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Let $\mathbb{N}^{\{0,1\}} :=\{f: \mathbb{N} \to \{0,1\}\}$ is uncountable

I have never heard of the table approach, and all the proofs say uncountability of $\mathcal {P}(\mathbb{N})$ I have seen so far do not use the table approach. There was a hint how to do this without listing stuff in a table: Is the set of all functions from $\mathbb{N}$ to $\{0,1\}$ countable or uncountable?

Suppose that $f_1,…$ is a countable sequence of functions in $\{0,1\}^\mathbb{N}$, we define the function $F(n)=1−f_n(n)$. Check that $F:\mathbb{N}\to\{0,1\}$, and now show that it differs from $f_1,f_2,…$. 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.

Couple of questions:

  1. Does $\{0,1\}^\mathbb{N} = \{f: \mathbb{N} \to \{0,1\}\}$, I maybe using conflicting notations

If that is the case, then the function $F$ makes sense

  1. What does it mean by "now show that it differs from $f_1,f_2,…$"?

Does it mean suppose $f_k(n) = F(n)$, then $F(n) = 1 - f_k(n) = 1 - F(n)$, so $F(n) = \dfrac{1}{2}$ so this is impossible.

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    By definition, $A^B$ is the set of functions from $B$ to $A$, for all sets $A$ and $B$. – Git Gud Jun 07 '16 at 21:37
  • @GitGud Oh that's a terrible convention, your pre-image should come before your image. Looking at $A^B$ makes me think of $f: A \to B$ – Shamisen Expert Jun 07 '16 at 21:39
  • Ad 1: This is right. Ad 2: It means that $F \neq f_n$ for all $n \in \mathbb N$. In particular $F(n) = 1-f_n(n) \neq f_n(n)$. –  Jun 07 '16 at 21:39
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    @TheSilenceoftheCows I disagree, since for finite sets, you have $|A^B| = |A|^{|B|}$. It feels more akward to have $|A^B| = |B|^{|A|}$ – JMoravitz Jun 07 '16 at 21:39
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    @TheSilenceoftheCows "Before" is a relative term. You're reading from left to right. Anyway, the motivation behind the notation is this. Consider the number $2^3$. Now take a set with two members, say ${0,1}$ and another one with three elements, say ${0,1,2}$. You can prove that the set of functions from the latter to the form has exactly $8$ elements (which equals $2^3$). This is true for any choice of finite sets, hence the generalization of the notation. – Git Gud Jun 07 '16 at 21:42
  • As an aside, the set of functions from ${0,1}$ to $\Bbb N$ is isomorphic to $\Bbb N\times \Bbb N$ and is known to be countable as it is the cartesian product of a finite number of countable sets. On the other hand, the set of functions from $\Bbb N$ to ${0,1}$ are in bijection with the binary expansions of the set of real numbers between $0$ and $1$ which we know to be uncountable. The proof for why those are uncountable is outlined already in your post. – JMoravitz Jun 07 '16 at 21:47
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    @TheSilenceoftheCows You couldn't be more mistaken about the definition of $A^B.$ It's perfectly natural. – zhw. Jun 07 '16 at 21:48

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