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I need to prove that two empty sets have the same cardinality, I know that proving equivalence is done by defining an onto and one-to-one function from one group to the other, but how can I define a function from an empty set to an empty set?

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Do you mean define a function from an empty set to an empty set? –  Git Gud Feb 6 '13 at 22:36
    
yes of course, sorry. –  Georgey Feb 6 '13 at 22:37
    
@amWhy That is correct. I will will delete my comment. –  Git Gud Feb 6 '13 at 23:49
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4 Answers

up vote 7 down vote accepted

Let $A=B=\varnothing$. A function from $A$ to $B$ is a subset $f$ of $A\times B$ with the property that if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$, then $b_1=b_2$. The only subset of $A\times B=\varnothing\times\varnothing=\varnothing$ is $\varnothing$. Does it have the function property? Yes: it contains no ordered pairs at all, so the hypothesis

if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$

is never satisfied, and the property

if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$, then $b_1=b_2$

never actually requires anything to be true $-$ it holds vacuously.

Thus, $\varnothing$ is a function from $\varnothing$ to $\varnothing$. Is it a bijection? Is it true that for each $a\in A=\varnothing$ there is a unique $b\in B=\varnothing$ such that $\langle a,b\rangle\in f$? Again the answer is yes, vacuously: there’s no $a\in A=\varnothing$ at all, so the requirement is never actually invoked.

However, all of that work is completely unnecessary, because it’s always true that if $A=B$, then $|A|=|B|$. And if $A$ and $B$ are both empty, then they have exactly the same members, so they are the same set, a fact that I implicitly used above when I wrote $A=B=\varnothing$.

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Nice answer, Brian. +1 –  Rick Decker Feb 7 '13 at 2:00
    
Thanks, @Rick. ${}$ –  Brian M. Scott Feb 7 '13 at 2:03
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Hint: Recall that there are no elements in an empty set. Show that if $A$ and $B$ are empty sets then for every $x\in A$ we have $x\in B$, and vice versa. Therefore $A\subseteq B$ and $B\subseteq A$ and so $A=B$.

If two sets are equal then certainly they have the same cardinality.

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Hint:

Use the definition of the (an) empty set - a set with no elements - to show, it is vacuously true that for all $x$ in the domain, $\;x \in \emptyset_1 \implies x \in \emptyset_2,\;$ so $$\emptyset_1 \subseteq \emptyset_2$$ and do similarly to show that it is vacuously true that $$\emptyset_2 \subseteq \emptyset_1$$

Hence, $\emptyset_1 = \emptyset_2$.

Hence, $|\emptyset_1| = |\emptyset_2|$

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The extension and specification axioms implies that there is only one set with no elements.

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