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I'm reading a book about Hilbert spaces, and in chapter 1 (which is supposed to be a revision of linear algebra), there's a problem I can't solve. I read the solution, which is in the book, and I don't understand it either.

Problem: Prove that the space of continuos functions in the interval (0,1): $C[0,1]$, has dimension $c=\dim(\mathbb{R})$.

Solution: The solution of the book goes by proving that the size of a minimal base of the space $B$ is first $|B|\leqslant c$ and $|B|\geqslant c$, and so $|B|=c$. the proof of it being greater or equal is simple and I understand it, the problem is the other thing. The author does this:

A continuos function is defined by the values it takes at rational numbers, so $|B|\leqslant c^{\aleph_0}=c$

I don't get that.

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Hmm, interesting that the book apparently assumes the continuum hypothesis. –  Tobias Kildetoft Feb 26 '13 at 20:30
    
@TobiasKildetoft Could you please develop that answer. I'm very interested in it. I just don't get that equality. –  MyUserIsThis Feb 26 '13 at 20:31
    
It was just an observation (the continuum hypothesis is that the cardinality of $\mathbb{R}$ is $\aleph_1$). It is strangely stated anyway, since it mentioned $\rm{dim}(\mathbb{R})$ but not over what field. –  Tobias Kildetoft Feb 26 '13 at 20:33
    
@TobiasKildetoft That was actually added by me under complete ignorance. I deleted it. –  MyUserIsThis Feb 26 '13 at 20:34
    
What is a minimal basis? –  1015 Feb 26 '13 at 20:36

3 Answers 3

Since the rationals are dense in the reals, a continuous function $f:\mathbb R \to \mathbb R$ is completely determined by its values on the rationals. Thus, the the function from the set of all continuous functions $f:\mathbb R \to \mathbb R$ to the set of all functions $g:\mathbb Q \to \mathbb R$ (continuous or not), given by restricting a function $f:\mathbb R \to \mathbb R$ to the rationals, is an injective function. Thus the cardinality of the former is less than or equal to the cardinality of the latter, which is $|\mathbb R|^{|\mathbb Q|}=c^{\aleph_0}=c$.

The question is concerned with the domain $[0,1]$, but there is no essential difference.

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Where can I see a proof of $c^{\aleph_0}=c$? –  MyUserIsThis Feb 26 '13 at 20:44

The point is that if $f$ and $g$ are two functions in $C[0,1]$ and the restrictions to $[0,1]\cap\mathbb Q$ of $f$ and $g$ are equal then $f$ and $g$ are actually equal.

This is a simple consequence of the density of $[0,1]\cap\mathbb Q$ in $[0,1]$ using the continuity of $f$ and $g$.

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Note that if $f,g$ are continuous and for every rational number $q$ it holds that $f(q)=g(q)$ then $f=g$ everywhere.

This means that $|B|\leq|\mathbb{R^Q}|=|\mathbb{R^N}|=|\mathbb R|$.

Also, $|\mathbb R|$ is not necessarily $\aleph_1$. This assumption is known as the continuum hypothesis.

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