I recently studied about Cartesian products and I thought that I understood its concept, Until I ran into this expression:   If  $S=\emptyset$, $\ne\emptyset$,  then $S\times T = \emptyset$ .   Is an empty set the same as zero? in the sense that in nullifies a non empty set?  Thanks!
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If we have a function $f:A \rightarrow B$, then one way to give meaning, I think, to this function, in terms of set theory, is to say, that $f$ is actually a binary relation $f=(A,B,G_f)$, where $G_f \subseteq A \times B$ is the graph of the function. Now my question is: what is $f$ if   $\bullet \ A=\emptyset, \ B\neq\emptyset$,?  $ \bullet \ B=\emptyset, \ A\neq\emptyset$ ?  $ \bullet \ B=\emptyset, \ A=\emptyset$ ?  (Another way to formulate this, I think, would be: How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset $ look like? Are they all $\emptyset$ ?)
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Are the following two assertions always true, always false or meaningless?  $\exists i \in \emptyset$  $\forall i \in \emptyset$  Because it seems that one encounters expressions of this kind fairly similar in mathematics, especially if we are dealing with degenerate cases of definitions. Let me give an example (in graph theory) of such a case. There, one can formalize the idea the  two vertices are connected in the following way: Let $G=(V,E)$ be a graph and let $v,w\in V $. We define $v$ and $w$ to be "connected" if: $\exists n \in \mathbb{N}, \ \exists \alpha: \left\{ 1,\ldots,n \right\} \rightarrow V, \ \alpha_0=v \ \&amp; \ \alpha_n=w \ \ \forall i\in \mathbb{N}, 0\leqslant i \ \&amp; \ i&lt;n:\  \left\{ \alpha_i, \alpha_{i+1} \right\} \in E$  Now, if we would ask, if every node is connected to itself (a fact which intuitively we would want to be true), we would have to exhibit an $n$ and a sequence $\alpha$, such that [bla bla...]. The obvious choice for $n$ is 0. But for this choice of $n$, no mattter which sequence $\alpha$ we would consider, the set of the $i$'s would be empty, because the set of the $i$'s is actually $\left\{ i\in \mathbb{N} |  0\leqslant j \ \&amp; \ j&lt;n \right\}$. But for $n=0$ this set is the empty set. Thus, because we quantify $\forall i \in \left\{ i\in \mathbb{N} |  0\leqslant j \ \&amp; \ j&lt;0 \right\}=\emptyset$, should the statement be true/false by definition?  EDIT: Sorry, for my unclear formulation and to everybody who on my fault interpreted it wrongly. The way Carl Mummert or Listing interpreted it, was what I meant.
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Give:  $fn(S)=\prod_{x\in S}x$  what is:  $fn(\emptyset)$  I can see reason that it would be defined as 1 or 0.    Note: I thought about restricting the domain of $S$ but that would make the problem less general. If there is no general answer then answers for restricted domains would be valid.
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A quick preface to the below question: this is my first post on math.se! I am excited to begin participating in such a wonderful community. Any feedback as to how I can improve this or subsequent posts is certainly feedback from which I could benefit.      (Exercise 3.2.2 of Tao's Analysis I) Use the  (and the singleton set axiom, which guarantees the existence of the singleton set) to show that if $A$ is a set, then $A\notin A{. \kern 0.05em}^\text{1}$ Furthermore, show that if $A$ and $B$ are sets, then $A\notin B$ or $B\notin A$.    Thoughts: Consider some set $A$ and the singleton $\{A\}$. By the axiom of regularity, the only element of $\{A\}$, namely $A$, is not a set or disjoint from $\{A\}$. Clearly, $\{A\}\cap A=\emptyset$ and thus, $A\notin A$.   Moreover, suppose $B$ is some other set such that $\neg(A\notin B \lor B\notin A)$, i.e. $A\in B\land B\in A$. This implies, in very vauge notation, that $A=\{\{A,\dots\},\dots\}$ and $B=\{\{B,\dots\},\dots\}$. Can I get a contradiction out of this? In the case of $A$, we get that $x\in\{A,\dots\}\lor x\in \{a\}$ for some $x\in A$ and all other objects $a\in A$. How can I prove that there exists no element of $A$ (and $B$) that is not a set unless it is disjoint from $A$. I would greatly appreciate any hints as to how I can complete my argument.     1 This first problem has been discussed on this site times before (see  and ). I include a solution only because the context demands it, but to avoid a duplicate question, I ask that it not be discussed here.
