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Knowing that there is no surjection from $\Bbb{N}$ to $\Bbb{R}$, I also know that there is no injection going the other way.

I know how to prove that there is no injection from the power set of $S$ to $S$, but how can I prove that there is no injection from $\Bbb{R}$ to $\Bbb{N}$?

(I would, of course, be happy with showing no injection from $(0,1)$, or any other interval to $\Bbb{N}$.)

[EDIT] I usually prove that cardinality of $\mathcal{P}(S)$ is greater than that of $S$ by showing that any function $f: S \to \mathcal{P}(S)$ fails to be surjective. This is done by showing that there is no $s \in S$ for which $f(s) = {\{ s\in S | s \notin f(s) \}}$.

It is possible instead to show that any function $f:\mathcal{P}(S) \to S$ fails to be injective by showing that the image of $T = \{s \in S| \exists U \left( f(U)=s \land s \notin U \right) \}$ must be the image of two elements of $\mathcal{P}(S)$.

I would like to replace the usual diagonal proof that any function $\Bbb{N} \to (0,1)$ fail to be surjective with a proof that any function $f:(0,1) \to \Bbb{N}$ fails to be injective. (Or that any function $f:\Bbb{R} \to \Bbb{N}$ fails to be injective.)

Jim H
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  • https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem If an injective map exists from $\mathbb{R}$ to $\mathbb{N}$, given the inclusion map from $\mathbb{N}$ to $\mathbb{R}$, we can construct a (surjective) bijection from $\mathbb{N}$ to $\mathbb{R}$. – Theo Bendit Sep 02 '17 at 04:50
  • @TheoBendit I'm sorry I wasn't clear. I would like to prove the uncountability of the reals by proving that there is no injection from $\Bbb{R}$ to $\Bbb{R}$ instead of the usual proof that there is no surjection from $\Bbb{N}$ to $\Bbb{R}$ – Jim H Sep 02 '17 at 04:59

2 Answers2

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Schröder–Bernstein theorem:

If there exist injective functions $f : A \to B$ and $g : B\to A$ between the sets $A$ and $B$, then there exists a bijective function $h : A \to B$.

Your $A$ and $B$ are $\Bbb{N}$ and $\Bbb{R}$.

Benjamin Dickman
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Bill Donald
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"I know how to prove that there is no injection from the power set of $S$ to $S$, but how can I prove that there is no injection from $\mathbb R$ to $\mathbb N$?"

Using binary decimal expansions of real numbers, you can show that there is an injection $(0,1)$ to $\mathcal P(\mathbb N)=2^\mathbb N$ with image whose complement is a countable set. Using the elementary fact you have above, this should be enough information to proceed directly, without appealing to the Cantor-Shroeder-Bernstein theorem.

UPDATE: I see your edit, and I think this answers your question better. Any injection $(0,1) \rightarrow \mathbb N$ will admit a partial inverse $M\rightarrow (0,1)$ where $M$ is some subset of $\mathbb N$. Then you can use that any infinite subset of $\mathbb N$ is in bijection with $\mathbb N$ (or prove from first principles of induction using the fact that $\mathbb N$ is well ordered). Then use your previous knowlegde of the diagonal proof that $\mathbb N$ does not inject into $(0,1)$, after eliminting the case that $M$ is finite.

Even if you want a direct proof, I believe some abstraction is probably helpful and the 'right' thing to do.

Victor Zhang
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    The injection of which you speak is slightly subtle; cf. the answer by Timothy Chow to MO 56633. – Benjamin Dickman Sep 02 '17 at 05:00
  • I mean nothing more than the fact that binary representation of a real number is unique except for strings that end in an infinite string of 1's. But there are only countably many of those. (this is considering $\mathcal P(\mathbb N)=2^\mathbb N$ as literally an infinite tuple of 1's and 0's. – Victor Zhang Sep 02 '17 at 10:31
  • In the end, this may be what I choose to do. What I am hoping to find is something like the diagonal proof that no function from $\Bbb{N}$ to $(0,1)$ can be surjective. That is, I hope to find something that can be shown prior to the more general proofs about power sets. – Jim H Sep 02 '17 at 13:03
  • see my edit, now I see what you mean better – Victor Zhang Sep 03 '17 at 06:53