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Is the set of all functions from $\mathbb{N}$ to $\lbrace 0,1\rbrace$ countable or uncountable ? I have no idea on how to start. Any hint or guide will be much appreciated.

Asaf Karagila
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Idonknow
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    Hint: Use Cantor's diagonal argument. Assume the set of all function form $\mathbb{N}$ to ${0,1}$ is countable, and derive the contradiction. – Hanul Jeon Jan 26 '13 at 17:21
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    Hint: Find a bijection with the powerset. For any set $S$ and any map $f\colon S\to P(S)$, $f$ fails to be onto because it misses ${x\in S\mid x\notin f(S)}$. – Hagen von Eitzen Jan 26 '13 at 17:22
  • @HagenvonEitzen: Find bijection between powerset and what ? – Idonknow Jan 26 '13 at 17:26
  • @Idonknow ... a bijection between the poswerset of $\mathbb N$ and the set of functions from $\mathbb N$ to ${0,1}$, cf. Asaf's answer. – Hagen von Eitzen Jan 27 '13 at 11:08

2 Answers2

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Hint: You can think of each function as a binary representation; then use Cantor's diagonal argument to show that it can't be countable.

Set up a table like $$\begin{align}&a_{11}a_{12}a_{13}\ldots\\ &a_{21}a_{22}a_{23}\ldots\\ &a_{31}a_{32}a_{33}\ldots \end{align}$$ where each $a_{ij}\in\{0,1\}$, and use the diagonal argument with the $a_{jj}$.

Clayton
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  • What is binary representation? – Idonknow Jan 26 '13 at 17:39
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    You can think of the function as assigning a position, so for example, you could have $$1_10_20_30_4\ldots$$ as possible representation (the binary says that only zeroes and ones are present). Here, the subscripts represent what number mapped to that image, i.e., $1\mapsto 1$, $2\mapsto 0$, and so on. – Clayton Jan 26 '13 at 17:43
  • I still don know how to proceed. – Idonknow Jan 26 '13 at 17:49
  • I've edited my answer for hopefully a more-clear hint. – Clayton Jan 26 '13 at 17:56
  • Sorry, I'm still very blur. – Idonknow Jan 26 '13 at 18:10
  • Idonknow, do you know how Cantor's Diagonal Argument works? Set $a_{jj}=1$ if it isn't $1$ already, otherwise set it to $0$, then you've constructed an element not in the set by definition, hence it can't be countable, otherwise you couldn't have constructed an element not in the set. – Clayton Jan 26 '13 at 18:17
  • Clayton, perhaps ditch the usual table approach. @Idonknow: Suppose that $f_1,\ldots$ is a countable sequence of functions in ${0,1}^\mathbb N$, we define the function $F(n)=1-f_n(n)$. Check that $F\colon\mathbb N\to{0,1}$, and now show that it differs from $f_1,f_2,\ldots$. If ${0,1}^\mathbb N$ was countable we could have enumerated it like that, but then $F$ would be somewhere in the list. This is impossible and therefore ${0,1}^\mathbb N$ is uncountable. – Asaf Karagila Jan 26 '13 at 18:47
  • @Clayton: okay , now I understand what you mean. But it seems you are not answering my original question. – Idonknow Jan 27 '13 at 06:45
  • @Idonknow: It shows the set cannot be countable.... Hence it is uncountable, which is exactly what you were asking. – Clayton Jan 27 '13 at 06:58
  • @Clayton: ooh, I fully understand now . – Idonknow Jan 27 '13 at 07:05
  • @Idonknow: I'm glad! :) Hopefully it is clear now. – Clayton Jan 27 '13 at 07:10
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Hint: Show that the function $F\colon\{0,1\}^\mathbb N\to\mathcal P(\mathbb N)$ defined by $F(f)=\{n\in\mathbb N\mid f(n)=1\}$ is a bijection.

Use Cantor's theorem to show that $\mathcal P(\mathbb N)$ is uncountable and deduce the same on $\{0,1\}^\mathbb N$.

Asaf Karagila
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