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Is "empty set" an element of a set? For example:  is this FALSE?:  0 ∈ {1,2,3}
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I Have a question. Given only the definition of equality of two sets, how can we prove that there is one and only one empty set. I mean by equality of two sets the following: $$A=B \iff  \forall x (x \in A \iff x \in B)$$
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In $Sets$, the initial object is $0 = \emptyset$. We have $C \times \emptyset = \emptyset$ for any set $C$. As there are no maps to the empty set, we don't get much of the universal property of the product.  I am now wondering what is special about arbitrary categories $\mathscr{C}$ with $C \times 0 \cong 0$ for all objects $C$ of $\mathscr{C}$, where $0$ is the initial object of $\mathscr{C}$.   From Sheaves in Geometry and Logic by MacLane (p.194), in any topos $\mathscr{C}$, any arrow $k:C \to 0$ is an isomorphism. The universal property of the product would then also not yield very much.   Are there any categories of higher interest?
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Can anyone explain why the predicate all is true for an empty set? If the set is empty, there are no elements in it, so there is not really any elements to apply the predicate on? So it feels to me it should be false rather than true.
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I was trying to find a proof of $a\times0 = 0$ by myself (assuming commutativity, associativity, distributivity, etc) and I came up with $$ a+0=a(1) \implies 1 = \frac{a+0}{a} = \frac aa + \frac 0a = 1 + \frac 0a \implies \frac 0a = 0 \implies a(0)=0$$ Is there anything wrong with this proof, or does it assume anything that shouldn't be assumed?
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I dont know the concept or how to answer the question.     Suppose there are two sets, $X=\{1,2\}$ and $Y=\{a,b\}$.         What are all the ordered pairs in the form of (element of $X$, element of $Y$)?   What are all the ordered pairs in the form of (element of $Y$, element of $X$)?
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I am trying to make sense of these. To me a is false because the set isn't empty. Is that correct? b is true because the empty set is an element of that set. c is false because the set the empty set isn't an element, it is a subset or proper subset. d is true because the set the empty set is a member of that set. e the set the empty set is false because it is not a proper subset I think f false for the same reason g false for the same  a) ∅ ∈ {∅} b) ∅ ∈ {∅, {∅}} c) {∅} ∈ {∅} d) {∅} ∈ {{∅}} e) {∅} ⊂ {∅, {∅}} f ) {{∅}} ⊂ {∅, {∅}} g) {{∅}} ⊂ {{∅}, {∅}}  I know not all these are correct, why is my logic wrong?
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Let $A$ to be a nonempty set and $B= \emptyset$; then $ A \times B$ is a set. And let $F$ be a function $A \to B$. Then $F \subseteq A \times B$. By the axiom of specification, $F$ must exists (if I didn't mess up something).   But the book I'm reading, Elements of Set Theory by Enderton, says that no function could have a nonempty domain and an empty range, and no more detail is given.  So my confusion arises. His statement against my proof. The only axiom as far as I know to prove such a set does not exist is the axiom of regularity. But I can't give such a proof. So I need help. I hope someone could clearly explain why such a function doesn't exist, and why this doesn't contradict the axiom of specification.    Let me explain my confusion in more detail:  First of all, I know that $A \times \emptyset$ is empty. But $ \emptyset \times A$ is also an empty set, yet there is no problem with functions with an empty domain.  The book I'm reading defines a function as:      "A function is a relation such that for each $x$ in $\operatorname{dom} F$ there is only one $y$ such that $x \mathop F y$."   and a relation as:     "A relation is a set of ordered pairs."   Now, let me define the function $F$ as: $$F =  \{\langle x,y \rangle \mid x \in A \text{ and } y \in B \text{ and ... other conditions} \}$$  so it's equal to: $$F = \{ \langle x,y \rangle \in A \times B \mid \text{... other conditions} \}.$$  Doesn't that mean that $F$ meets the conditions of the axiom of specification, and therefore exists?    What I would specifically like to ask is:   How do you define a function $F$ (as precisely as possible)? Why does the argument above not imply that a function $F: A \to B$ always exists? Does $F \subseteq A \times B$ still hold when $B = \emptyset$?   (The answer to #2 cannot simply be that $A \times B = \emptyset$, because $ B \times A = \emptyset$ as well, but a function $B \to A$ does exist.)
