1

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.

Martin Sleziak
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Gustavo Louis G. Montańo
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  • I am sure you can find these questions answered on this site several times (maybe in some equivalent form). For example: http://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions and http://math.stackexchange.com/questions/17914/cardinality-of-the-set-of-all-real-functions-of-real-variable or http://math.stackexchange.com/questions/478294/the-set-of-all-real-functions-defined-on-0-1-has-cardinal-number-2c (Some of them were listed among related questions on the right.) – Martin Sleziak Mar 15 '14 at 09:05

1 Answers1

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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) $|[0,1]|=|\mathbb R|$.

  • To construct the injection for the second one you can generalise the proof I gave in http://math.stackexchange.com/questions/707705/what-is-the-cardinality-of-the-set-of-all-infinite-sequences/710825#710825 – Tom Collinge Mar 15 '14 at 08:56
  • 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). – Tom Collinge Mar 15 '14 at 09:00
  • @TomCollinge: There's an injection from ${f:\mathscr D \to \mathbb R}$ to $P(\mathscr D) \times \mathbb R$. –  Mar 15 '14 at 09:19