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I know that on a closed interval, for example, $C[a,b]$ has the same cardinality as $R$. However, can we know the cardinality of functions that are continuous everywhere but differential nowhere($S$) on an interval $[a,b]$? By Baire's theorem, we know that these functions form a second category. Can we construct a one-to-one mapping showing that $S$ has the same cardinality as $R$?

Any materials to refer to are highly appreciated.

Andrew_Ren
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    The class of continuous functions altogether has cardinality $\lvert\Bbb R\rvert$, and given any function in your set there are $\lvert\Bbb R\rvert$ many such functions by considering $\lambda f$ for $\lambda\ne0$. – Sassatelli Giulio Nov 27 '22 at 11:50
  • So it has nothing to do with Baire's theorem. We can build a one-to-one mapping from this set to $R \times R$? – Andrew_Ren Nov 27 '22 at 11:56
  • The map $f\mapsto \left.f\right\rvert_{\Bbb Q}$ is an injection into $\Bbb R^{\Bbb Q}\cong 2^{\Bbb N\times\Bbb N}\cong\Bbb R$. You can do everything with Cantor-Bernstein, if that qualifies as "constructing". – Sassatelli Giulio Nov 27 '22 at 12:01
  • @Sassatelli Giulio: Perhaps simpler, at least it seems to me, is that given any such function $f,$ each of the continuum many functions $f + \lambda$ is such a function, where $\lambda$ ranges over the real numbers (and $\lambda = 0$ is also OK). Indeed, by restricting $\lambda$ to belong to the open interval $(-\epsilon, \epsilon),$ we get continuum many functions within an $\epsilon$-neighborhood of $f$ (sup norm), and with a bit more work (such as is done here), we can get continuum many functions within any $\epsilon$ ball in $C[a,b].$ – Dave L. Renfro Nov 27 '22 at 15:56

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