Let $C[0,1]$ be the set of all continuous function from $[0,1]$ to $\mathbb R$. Is the cardinality of $C[0,1]$ larger than or equal to that of $\mathbb R$?
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It's equal. As a hint, consider the set of functions from $\Bbb Q\cap [0,1]$ to $\Bbb R$. – Ben Grossmann Nov 25 '15 at 03:48
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It is equal, because of continuity.
Specifically, let $X\subset [0,1]$ be a countable dense subset (for example, you could take $X$ to be the set of rational numbers in $[0,1]$).
Then since every $f\in C[0,1]$ is continuous, it is determined by its restriction to $X$. Thus, restriction determines an injection $C[0,1]\hookrightarrow C(X)$. To see this, suppose we know the values of $f$ at every $x\in X$. Now let $a\in [0,1]$, then we know by density that there is some sequence $(x_n)_n$ with $a = \lim_n x_n$. But then, we have $$f(a) = f(\lim_n x_n) = \lim_n f(x_n)$$ Here, it is continuity that lets us say that "$f$ of a limit is the limit of the $f$'s" (this is a straightforward consequence of the definitions). Thus, $f(a)$ is determined by the values of $f$ at $X$. Hence, any two functions in $C[0,1]$ which agree on $X$ must actually be equal.
On the other hand, if $\mathfrak{c}$ is the cardinality of $\mathbb{R}$, then the cardinality of $C(X)$ is $\mathfrak{c}^{\aleph_0} = \mathfrak{c}$
oxeimon
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1I understand your idea, but can please you elaborate "it is determined by its restriction to X" a bit more rigorously? – JSCB Nov 25 '15 at 03:57
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