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Given $\mathbb{A}$ is a countable set. Cantor's theorem claim that, the set of all subsets $2^{\mathbb{A}}$ is not countable. Prove that the set of all subsets including (created from) $3$ elements is countable.

I did try to find a bijection map from $\mathbb{A}$ - a countable set, to $\mathbb{B}$ - set of all subsets from $3$ elements. But I was wrong, my way of doing it, because I choose a subset of A to be an element in A, did not make sense.

cqfd
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This is Able
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Since $A$ is countable, there is a bijection $f:A\to\mathbb N$.

Consider the map $g:A^3\to\mathbb N$ given by $g((a,b,c))=f(a)+f(b)+f(c)$.

I claim that for any $n\in\mathbb N$, $g^{-1}(n)$ is finite.

Note that for any $k>n$, and $b,c\in A$, $$g((f^{-1}(k),b,c))=g((b,f^{-1}(k),c))=g((b,c,f^{-1}(k)))=k+f(b)+f(c)>n$$which means that $g^{-1}(n)\subset f^{-1}(\{1,...,n\}^3)$, and the set on the right is finite.

And, $A^3=\bigcup\limits_{i=1}^\infty g^{-1}(n)$, and the countable union of finite sets is at most countable.

Rushabh Mehta
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  • It’s also possible to set up relatively simple injections in each direction, which proves the existence of a bijection without having to do all the work to exhibit one explicitly. – David K Feb 18 '20 at 23:10