Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a Polish space. Assume that $(C_m)_{m\in\mathbb{N}}$ is an increasing sequence of compact subsets of $X$ and denote $C=\bigcup_{m}C_m$. Let $\{f_n:n\in\mathbb{N}\}$ be a family of functions from $X$ in $\left[0,1\right]$, which is equicontinuous on compact subsets of $X$.

By the Arzelà-Ascoli theorem we can find a subsequence $(f_{k_n})_{n\in\mathbb{N}}$ convergent to a function $f$ uniformly on the set $C_m$, for any $m \in \mathbb{N}$.

Naturally $f$ is continuous on each set $C_m$, but a function with this property need not to be continuous on the set $C$.

Can one choose a subsequence in such a way that the limit be a continuous function on $C$?

As for me, this concept is too optimistic, but it was used in the paper Remarks on Ergodic Conditions for Markov Processes on Polish Spaces by Stettner (page 110, step 3).

share|cite|improve this question

Your function $f$ is in fact continuous on $C$. The reason is that a function defined on a metric space is continuous iff its restriction to any compact set is continuous. Here are some more details.

Let $(x_i)$ be a sequence in $C$ converging to $x\in C$. Then $K=\{ x\}\cup \{ x_i;\; i\in\mathbb N\}$ is a compact subset of $X$. So $(f_n)$ is equicontinuous on $K$, and since $(f_{k_n})$ converges pointwise to $f$ on $K$ (because $K\subset C$), this implies that $f(x_i)\to f(x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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