Let $\{f_n\}$ be a sequence of continuous complex functions on a (nonempty) complete metric space $X$, such that $f(x)=\displaystyle \lim_{n\to\infty} f_n(x)$ exists for every $x\in X$.

a) Prove that there is an open set $V\neq\emptyset$ and a number $M<\infty$ such that $|f_n(x)|<M$ for all $x\in V$ and for $n=1,2,3,\ldots$

b) If $\epsilon>0$, prove that there is a open set $V\neq\emptyset$ and an integer $N$ such that $|f(x)-f_n(x)|\leq \epsilon$ if $x\in V$ and $n\geq N$.

$\textbf{b)}$: For $N=1,2,3\ldots$, put

$$A_N=\{x:|f_n(x)-f_m(x)|\leq \epsilon \ \ \mbox{if} \ \ m,n\geq N\}$$

then $X=\bigcup A_N $, by Baire's theorem there exists some $N$ such that $\overline{A_N}$ has a nonempty interior e.g. $\emptyset\neq V$ open and $V\subset \overline{A_N}$

Hence in $V$ we get

$$|f(x)-f_n(x)|\leq \epsilon$$

for all $x\in V$ and $n\geq N$, because $f_m(x)\to x$ if $m\to \infty$.

Is correct b)?

Thank you, for any suggestion for a).


Your argument for (b) is correct. You can use the same idea for (a); this time let

$$A_N=\{x\in X:|f_n(x)|\le N\text{ for all }n\in\Bbb Z^+\}\;.$$

  • $\begingroup$ Thank you, Brian M. Scott. $\endgroup$
    – user126033
    Dec 15 '14 at 1:23
  • $\begingroup$ @user126033: You’re welcome. $\endgroup$ Dec 15 '14 at 1:26
  • $\begingroup$ How to use the continuity of each $f_n$?. $A_N=\overline{A_N}$ because each $f_n$ is continuous? $\endgroup$
    – user126033
    Dec 15 '14 at 3:39
  • $\begingroup$ @user126033: I was a little sloppy: I should have written a non-strict inequality. I've fixed it; now $A_N$ is an intersection of closed sets, so it's closed. $\endgroup$ Dec 15 '14 at 3:56
  • $\begingroup$ The continuity of $f_n$ is relevant in "$A_N$ is closed"?. $\endgroup$
    – user126033
    Dec 15 '14 at 3:58

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