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Here's the problem statement:

Let $X = \{ x:\mathbb{N}\to\{0,1\}:\{i\in\mathbb{N}: x(i) = 1\}$ is finite $\}$.

The function $b_\infty: X \to \mathbb{N}$ defined by $b_\infty(x)=\sum_{i=0}^{\infty} x(i) \cdot 2^i$ is well-defined and bijective.

First let's prove injectivity.

Let $x_1,x_2 \in X$

Let $j\in\mathbb{N}$ be the largest $i$ s.t. $x_1(i) = 1$. Define $k \in \mathbb{N}$ similarly for $x_2$.

Let $x_1' = \{x_1(0), x_1(1), ...,x_1(j)\}$ and $x_2' = \{x_2(0), x_2(1), ...,x_1(k)\}$.

Then we can rewrite $b_\infty(x_1')=\sum_{i=0}^{j} x_1' \cdot2^i$ and $b_\infty(x_2')=\sum_{i=0}^{k} x_2' \cdot2^i$.

So, in general $b_\infty(x')=\sum_{i=0}^{n} x' \cdot2^i$ which is the decimal representation of a binary expansion and demonstrably injective.

I'm now stuck trying to prove $b_\infty$ is surjective.

$X$ must have the same cardinality as $\mathbb{N}$, correct? If so, I can use the fact that $b_\infty$ is injective to conclude that $b_\infty$ is surjective using the pigeonhole principle. But how can I show this mathematically?

Thanks in advance!

alwaysiamcaesar
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    "The function $b_\infty=\dots$ is well defined and bijective"... bijective with what domain and codomain? These are quite important. Presumably the domain is $X$, but you made no mention about codomain. It seems obvious that you intend to use $\Bbb N$ as the codomain, but you didn't explicitly say it, leading to silly answers such as "if you assume the codomain is $\Bbb R$, then it can't possibly be bijective as $X$ is countable." – JMoravitz Aug 19 '18 at 03:30
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    I do not at all follow your attempt for proving injectivity, and suspect that it is incorrect. For injectivity, it helps to assume $x_1<x_2$ and instead let $i\in\Bbb N$ be the largest $i$ for which $x_1(i)\neq x_2(i)$. For surjectivity, induction is the usual tool to use for this problem. As for using that $X$ has the same cardinality as $\Bbb N$ and invoking some form of a pigeonhole principle, first pigeonhole principle doesn't work for infinite sets. Second, this exercise leads to a proof that $X$ has the same cardinality as $\Bbb N$ so it seems like a circular argument. – JMoravitz Aug 19 '18 at 03:35
  • @JMoravitz Totally skipped over that part. My apologies. Edited now. – alwaysiamcaesar Aug 19 '18 at 03:35
  • @JMoravitz I think that's exactly the procedure in the linked proof. In this example, I was trying to show that the function we're dealing with is the same one, namely $\sum_{i=0}^{n} x \cdot2^i$, and therefore injective. – alwaysiamcaesar Aug 19 '18 at 03:39
  • As you noted, this seems to just be mapping between finite binary sequences and natural numbers, which can be done using a base 2 representation. I think you have the right ideas, and just need to arrange them correctly. – theyaoster Aug 19 '18 at 04:26

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