empty set which contains finite elements. However, can anyone give any hint for the general case?  Suppose $A \in A$   Then eixsts $x$ such that either $x$ is not a set or $x \cap A$=$\emptyset$  Then it is clear thar $x \ne A$ (if $x = A$, then $x \cap A \ne \emptyset$)  We observe that as an element, $A$ $\ne$ $x$, $A$ $\in$ $A$  Then $A$ $\in$ $A-x$   Define the new set $A-x$ as $A_1$  then exists $x_1$ such that $A \in A_1-x_1$  After finite steps  we find that $A \in \emptyset $  which leads to contradiction.
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I am trying to understand   How can a set contain itself? Can you show example of set which is not a set of all sets and it contains itself.
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We define the term "Set" as,     A set is a collection of objects.   And an "Empty set" as,     An empty set is a set which contains nothing.    First problem I encountered:  How the definition of "Empty set" is consistent with the definition of "sets" if "Empty set" contains nothing and a "set" is a collection of objects.  Further we discovered in set theory that every set has a subset that is the Null set.  Such as,  If $A=\emptyset$ and $B=\{1,2,3\}$ then,  $$A \subset B$$  Second problem I encountered:   How and why "No element" is referred and considered as an element as we do in case of null set that is, when we say that every set has a subset that is the Null set?  Third and Last one:  How can a set possess "some thing" and "nothing" simultaneously that is when we say that every set (containing objects) has a subset which is Null set (contains nothing)?
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Where is axiom of regularity actually used? Why is it important? Are there some proofs, which are substantially simpler thanks to this axiom?   This question was to some extent provoked by 's : Would regularity ever be used in a formal development of, say, number theory or real analysis? I can't imagine it.  I have to admit that I do not know other use of this axiom than the proof that every set has  in cumulative hierarchy, and a few easy consequences of this axiom, which are mentioned in .  I remember seeing an introductory book in axiomatic set theory, which did not even mention this axiom. (And that book went through plenty of stuff, such as introducing ordinals, transfinite induction, construction of natural numbers.)  Wikipedia article on  links to Metamath page for  and says: Scroll to the bottom to see how few Metamath theorems invoke this axiom.  Based on the above, it seems that quite a lot of stuff can be done without this axiom.  Of course, it's quite possible that this axiom becomes important in some areas of set theory which are not familiar to me, such as forcing or working without Axiom of Choice. (It might be difficult to define cardinality without AC and regularity, as mentioned .) But even if the this axiom is important only for some advanced stuff - which I will probably never get to - I'd be glad to know that.
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I am having difficulty understanding  :     Every non-empty set $\rm A$ contains an element $\rm B$ which is disjoint from $\rm A.$   So from my understanding if I have a set like:   $$\{1, 2, 3, 4, 5\}$$  Then one of $[1,2,3,4,5]$ is not an element in $\rm A?$ huh?
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I'd like to know if my proof is correct or incorrect in the following.  Given definition of Axiom of Choice (AoC): The direct product of a family of non-empty sets indexed by a non-empty set is non-empty.  Show that the Axiom of Choice is equivalent to the statement that every set has a choice function.  (Choice Function => AoC):   Suppose every set has a choice function.  Let S be a set, and fix a natural number n.   Let $$ I = \{1,2,....,n\}  $$ and let $\mathbf A$ be a family of non-empty subsets of S sets indexed over I.  $$ \mathbf A = \{A_1, A_2, ... , A_n\} $$  The task is to show that the direct product over sets in $\mathbf{A}$ is non-empty. This will be shown if an element of this set can be produced.   Let $$ D = A_1 \times A_2 \times ... \times A_n $$  By supposition, this set has a choice function $f$ defined over non-empty subsets $X$ of $D$ with the following:  $f: X \to D$ such that $f(X)∈X$ for all subsets $X$.  So, $f(X) ∈ X$ $\Rightarrow$ $f(X) ∈ D$.   So, $D$ is non-empty.  (Show AoC => Choice Function):  Suppose the Axiom of Choice, let S be a set, and fix a natural number n.  Let  $$ I = \{1,2,....,n\}  $$ and let   $$ \mathbf{A} = \{A_1,A_2,....,A_n \}  $$  be a family of non-empty subsets of S indexed over I.  Let $$ D = A_1 \times A_2 \times ... \times A_n $$  By the Axiom of choice, $D$ is non-empty.  Let $y ∈ D$ and define a function $f: D \to D$ by $f(x) = y$ for all elements $x ∈ D$.  (So, every element of D is mapped to this fixed element $y$.)  So, $f(x)∈D$ for all $x∈ D$.  In other words, $f$ defines a choice function on $D$.   So, assuming the Axiom of Choice implies that $S$ has a choice function defined on its non-empty subsets.
