my question reads as:

Let $\mathcal R[0,1]$ denote the set of all real-valued functions from $[0,1]$ to $\mathbb R$ and let $\mathcal C[0,1]$ denote the set of continuous functions on $[0,1]$.
i) Prove that $|\mathcal C[0,1]|=|\mathbb R$|.
ii) Prove that $|\mathcal R[0,1]|>|\mathbb R$|.

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I've been fairly confident in my work with cardinality, but with a set containing functions, I am finding it hard to get a lead.

Thank you in advanced for your help.


I would give you some hints here.

For the first one, you can use the fact that if $f, g\in C [0,1]$ satisfies $f(x) = g(x)$ for all $x\in \mathscr D$, where $\mathscr D$ is a dense subset of $[0,1]$, then $f\equiv g$. Here you can choose $\mathscr D$ to be countable. Then think of $|\{ f:\mathscr D \to \mathbb R\}|$. (The set of all real-valued function from $\mathscr D$).

For the second one, try to construct an injection $P([0,1]) \to R[0,1]$ where $P([0,1])$ is the power set of $[0,1]$. Then use (or prove) $|[0,1]|=|\mathbb R|$.

  • $\begingroup$ To construct the injection for the second one you can generalise the proof I gave in math.stackexchange.com/questions/707705/… $\endgroup$ – Tom Collinge Mar 15 '14 at 8:56
  • $\begingroup$ For the first one you can choose D as the rationals and use a simple $\epsilon, \delta$ proof to show that continuous functions from R -> R are equivalent to functions from Q -> R. But I'd like to see how they are then equivalent to functions from Q -> Q (which I believe is the answer). $\endgroup$ – Tom Collinge Mar 15 '14 at 9:00
  • $\begingroup$ @TomCollinge: There's an injection from $\{f:\mathscr D \to \mathbb R\}$ to $P(\mathscr D) \times \mathbb R$. $\endgroup$ – user99914 Mar 15 '14 at 9:19

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