Why is $\{0,1\}^{\Bbb N}$ uncountable?

Question

In the book Measure and Integral : An introduction to real analysis, in chapter 8 Lp spaces, theorem 8.18, the authors give a counterexample to show that $l^\infty$ is not separable:

Question: why is this set of sequences uncountable? I am given an answer that it is related to Cantor's diagonal argument. But how can I write this formally?

We know the interval $[0, 1]$ is uncountable. You can think of the binary expansions of all numbers in $[0, 1]$. This will give you an uncountable collection of sequences. — Debarghya, May 16 '24 at 13:47
so is it enough if i just repeat the proof in Cantor's diagonal argument? — Bully Gary, May 16 '24 at 13:50
Yes exactly. That argument gives you the existence of uncountable binary sequences. Now each element of the sequence is a binary expansion for the different numbers on $[0, 1]$. Therefore, they must differ at some index, i.e. there must be some index where one sequence is 1 and the other one is 0. Therefore, their $l_\infty$ norm is 1. — Debarghya, May 16 '24 at 13:52
Simpler and more direct argument than @Debarghya's suggestion (avoiding the trouble that some numbers have two binary expansions, one proper and one improper i.e. 'ending' with an infinite tail of ones): you can easily define a bijection between your set of binary sequences and the powerset $\mathcal P(\Bbb N)$ of $\Bbb N$. Now, the proof that $|\mathcal P(X)|>|X|$ for every set $X$ is standard. — Anne Bauval, May 16 '24 at 14:13

score 0 · Answer 1 · answered May 16 '24 at 15:31

Big thanks to @Debarghya and @AnneBauva

Conclusion:

1)Lemmas and theorems (I looked it up in my textbook):

(1)If E is a set, then there does not exist a surjective mapping $E \rightarrow \mathcal{P}(E)$

(2)$\mathcal{P}(\mathbb{N})$ is uncountable. (Implied by (1), there is no surjection $\mathbb{N}\rightarrow \mathcal{P}(\mathbb{N})$ )

(3)For every set $E$ , $|E| < |\mathcal{P}(E)|$ (again , thank you @AnneBauva for reminding me this)

(again , thank you @AnneBauva for reminding me this)

From @AnneBauva (the post Prove that $\mathcal{P}(A) \sim \{0,1\}^A$, but i think that this post Finding a correspondence between $\{0,1\}^A$ and $\mathcal P(A)$, the rest is actually a "modified" answer from the latter) we can construct a bijective function $\{0,1\}^{\mathbb{N}} \rightarrow \mathcal{P}(\mathbb{N})$ as follow:

For a function $f$ from $\mathbb{N}$ to $\{0,1\}$, let $\mathbb{N}_f$ be the set of elements of $\mathbb{N}$ that are mapped to 1 by $f$. That is, $n \in \mathbb{N}_f$ if and only if $f(n)=1$.

Consider the map $\Phi(f) =\mathbb{N}_f$.

Now if $f\ne g$, there is an $n\in \mathbb{N}$ with $f(n)=0$ and $g(n)=1$ (or $f(n)=1$ and $g(n)=0$).

Then $\mathbb{N}_f \ne \mathbb{N}_g$. So $\Phi$ is one-to-one.

Now let $B\in{\cal P}(\mathbb{N})$. Define $f(x)=\cases{1,&x $\in$ B\cr 0,&x $\notin$ B }$

Then $\Phi(f)=B$. This shows that $\Phi$ is onto ${\cal P}(\mathbb{N})$

So $|\{0,1\}^{\mathbb{N}}|=|{\cal P}(\mathbb{N})|$, and by lemma (2), $\{0,1\}^{\mathbb{N}}$ is uncountable.

Why is $\{0,1\}^{\Bbb N}$ uncountable?

1 Answers1