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I'm not sure how to approach this. I've seen a proof how to prove that $[0,1]$ is uncountable. I thought of doing this by contradiction, and assuming that $X$ was countable, but I can't really go anywhere with it.

How would I approach this?

John
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  • Hint: given a real number $x = 0.x_1x_2x_3\dots \in (0,1]$, define $f_x$ by $f_x(n) = x_n.$ – BaronVT Jan 16 '15 at 02:13
  • @BaronVT: For $\frac12$ do you define $f_{\frac12}(1)=5$ and the rest $0$; or do you define $f_{\frac12}(1)=4$ and the rest $9$? – Asaf Karagila Jan 16 '15 at 02:14
  • Also, the powerset of $\mathbb{Z}_+$ is a subset of your set and is clearly uncountable, if you know some set theory. –  Jan 16 '15 at 02:14
  • @AsafKaragila Good point, a full solution should probably take this into account – BaronVT Jan 16 '15 at 02:28

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Edit: initially, the OP defined $X$ as the set of functions from $\mathbb{Z}_+$ to itself.

Suppose that $X$ is countable so that we can enumerate $X$ as $f_1,f_2,\ldots$ Now consider a function $f_0$ defined as $f_0(n)=f_n(n)+1$. Then, $f_0$ is clearly an element of $X$ and $f_0$ differs from $f_n$ for all $n>0$. But this is a contradiction because we have assumed that we can completely enumerate $X$ as $\{f_1,f_2,f_3,\ldots\}$.

If $X$ is the set of functions from $\mathbb{Z}_+$ to $\{0,1\}$, then the above argument still works with a slight modification: $f_0(n)=1-f_n(n)$.

yurnero
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