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I was attempting this question:

Consider $A \ne \varnothing$, where the number of subsets of $A$ is the same as the number of elements of $A\times A$. Determine the number of elements in $A$.

Because $\varnothing$ is naturally a subset of $A$ and in this question, it is clearly specified $A \ne \varnothing$, therefore can it be concluded that the number of elements is $A^n$?

Thank you in advance!

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Hm ... if $|A| = n$, then $A$ has $2^n$ subsets and $|A \times A| = n^2$ ... so $n=?$. – martini Apr 14 '12 at 9:39
The number of elements in A is 2 – Prasad G Apr 14 '12 at 10:12
up vote 1 down vote accepted

$A$ is a set, not a number, so $A^n$ doesn’t really make sense. Moreover, you’ve not said what $n$ is. Let’s back up.

Let $n=|A|$. Then $|A\times A|=n^2$, and $A$ has $2^n$ subsets. (Why?) Thus, $2^n=n^2$. For what value of $n$ is that true?

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A has 2^n subsets? as in the ∅ and it self? I can see why it is n^2 as A= AxA so A can = n^2. Thanks – Xabi Apr 14 '12 at 9:49
@Fatz: Yes, if a set $A$ has $n$ elements, it has $2^n$ subsets. This is true even if $A=\varnothing$ and has no elements, so that $n=0$: $2^0=1$, and $\varnothing$ has one subset, namely, itself, $\varnothing$. – Brian M. Scott Apr 14 '12 at 9:52
Thank you sooo much! :D – Xabi Apr 14 '12 at 9:54

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