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Proposition 21. Let $\varphi$ be the Cantor-Lebesgue function and define the function $\psi$ on $[0,1]$ by $$ \psi(x) = \varphi(x) + x \quad \text{for all $x \in [0,1]$}. $$ Then $\psi$ is a strictly increasing continuous function that maps $[0,1]$ onto $[0,2]$,
(i) maps the Cantor set $C$ onto a measurable set of positive measure and
(ii) maps a measurable set, a subset of the Cantor set, onto a nonmeasurable set.

Proof. The function $\psi$ is continuous since it is the sum of two continuous functions and is strictly increasing since it is the sum of an increasing and a stricly increasing function. Moreover, since $\psi(0) = 0$ and $\psi(1) = 2$, $\psi([0,1]) = [0,2]$. For $\mathcal{O} = [0,1] {\sim} C$, we have the disjoint decomposition $$ [0,1] = \mathbf{C} \cup \mathcal{O} $$ which $\psi$ lifts to the disjoint decomposition \begin{equation} \tag{18} [0,2] = \psi(\mathcal{O}) \cup \psi(\mathbf{C}). \end{equation} A strictly increasing continuous function defined on an interval has a continuous inverse. Therefore $\psi(C)$ is closed and $\psi(\mathcal{O})$ is open, so both are measurable. We will show that $m(\psi(\mathcal{O})) = 1$ and therefore infer from $(18)$ that $m(\psi(C)) = 1$ and therefore prove (i).

(Original image here.)

This is from Real Analysis by Royden. I just have a small question in this proof. I don't really understand the red line. Does this mean that since $C$ and $O$ are closed and open, respectively, the function of each set should be closed and open as well? If it is, why is that? I think that it is something related to the inverse of this function, but I don't get it.

I will appreciate for any comment.

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No, in general continuous maps need not be neither open nor closed. That is, they need not map open sets to open sets. In this case however, Royden tells you that the function has a continuous inverse. In this special case, the map will be open, because the inverse image of $\psi^{-1}$ of an open set is open, by continuity of the inverse, which exactly means that the map itself is open. Similarly for closed.

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A homeomorphism is both open and closed. So by the following result, the answer to your question is yes.

Thm: A continuous bijection $f:X\to Y,$ where $X,Y$ are compact metric spaces, is a homeomorphism.

Proof: We only need to show $f^{-1}$ is continuous. So let $E\subset X$ be closed. We need to show $(f^{-1})^{-1}(E)$ is closed in $Y.$ But this is just the set $f(E).$ Since closed subsets of a compact metric space are compact, $E$ is compact. Thus $f(E)$ is compact in $Y.$ We're done, because compact subsets of $Y$ are closed in $Y.$

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