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I'm currently in my first year of studying mathematics. In one of the subjects we learnt about the cardinality of different sets, among which $\Bbb{N}$ and $\Bbb{R}$. We also discussed the Power Set $P(A)$ as the set of all subsets of $A$ and proved that $P(A) > A$ for all $A$.

Today I asked myself: What is the cardinality of the set $X$ of continuous functions $f: D \rightarrow \Bbb{R} $ where $D \subseteq \Bbb{R}$. So: $$X=\{f: D \rightarrow \Bbb{R} \space \space where f \space is \space continuous\}$$

After some time I got the idea to construct a subset $Y$ of $X$ in the following way:

$$Y = \{f(x) \space where \space f(x) = \sum_{\alpha\in I}x^\alpha \space \space with \space I \subseteq \Bbb{R} \}$$

Is it allowed to do this, or can I only sum over a subset with countably many elements? And are these functions continuous? Because if this is true, then I have shown that $X \le P(\Bbb{R})$, because there is an element in $Y$ for each subset of $\Bbb{R}$.

Dion Leijnse
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The cardinality of the set $C(D,\mathbb{R})$ of all continuous functions $D \to \mathbb{R}$, where $D \subset \mathbb{R}$, is actually the same as the cardinality of $\mathbb{R}$.

Obviously (assuming $D$ has at least one point) we have $|C(D,\mathbb{R})| \geq |\mathbb{R}|$, since we can consider the constant functions. To see the reverse inequality takes a bit of knowledge of analysis and set theory. The key fact from analysis is

Every subset $D \subset \mathbb{R}$ is separable meaning there is a countable subset $D_0 \subset D$ which is dense in $D$.

Since any continuous function on $D$ is determined by its values on $D_0$, restriction gives an injective map $$C(D,\mathbb{R}) \to C(D_0,\mathbb{R})$$ It remains to know that $C(D_0,\mathbb{R})$ has the cardinality of $\mathbb{R}$. In fact, this will still be true if we drop the continuity. The key thing from set theory to know is

The cardinality of the set of functions from a countable set to $\mathbb{R}$ is the same as the cardinality of $\mathbb{R}$. In other words, $\mathbb{R}^{D_0}$ has the cardinality of $\mathbb{R}$ when $D_0$ is countable.

You shouldn't have trouble looking up either of these facts, if you want more details.

Mike F
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  • Very nice proof! –  Nov 03 '16 at 15:43
  • I like this proof. I don't yet fully understand it, but I get the idea. But my sketch of a proof implies something else, so what is wrong with it? – Dion Leijnse Nov 04 '16 at 07:31
  • @DionLeijnse: There are definitely some problems with taking uncountable sums of real numbers; such a sum can only coverge when all but countably many terms of the sum are zero. But before getting into any of that, do you mean to write "then I have shown $X \geq P(\mathbb{R})$" instead of the reverse inequality? – Mike F Nov 05 '16 at 14:07
  • @MikeF with that $X \leq P(\Bbb{R})$ I mean that I made an injection from $P(\Bbb{R})$ to $X$. – Dion Leijnse Nov 05 '16 at 16:18
  • So it should go the other way. For example, I can inject $P = {1,2,3}$ into $X = {1,2,3,4}$, and therefore $4 = |X| \geq |P| = 3$. – Mike F Nov 05 '16 at 16:21