Let $f_n:X\to\Bbb R$ be a sequence of continuous functions with $f_n(x)\to f(x)$ for each $x\in X$. If $f$ is continuous and if the sequence $f_n$ is monotone increasing and if $X$ is compact, then the convergence is uniform.

Proof. For each $x\in X$, by the continuity of $f_n$, there is $U_x$ such that for any $y\in U_x$, $|f_n(x)-f_n(y)|<\epsilon/3$. Also by the continuity of $f$, there is $V_x$ such that for any $y\in V_x$, $|f(y)-f(x)|<\epsilon/3$. Let $W_x:=U_x\cap V_x$ which is nonempty open set as $x\in W_x$. Then $\{W_x\}_{x\in X}$ is an open covering of $X$ so by compactness, there is a finite subcover $\{W_{x_1},...,W_{x_n}\}$. Note that for each $x\in X$, there is a number $N_x$ such that $n\geq N_x$ implies $|f_n(x)-f(x)|<\epsilon/3$. So define $N = \max\{N_{x_1},...,N_{x_n}\}$.

Now let $y\in X$ be given. Then there is $i$ such that $y\in W_{x_i}$. For $n\geq N$, $$|f_n(y)-f(y)|\leq |f_n(y)-f_n(x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i)-f(y)|<\epsilon.$$ I think I didn't use monotone increasing condition. I can't find the error. Could you help?


1 Answer 1


The error lies in your first sentence. That set $U_x$ depends not only on $x$ but on $n$ too.

  • $\begingroup$ Thanks. But it seems monotonicity does not fix that problem directly. $\endgroup$ Jan 6, 2022 at 7:50
  • 2
    $\begingroup$ @love_sodam maybe a totally different approach? Have you done a search on this site? It’s an old problem…. $\endgroup$ Jan 6, 2022 at 8:37
  • $\begingroup$ JSC means, these sets $U_x$ are not the sets that you should be concerned with covering $X$ with. Perhaps consider that for a fixed $\epsilon \in (0, \infty)$ the fact that $\lim_{n \to \infty} f_n(x)=f(x)$ guarantees that the sets $\{x: |f_n(x)-f(x)|<\epsilon\}$ ($n=1,2,\ldots$) cover $X$. $\endgroup$
    – M A Pelto
    Jan 6, 2022 at 8:43
  • $\begingroup$ @HennoBrandsma Yeah I know this is an old problem. I just wonder why the above proof is wrong. Also I wanted to conclude the statement that frequently appears in analysis proofs $|-|\leq |-|+|-|+|-|<3\cdot\frac{\epsilon}{3} = \epsilon$ which failed eventually though. $\endgroup$ Jan 6, 2022 at 9:27
  • $\begingroup$ @love_sodam See here for a better proof. The linked question in it is also relevant, as it gives examples where this fails if we drop assumptions. $\endgroup$ Jan 6, 2022 at 11:24

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