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How can I prove that the set of all functions from $\mathbb{N} \to \{0, 1\}$ is uncountable?

Edit: This answer came to mind. Is it correct?

This answer just came to mind. By contradiction suppose the set is $\{f_n\}_{n \in \mathbb{N}}$. Define the function $f: \mathbb{N} \to \{0,1\}$ by $f(n) \ne f_n(n)$. Then $f \notin\{f_n\}_{n \in \mathbb{N}}$.

user62268
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4 Answers4

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Hint: Show that $\{0,1\}^\mathbb N$ is equinumerous with $\mathcal P(\mathbb N)$ and use Cantor's theorem to conclude there is no bijection between $\mathbb N$ and $\mathcal P(\mathbb N)$.

Asaf Karagila
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You can proof it by contraposition. I will identify the functions with sequence so $a_{n}$=$a(n)$. Now lets say it's countable, now let $a_{nk}=a_k(n)$ be the $k$-th function. Now we construct the function $$b(k)=\left\{ \begin{array}{rl} 1 & a_{kk}=0\\ 0& a_{kk}=1\end{array}\right.$$ You should be able to do the rest.

Dominic Michaelis
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    You don't need contraposition. You can do a direct proof that goes like this: Let $f$ be a mapping from $\mathbb N$ to $2^{\mathbb N}$. We can construct an element $b $ of $2^{\mathbb N}$ just as you did: $b(k) = 1-f(k)(k)$, and show that $b$ is not in the range of $f$. Therefore $f$ is not a bijection. But $f$ was completely general, so we have just proved that no mapping from $\mathbb N$ to $2^{\mathbb N}$ is a bijection, QED. – MJD Feb 14 '13 at 22:25
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Hint : use the diadic developpement of elements of $[0,1]$.

Damien L
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Hint: Each function from $\mathbb{N} \to \{0, 1\}$ is isomorphic to a subset of $\mathbb{N}$. Simply count $n \in \mathbb{N}$ to be part of the subset of $f(n) = 1$.

Joe Z.
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