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Is it true to say that $\forall A,B$ (set)
$\emptyset \subseteq A\times B $

Or: The empty set is contained in every Cartesian Product? (As it is contained in every set).

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The product is still a set, so this is true. – awllower Feb 12 '13 at 10:52
Are you able to find an element of the empty set which is not an element of the product? – Sigur Feb 12 '13 at 10:54
@Sigur: Why didn't you post your comment as an formal answer. I think, if you did, then the OP would remind something important from logic. +1 for you. – Babak S. Feb 12 '13 at 13:09
you asked "contained" in the title but symbolically wrote "is a subset f" in your question body – user58512 Feb 12 '13 at 13:24
up vote 2 down vote accepted

If the product of two sets is a set then the answer is yes.

Now it depends on how you were taught:

  • It could be that you were taught naively, and then you were just given the axiom that the product of two sets is a set. Maybe it wasn't even called an axiom, just a definition which said that $A\times B$ is a set.

  • It could be that you were taught axiomatically and were given the axiom of pairing, i.e. for every two elements $x,y$ there is an ordered pair $\langle x,y\rangle$ and then you construct $A\times B$ using the replacement axiom, $A\times B=\{\langle a,b\rangle\mid a\in A\land b\in B\}$.

  • It could be that you were taught that $\langle a,b\rangle=\{\{a\},\{a,b\}\}$ (or some other encoding of ordered pairs via sets), and then using subset axiom (or replacement, again) you can deduce that $A\times B\subseteq\mathcal P(\mathcal P(A\cup B))$ (in the case given here, but you may have $\mathcal P(A\cup\mathcal P(B))$ instead if you encode ordered pairs differently).

In all these cases the Cartesian product of two sets is a set, and therefore $\varnothing$ is a subset of this product. Why is this point so important? Because we define $\subseteq$ as a relation between sets, and then we can extend it to larger collections by one trick or another.

Of course you can ask yourself generally whether or not the definition of $\subseteq$ holds, i.e. every element of the empty set is an element of $A\times B$. Of course this holds vacuously, there is no counterexample. But then there is a delicate issue about how to apply the definition of $\subseteq$ to non-sets. But once you know that $A\times B$ is a collection of objects from your universe it should suffice to conclude that $\varnothing\subseteq A\times B$ is a legitimate and meaningful sentence.

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For every set $S$ one has $\emptyset\subseteq S$.

Given sets $A$ and $B$, let $S=A\times B$.

$S$ is a set, so ...

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And why might $S$ be a set? – peoplepower Feb 12 '13 at 13:21
@peoplepower, just make use of the Axiom of pairing – Sigur Feb 12 '13 at 13:44
@Sigur and, of course, the axiom of power set as well as the axiom of replacement. – peoplepower Feb 12 '13 at 14:10

[Following the suggestion from @Babak Sorouh]

Are you able to find an element of the empty set which is not an element of the product?

This is the reason!

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In fact, $p\to q$ is always a true statement when $p$ is always a wrong term. – Babak S. Feb 12 '13 at 13:39
Are you able to find an element of the empty set which is an element of the product? – alancalvitti Feb 12 '13 at 15:01
@alancalvitti, your statement does not solve the problem since even if I'd be able to find one element it would not be sufficient to obtain the inclusion. – Sigur Feb 12 '13 at 15:06
Not meant to solve the original problem, just a comment on constructive versus nonconstructive math. – alancalvitti Feb 12 '13 at 15:26
@alancalvitti, since you just copied the phrase I thought that you were correcting. – Sigur Feb 12 '13 at 15:28

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