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Let $F$ denote the set of all functions $f : \mathbb{R} \to \mathbb{R}$, and let $C \subset F$ denote the subset of all continuous functions. Prove that $|\mathbb{R}|=|C| < |F|$. Hint: use the fact that a continuous function on $\mathbb{R}$ is determined by its values on the rational numbers $\mathbb{Q} \subset \mathbb{R}$.

Here is my attempt. There are several preliminary lemmas, such as $|\mathbb{R}| = |\mathbb{R}^{\mathbb{N}}|$, that I am still trying to work out the details for, but I want to just be sure that the general sketch of the proof is correct.

We use the Schroder-Bernstein theorem. Given $r \in \mathbb{R}$, the map sending $r \longmapsto f(x) = r$ is an injection $\mathbb{R} \hookrightarrow C$, so we have $|\mathbb{R}| \leq |C|$. Given $f \in C$, $f$ is uniquely determined by its values on rational inputs. In particular, the mapping $f \longmapsto f\mid_{\mathbb{Q}}$ embeds $C$ into $\mathbb{R}^{\mathbb{Q}}$. Since $\mathbb{Q} \cong \mathbb{N}$, we have $\mathbb{R}^{\mathbb{Q}} \cong \mathbb{R}^{\mathbb{N}}$, and we know $\mathbb{R}^{\mathbb{N}} \cong \mathbb{R}$. So we have $$|C| \leq |\mathbb{R}^{\mathbb{Q}}| = |\mathbb{R}^{\mathbb{N}}| = |\mathbb{R}|,$$ so, composing maps, we get an injection $C \hookrightarrow \mathbb{R}$, so by Schroder-Bernstein, $|R| = |C|$. Now, it suffices to show that $|F| > |\mathbb{R}|$. By Cantor's theorem, we have $|\mathbb{R}| < |\mathcal{P}(\mathbb{R})|$. Furthermore, we know $\mathcal{P}(\mathbb{R}) \cong 2^{\mathbb{R}}$, where $2^{\mathbb{R}}$ is the set of functions $\mathbb{R} \to \{0,1\}$. But given such an $f: \mathbb{R} \to \{0,1\}$, since $\{0,1\} \subset \mathbb{R}$, we can identify $f$ with $f': \mathbb{R} \to \mathbb{R}$ where $f'(x) = f$ for all $x \in \mathbb{N}$. So we have $$ |C| = |\mathbb{R}| < |\mathcal{P}(\mathbb{R})| = |2^{\mathbb{R}}| \leq |\mathbb{R}^{\mathbb{R}}| = |F|, $$ so we conclude $$ |C| = |\mathbb{R}| < |F|. $$

How does this sketch look?

Stanley Smith
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  • A reference: https://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions – Robert Z May 06 '21 at 08:06
  • This helps, but it doesn't cover (unless I missed it) the three facts I am still struggling to prove. – Stanley Smith May 07 '21 at 00:25
  • The map that sends a real number $r$ to the constant function $f(x)=r$ is not an injection from $C$ to $\mathbb{R}$, it’s an injection from $\mathbb{R}$ to $C$. So it does not prove that $|C|\leq|\mathbb{R}|$, it proves that $|\mathbb{R}|\leq|C|$, – Arturo Magidin May 07 '21 at 00:34
  • You don’t need an ennumeration. The map from $C$ to $\mathbb{R}^{\mathbb{Q}}$ given by restriction is an injection, which proves $|C|\leq|\mathbb{R}^{\mathbb{Q}}| = |\mathbb{R}|^{|\mathbb{Q}|} = |\mathbb{R}|^{\aleph_0}$. If you’ve shown $|\mathbb{R}^{\aleph_0}| = |\mathbb{R}|$, that’s all you need. – Arturo Magidin May 07 '21 at 00:38
  • The restriction map is much nicer, but I don't have any experience with cardinality arithmetic, so I'm not able to follow the steps $|\mathbb{R}^{\mathbb{Q}}| = |\mathbb{R}|^{\mathbb{Q}|$. Could you explain a bit more about what the underlying bijections are? – Stanley Smith May 07 '21 at 02:06
  • Given sets $A$ and $B$, $A^B$ is the set of all functions $B\to A$. The definition of cardinal exponentiation is precisely that $|A^B| = |A|^{|B|}$, so by definition, $|\mathbb{R}^{\mathbb{Q}}| = |\mathbb{R}|^{|\mathbb{Q}|}$. – Arturo Magidin May 07 '21 at 02:23
  • I don't fully understand what $|A|^{|B|}$ means In the case where either $A,B$ are not finite, though. Is this just a notation? – Stanley Smith May 07 '21 at 05:57

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