$\textbf{Proposition: }$Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of continuous functions from $X$ to $Y$ which is pointwise convergent to some $f:X\to Y$. If $X$ is not compact, but the convergence $f_n \to f$ is uniform on every compact subset of $X$, then $f$ is continuous.

$\textbf{Proof:}$ We know that if the convergence $f_n\to f$ is uniform on $K\subset X$, then $f$ is continuous on $K$.

Choose an $a\in X$. Let $x_n\to a$. We will show that $f(x_n) \to f(a)$.

Because $A=\{x_n\}_{n\in\mathbb{N}} \cup \{a\}$ is compact, $f$ will be continuous on $A$. Hence $f(x_n)\to f(a)$.

Is this proof correct?

  • $\begingroup$ I added my question. $\endgroup$ Aug 22, 2019 at 18:39
  • 1
    $\begingroup$ I see. You might want to add the proof-verification tag :) Yes, indeed, your proof is correct. $\endgroup$ Aug 22, 2019 at 18:41
  • $\begingroup$ Oh, allright. Thank you :) I thought it was too simple to be correct and that I'd missed something. $\endgroup$ Aug 22, 2019 at 18:42
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    $\begingroup$ Nicely done. ................. $\endgroup$ Aug 22, 2019 at 19:19

1 Answer 1


Yes, I agree that the proof is correct.


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