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 $\Bbb{R}$.
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
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
You shouldn't have trouble looking up either of these facts, if you want more details.