Summation over uncountable set while proving that cardinality of functions is bigger than $\Bbb{R}$ 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 $\Bbb{R}$.
 A: 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.