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Given set $ A = \{  \emptyset   \}$ what is $|A|$? I just learned about sets today in class and I'm unsure how to answer this.
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The question goes: Let $A,B,C,D$ be sets. Prove that $\big(A\times B\big)\bigcup \big(C\times D\big)=\big(A\bigcap C\big)\times \big(B\bigcap D\big)$ I started with the definition of the cartesian product $A \times B=\left\{\big(a,b\big): a\in A \wedge b \in B\right\}$ The union of $A\times B$ and $C \times D$ would be a set of coordinates where $\big(x,y\big): x\in A\bigcup C \wedge y\in B\bigcup D$ I am stuck here because the union of the two cartesian products on the left doesn't seem to imply that $x\in A\bigcap C$ and $y \in B\bigcap D$. Could anyone give me a hint or a possible step I might not be seeing? Please and thank you.
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I have some trouble trying to understand Halmos' explanations about generalized Cartesian products. I've read these entries already:    This also has the page I'm stuck on.      Is his generalization the same as a Cartesian product of several sets? As here:   What puzzles me is $\{a,b\}$. Where does it come from, what does it do? I know that it is supposed to be arbitrary. How does it identify distinct elements of $X \times Y$? Does some function map each $a \in A$ to an element in $X$? Or is it always the same $a$?  What does he mean by      the set $Z$ of all families $z$, indexed by $\{a,b\}$, such that $z_a \in X$  and $z_b \in Y$   ? What does does an element of $Z$, namely one family $z$, consist of? By family, does he mean an indexing function or the indexed elements? This comes to a very important question for me: Does family mean a function or the set of all indexed elements?   I will try to explain how I understand it thus far: $Z$ contains elements of the form $\{(a,x),(b,y)\}$. The choice of $x$ depends on $a$, so an $x$ in a $z \in Z$ equals the $z_a$ of $X$. Now, we want to map these pairs to $X \times Y$, so we have a function $f$ that takes $z \in Z$ and connects it to the corresponding ordered pair $(z_a, z_b)$, where $z_a \in X$. $X \times Y$ is not the same as $Z$; they only, effectively, contain equal elements.  He goes on:     If $\{X_i\}$ is a family of sets ($i \in I$), the Cartesian product of the family is, by definition, the set of all families $\{x_i\}$ with $x_i \in X_i$ for each $i$ in $I$.   How I understand it: $\{X_i\}$ contains, as elements, several sets, each single one a set $X$ identified by an index/element of $I$. Then, we go over all indeces $i$ in $I$, take from each set $X$, namely $X_i$, an element $x$ and add it to current set of the newly created sets/families. So, the lower index $i$ does not refer to an element of $X_i$ indexed by $i$, but rather calls all elements of $X_i$ $x_i$. We end up with a set, defined as Cartesian product, of sets, in which each element (set) contains ordered elements, each from a different $X_i$. How does the ordering occur?  $\prod_i X_i$, the Cartesian product, becomes equal to $X^I$, if all $X_i$ are the same set $X$. $X^I$ means all possible combinations of mapping from $I$ to $X$. We randomly take an elements $x$ from some $X_i$, put it in a set/family with other randomly obtained $x$, put the entire set into a set denoted $X^I$, and continue doing so, till we have no more options. Then, each set in the new $X^I$ contains a mapping of some $i$ to some $x$. Since we exhausted the options, they are indeed all combinations.  Where did I go wrong? Someone needs to spoonfeed my the answer; I am completely confused by now.
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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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I believe that this is a pretty simple question in terms of combinatorics.  Assuming we have the following set $\{ {1,2,3...n} \}$  In how many ways can we choose 3 subsets (A,B,C) such that the intersection between all them is empty?   $ A \cap B \cap C = \emptyset $  Well, what I have so far is the following:  For A we have $ 2^n $ options (where n is the size of the set) Then, for B, the remaining options are exactly $ 2^{n-|A|} $ since we can choose only from subsets who does not have elements in common with A. And finaly, for C, since we already have an empty intersection, we can choose whatever subset we want and get the expected result $ \emptyset $.  But for some reason it feels like the numbers are too big. If we take for example the set {1,2}, then, according to my "formula", the number of options should be:  $ \binom{4}{1} * \binom{2}{1} * \binom{2}{1} = \frac{4!}{3!} * \frac{2!}{1!} *  \frac{2!}{1!} = 16 $  Does it make sense?