Let $(X, \rho_{X})$ and $(Y, \rho_{Y})$ be two compact metric spaces. Consider the metric space $(M_{XY}, \rho)$, where $M_{XY}$ is the set of any mappings from X to Y and $\rho(f,g) := \sup_{x \in X}\{\rho_{Y}(f(x), g(x))\}$. Suppose $C_{XY}$ is the set of continuous mappings from X to Y. Show that $C_{XY}$ is closed in $M_{XY}$.

I have shown that $(M_{XY}, \rho)$ is indeed a well defined complete metric space. To show the closedness of $C_{XY}$, I suspect that I should use the generalised Arzela-Ascoli theorem stating that "Given $(X, \rho_{X})$ and $(Y, \rho_{Y})$ be two compact metric spaces, and $D = \{f_{\alpha}\}_{\alpha \in I} \subset C_{XY}$ be a set of continuous mappings from $X$ to $Y$. Then $D$ is relatively sequentially compact if and only if $D$ is equicontinuous." But I really don't know how to show that $C_{XY}$ is equicontinuous.

Appreciate any help. Thanks.


We can show that the complement of $C_{XY}$ is open: if $f$ belongs to this set, then there is some $x_0$ such that $f$ is not continuous at $x_0$. Hence, for some positive $\delta$ and some sequence $(x_n)_{n\geqslant 1}$ converging to $x$, we have $$\rho_Y(f(x_n),f(x_0))\gt 3\delta.$$ We can show that each function $g$ such that $\rho(f,g)\lt\delta$ is discontinuous at $x_0$. Indeed, since $$3\delta\lt \rho_Y(f(x_n),f(x_0))\leqslant \rho_Y(f(x_n),g(x_n))+\rho_Y(g(x_n),g(x_0))+\rho_Y(g(x_0),f(x_0))\leqslant 2\rho(f,g)+\rho_Y(g(x_n),g(x_0)),$$ we get $\rho_Y(g(x_n),g(x_0))\gt \delta$ for each $n$.

| cite | improve this answer | |
  • $\begingroup$ That's elegant. Thank you very much. $\endgroup$ – theflyingwolves Nov 9 '14 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.