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This question was inspired by the question on .  From the discussion in the comments there, it seems there does not exist a set of all sets of a given cardinality.  To me, it seems easy to see that this is true for any nonzero finite cardinality $n$. Given any set $x$, you can always make a set of size $n$ by adding $n-1$ other sets to $x$ to make a set of size $n$, which exists by repeated use of the pairing axiom.   But how would you do this for infinite cardinalities? For suppose $\kappa$ is some given infinite cardinality. To show that the set of all sets of cardinality $\kappa$ does not exist, it seems you would have to show that for any set $x$, there exists a set of that size with $x$ as an element, and then you could take the union of that set to find the set of all sets, and thus a contradiction. But it doesn't seem reasonable to simply say, for a given set of size $\kappa$ if $x$ is in the set, we have no problem. If not, just take an element out of the set and put $x$ in. Something about that seems like it would not be allowed.  So how would you do this for infinite cardinalities?
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I am trying to solve the following problem.     Let $X$ be a compact Hausdorff space and let $f:X\to X$ be continuous.   Show that there exists a non-empty set $A\subset X$ such that   $f(A)=A$.   There is a hint to define $A=\cap_{n\ge 0} A_n$ with $A_{n+1}=f(A_n)$ and $A_0=X$.  Then $x\in A_{n+1}$ implies that $f^{-1}(x)\cap A_n\ne\emptyset$. Right? So if $f^{-1}(x)\cap A_n=\emptyset$ then $x\notin A_{n+1}$. Is this correct?  First I need to show that for $A$ defined in the hint it holds $f(A)=A$. For this to happen it is enough to have $f(A)=\cap_{n\ge 0} f(A_n)$ Because then $f(A)=\cap_{n\ge 1} A_n=\cap_{n\ge 1} A_n\cap X=A$. But got stuck proving the equality.  What I did for that is to assume that there exists $x\in A$. This means that $\exists y_n\in A_n$ such that $f(y_n)=x$ for all $n$. Is this enough to deduct that $f(x)\in A$?  Then I need to show that $A$ is not empty. For this I think it is enough to show that $A_{n+1}\subset A_n$, because $A_n$ are compact and then their intersection is not empty. However I am stuck there too. I tried by contradiction assuming $\exists x\notin A_n$ with $f(x)\in A_{n}$, but I cannot work from there.  Initially I thought of something else. Choose $x_n\in A_n$ and create the sequence $\{x_n\}$. This has a convergent subsequence $\{x_{n_k}\}$. Then I create a sequence $\{x'_m\}$ by defining $x'_m=x_{n_k}$ with $n_k=\max\{j\in\{n_k\}|j\le m\}$. So basically i repeat the terms of the subsequence to get a convergent sequence. However this again requires that $A_{n+1}\subset A_n$ so I have $x'_m\in A_m$. Then I think that the limit $x'_m\to x$ is in $A$ but I was not able to show that also.  Thanks in advance.
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Claim. Let $a$ and $b$ be real numbers. If $a\geq b$  then $(a,b), [a,b)$ and $(a,b]$ are empty set.   My is question is that why?
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Ever since day one of of my Mathematical Logic course, this fact has really bothered me. I cannot wrap my head around how an empty set is a subset of every possible set. Could someone kindly explain how this is true? Any help is appreciated!
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From axiom of regularity, how can I prove following statement? $$x \notin y \lor y \notin x$$  I’ve been reading about ZF theory.  So far, I understood the axiom of regularity as follows:  $$\forall x (x \neq \phi \Rightarrow  \exists z. z \in x \land z \cap x = \phi)$$  From this, (and with other axioms in ZF), I should be able to derive the first statement, but have not managed to do so.  Question:   How can I derive the statement within ZF?
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I'm reading Guerino Mazzolla's Comprehensive Mathematics for Computer Scientists 1:     Axiom 3 (Axiom of Union) If $a$ is a set, then there is a set      $$\{x| \text{ there exists an element } b \in a \text{ such that } x\in b \}$$      This set is denoted by $\cup a$ and is called the union of $a$.    This axiom seems to take elements from inside of a set of sets, for example in the set: $a=\{\{u_1,u_2 \},\{u_3,u_4 \} \}$, there are two elements $b\in a$, then the mentioned set created by the axiom is $\{u_1,u_2,u_3,u_4 \}$.           The axiom seems to imply that the set $a$ must have a set inside of it for it to work, what if it  have an element that is not a set? (I'm not sure if this is possible, but I remember of having read something about the difference of sets and elements).         There's also an extension of my proposal, In the same book (and also in others) I've seen that the natural numbers are defined as:   $$\begin{eqnarray*}   {0}&amp;=&amp;{\emptyset} \\    {1}&amp;=&amp;{\{\emptyset\}} \\    {2}&amp;=&amp;{\{ \{\emptyset\}\}} \\    {3}&amp;=&amp;{\{ \{ \{\emptyset\}\}\}} \\   {\vdots}&amp;=&amp;{\vdots} \end{eqnarray*}$$  Then the singleton $\{ 3\}$ could be seen as $\{ \{ \{ \{\emptyset\}\}\}\}$, which could mean that   $$\cup \{ 3 \}=\cup\{ \{ \{ \{\emptyset\}\}\}\}= \{ \{ \{\emptyset\}\}\}=\{2 \}$$          Then the union could also be seen as the antecessor of a number?            Using the framework I provided, I see some ambiguity, look:   $$\{ \{ \{\emptyset\}\}\}=\{2 \}$$          I decided to interpretate it this way, but from the notation it's not clear in which pair of brackets I should stop, I guess that $\{ \{ \{\emptyset\}\}\}=\{2 \}$ but it could also be the case that $\{ \{ \{\emptyset\}\}\}=3$. I could want a singleton with a $2$ inside of it and a $3$ outside a set, no?
