# Munkres topology Exercise 26.10

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?

The error lies in your first sentence. That set $$U_x$$ depends not only on $$x$$ but on $$n$$ too.
• 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 $X$. Jan 6, 2022 at 8:43
• @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. Jan 6, 2022 at 9:27