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It is well known that uniform convergence implies pointwise convergence. I have checked on this page (see for example 1 and 2), that locally uniform convergence does not necessarily imply uniform convergence. So, I am wondering if locally uniform convergence implies pointwise convergence in general. Somehow it seems trivial, I still need to convince myself.

Here is what I thought: let $(f_n)_{n\geq 1}$ be a sequence of functions defined on a subset $A$ of $\mathbb{C}$. Assume there exists $f:A\to \mathbb{C}$ such that $f_n\to f$ locally uniformly on $A$. (*) Then $f_n\to f$ uniformly on every compact subset of $A$. Then $f_n\to f$ pointwise on every compact subset of $A$. Does it imply that $f_n\to f$ pointwise on $A$?

Edit: After having read the comments: [...] (*) Then $f_n\to f$ uniformly on every compact subset of $A$. Let $x \in A$ be given. Then $\{x\}$ is a compact subset of $A$. Then $f_n\to f$ pointwise on $\{x\}$. Since $x$ is arbitrary, $f_n\to f$ pointwise for all $x\in A$.

But this is still inconsistent. See my comment below.

Hopeless
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    For each $x\in A$, ${x}$ is a compact subset of $A$... – Maxime Ramzi Feb 26 '18 at 21:55
  • @Max So the answer is "yes"? An infinite unions of compact sets is not compact. It is not known if $A$ is compact or not. – Hopeless Feb 26 '18 at 21:58
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    as ${x}$ is compact we have uniform convergence on that set and so in particular pointwise convergence. – Jonas Lenz Feb 26 '18 at 22:01
  • @JonasLenz I have edited my post. How does my reason sound? – Hopeless Feb 26 '18 at 22:07
  • @Hopeless sounds gut – Jonas Lenz Feb 26 '18 at 22:09
  • @JonasLenz Hold on a moment. Before reaching to the conclusion, one would have: "Then $f_n\to f$ uniformly on every compact subset of $A$. Let $x \in A$ be given. Then ${x}$ is a compact subset of $A$. Then $f_n\to f$ uniformly on ${x}$, and so $f_n\to f$ uniformly on $A$, since $x$ is arbitrary." But is this not a contradiction, because locally uniform convergence does not imply uniform convergence in general? – Hopeless Feb 26 '18 at 22:17
  • The part and so $f_n\to f$ uniformly on A is false. – Jonas Lenz Feb 26 '18 at 23:10

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