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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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Proof:- From (c) $X\in \mathscr T$. $X\subset cl(X)[\because by (a)]$ and also we have $cl(X)\subset X$. Hence $cl(X)=X$ and $\emptyset \in \mathscr T$.   Let $\{U_i\}_{i=1}^n$ be a collection of open sets in $\mathscr T$. For each $i$ there exists $C_i$:$cl(C_i)=C_i$ and $U_i=X\setminus C_i$ then $\bigcap_{i=1}^{n}U_i=X\setminus \bigcup_{i=1}^n C_i$. We have $cl(C_i)=C_i,i=1,2,3,..,n$. By (d), we have $\bigcap_{i=1}^{n}U_i \in \mathscr T$. How do I prove arbitrary union axiom? If there is an axiom $\bigcap_{\alpha \in \Lambda}cl(A_\alpha)=cl(\bigcap_{\alpha \in \Lambda}A_\alpha)$. I could prove. Can you please help me?
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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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$\{\emptyset\}$ is a set containing the empty set. Is $\{\emptyset\}$ a subset of $\{\{\emptyset\}\}$?  My hypothesis is yes by looking at the form of "the superset $\{\{\emptyset\}\}$" which contains "the subset $\{\emptyset\}$".
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This is the first part of the first half of the a proof in the Axiom of Choice. The proof that the Axiom of Choice is equivalent to the choice function. REsources: Keith Devlin's Joy of Sets: pg. 58. F.R. Drake and Singh Intermediate Set Theory: pg. 110).   I am confused on proving disjointness and will ask further questions of proof once I get this first part v. clear for myself.   AC1: Let $\mathcal F$ be a family of pairwise disjoint, non-empty sets. Then there is a set $\mathcal M$ that consists of precised one element from each member of $\mathcal F$  AC2: For every family $\mathcal F$ of non-empty sets, there is a choice function $\mathcal f$. (I.e. $\mathcal f(X)$ is a member of X, for each X in $\mathcal F$.   Disjointness: Two sets are disjoin if Q $\cap$ R = $\emptyset$. Meaning that they have zero elements in common. These sets do not overlap.   Show: AC1 $\leftrightarrow$ AC2   (1) ($\rightarrow$) Let $\mathcal F$ be a set of nonempty sets.   (2)For each X $\in$ $\mathcal F$, let $X^*$ = (X $\times$ {X}) = {(x, X) | x $\in$ X})  (3) Show Disjointness.  An outline of how book proves  disjointness:  (a) if u = {(X $\times$ {X}) | x $\in$ X} then u is of the form $\lt  z, t \gt$ where z = X and t=x, i.e. x $\in$ X.   (b) then if u $\in$ ({X} $\times$ x) $\cap$ ({Y} $\times$ Y) $\rightarrow$ u = $\lt z, t \gt$ with z = X and z=Y, i.e. X=Y. Which means ({X} $\times$ x) = ({Y} $\times$ y).  Questions: (1) Apparently we cannot prove disjointness in $\mathcal F$ but must prove in a different set. If so, why is that?  (2) why is the case that if u $\in$ ({X} $\times$ X) $\cap$ ({Y} $\times$ Y) then z =X and z=Y? I am not following that step at all.   (3) why does the axiom of choice have to do with ordered pairs? Why are we setting X $\in$ $\mathcal F$, let $X^*$ = (X $\times$ {X})?    Thank you!
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Prove or disprove:  The set: $\{ \emptyset\}^{\Bbb N}$ has only one element.  The set $\emptyset^{\Bbb N}$ is empty.    I'm pretty sure both are true, for 1. all the natural numbers go to the empty set and that is the only element. Where as 2. all the natural numbers go to nothing so that's why it's empty. (Same as $ {\Bbb N}^{\emptyset} $ nothing goes to something is nothing (I think)). The problem is I have no idea how to show this so advice would be appreciated.  Edit: My defintion of a function is: $f:B\to A$ if  $(a_1,b),(a_2,b)\in f \ \Rightarrow \ a_1=a_2$ So for 2. I can simply say that: $(1,\emptyset),(2,\emptyset)\in f \Rightarrow a_1\neq a_2$  ?