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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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  • $\begingroup$ You can't have an ordered pair whose first component is an element of $S$ if $S = \varnothing$. Since $S\times T$ is the set of ordered pairs whose first component is an element of $S$ and whose second component is an element of $T$, you have $S = \varnothing \Rightarrow S\times T = \varnothing$. $\endgroup$
    – Daniel Fischer
    Dec 4, 2013 at 20:50

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Yes, for any sets $A,B$ we have $|A\times B|=|A|\cdot|B|$ where $|X|$ means the cardinality of set $X$.

So, Cartesian product of sets correspond to multiplication of (natural) numbers.

Also, for the specific example $\emptyset\times\emptyset$, its elements would be exactly the ordered pairs $\langle x,y\rangle$ such that $x\in\emptyset,\ y\in\emptyset$. As there are no such pairs, we have $\emptyset\times\emptyset$ is empty.

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  • $\begingroup$ Wasn't there some crazy bijection from $\mathbb R \to \mathbb R^2$?... For the finite case this is obvious ofc.. $\endgroup$
    – AlexR
    Dec 4, 2013 at 20:52
  • $\begingroup$ Yes, there is, and $|\Bbb R^2|$ is still continuum, even $|\Bbb R^{\Bbb N}|$ is so. $\endgroup$
    – Berci
    Dec 4, 2013 at 20:53
  • $\begingroup$ Ah yeah, I remember $|\mathbb R|^2 = |\mathbb R| = |\mathbb R^2|$^^ $\endgroup$
    – AlexR
    Dec 4, 2013 at 20:54
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$S\times T$ is the set of ordered pairs where the first element is in $S$ and the second one in $T$. As $S$ has no elements, there is nothing to be put in the first coordinate, so no order pairs exists, i.e. $S\times T=\emptyset$.

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Here's one way to think about it:

$S \times T$ is the set of all pairs taken from $T$ and $S$ - i.e. take one thing from $S$ and one thing from $T$ to get an element of $S \times T$. However, if there are no elements in $T$, then there are no such pairs!

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