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Question: Let $C[0,1]$ be the set of all continuous function from $[0,1]$ to $\mathbb R$. Prove that the cardinality of $C[0,1]$ is equal to that of $\mathbb R$.

My attempt: Let $X$ be a countable dense subset of $[0,1]$.

  • $f$ is continuous. So, $f$ is determined by its restriction to $X$.
  • Also the restriction to $X$ determines an injection from $C[0,1]$ into $C(X)$.
  • #$C(X) =$#$\mathbb{R}$ implies $C[0,1] \leq$ #$\mathbb{R}$.

Now we want to prove #$\mathbb{R} \leq$ #$C[0,1]$.

Now for each $a \in [0,1]$ the map $a \mapsto f_a$ ($f_a(x) = a\ \forall\ x\in [0,1]$) defines an injection from $[0,1]$ into $C[0,1]$. Since #$[0,1] =$#$\mathbb{R}$, we get #$\mathbb{R} \leq$ #$C[0,1]$.

Is it correct? I am confused about the second inequality.

user1234
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    This all looks correct to me! Of course we should note that it's still necessary to prove that $#C(X)=#R$ to finish the problem. – Greg Martin Jul 20 '22 at 16:17
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    Another way to prove this uses the Weierstrass Theorem. This shows that $C([0, 1])$ with the $\sup$ norm has a countable dense subset (namely, polynomials with rational coefficients). We can show that any metric space with a countable dense subset has cardinality at most $2^{\aleph_0}$. – Mark Saving Jul 20 '22 at 16:51

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