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I'm trying to prove that the real numbers are uncountable using two simple facts which I've already proved:

Fact 1

There aren't any surjection $f:\mathbb N\to P(\mathbb N)$

Fact 2

There exists an injective function $g:P(\mathbb N)\to \mathbb R$.

I would like to know if with only these facts I can deduce that the reals aren't countable.

Thanks in advance

EDIT

Following the comments, can I say the statement is true because of $\mathbb N\lt P(\mathbb N)\le \mathbb R$, where the first inequality comes from fact 1 and second inequality comes from fact 2.

user42912
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3 Answers3

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Suppose there exists a bijection $\varphi : \mathbb{N} \rightarrow \mathbb{R}$, and let $g : P(\mathbb{N}) \rightarrow \mathbb{R}$ be an injective map. Choose a left inverse of $g$, say $h : S \rightarrow P(\mathbb{N})$ where $S \subset \mathbb{R}$. Then we have a surjective map $h \circ \varphi : T \rightarrow P(\mathbb{N})$, where $T \subset \mathbb{N}$, $T = \varphi^{-1}(S)$, a contradiction.

RghtHndSd
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Yes, your two facts are sufficient.

You can prove that for any two sets $A$ and $B$, there exists an injection $A\hookrightarrow B$ if and only if there is a surjection $B\twoheadrightarrow A$. You can also prove (the Cantor-Bernstein-Schroeder theorem) that if there are injective maps $A\hookrightarrow B$ and $B\hookrightarrow A$, then there is a bijection $A\leftrightarrow B$. From this, you can use the existance of an injection $A\hookrightarrow B$ to define $|A|\le|B|$, with $|A|=|B|$ if $|A|\le|B|$ and $|B|\le|A|$ (i.e. if there is a bijection between them), and defined $|A|<|B|$ as $|A|\le|B|$ but not $|A|=|B|$.

Finally, for any two sets $A$ and $B$, you can always find an injection $A\hookrightarrow B$ or $B\hookrightarrow A$.

Thus, your two facts mean that $|\mathbb{N}|<|P(\mathbb{N})|$ and $|P(\mathbb{N})|\le|\mathbb{R}|$.

Einar Rødland
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    Cantor-Bernstein requires absolutely no trace of the axiom of choice, nor does Cantor's theorem, nor does $|\cal P\Bbb{(N)|=|R}|$. – Asaf Karagila Apr 26 '14 at 01:19
  • @AsafKaragila: You're right. I'll correct the answer and add a link as well. However, how about the second part about always being able to find an injection $A\hookrightarrow B$ or $B\hookrightarrow A$? I can't see how to get this without AoC. Same with the correspondence between injection in one direction and surjection in the other. – Einar Rødland Apr 26 '14 at 02:11
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    That is equivalent to the axiom of choice. The assumption that if there is a surjection, then there is an injection follows from the axiom of choice, not provable without some choice, and it's still open if it's equivalent to the axiom of choice (it's called the "Partition Principle" and I have minced many words on the subject at least once or twice over this site). – Asaf Karagila Apr 26 '14 at 02:12
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Proposition: $\mathbb{R}$ is uncountable.

Proof: Let define the map $f: \mathcal{P}(\mathbb{N}) \to \mathbb{R}$ by the formula $f(A):= \sum_{n\in A}10^{-n}$. Note that the map is well-define since $\sum_{n=0}^\infty 10^{-n}$ is absolutely convergent.

We will show that $f$ is one-to-one. Suppose to the contrary that there exists $A, B \in \mathcal{P}(\mathbb{N}) $ such that $f(A)=f(B)$ but $A \not=B$. Then the set $(A\setminus B )\cup (B\setminus A)$ is non-empty. Let $n_0$ be the least element and for sake of definiteness assume that $n_0 \in A\setminus B $. So we have

\begin{align}0=f(A)-f(B)=\sum_{n\in A}10^{-n}-\sum_{n\in B}10^{-n}=\sum_{n\in A:n<n_0}10^{-n}+10^{-n_0}+\sum_{n\in A:n>n_0}10^{-n}\\-\sum_{n\in B:n<n_0}10^{-n}-\sum_{n\in B:n>n_0}10^{-n}\\ =10^{-n_0}+\sum_{n\in A:n>n_0}-\sum_{n\in B:n>n_0}10^{-n}\\ \ge 10^{-n_0}-\sum_{n>n_0}10^{-n}\\ =10^{-n_0}-\frac{10^{-n_0}}{9}>0 \end{align}

a contradiction. Hence $f$ is an injection.

Then $\mathcal{P}(\mathbb{N})\simeq f(\mathcal{P}(\mathbb{N}))$, i.e., $f(\mathcal{P}(\mathbb{N}))$ has the same cardinality as $\mathcal{P}(\mathbb{N})$ and by the Cantor's theorem $\mathcal{P}(\mathbb{N})$ is uncountable. Since $f(\mathcal{P}(\mathbb{N}))$ is a subset of $\mathbb{R}$ then this forces $\mathbb{R}$ to be uncountable.

Jose Antonio
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