**Note:** I asked this question before but it wasn't well written, So I deleted my previous question and re-wrote it.

According to **Dini's theorem**:

If $X$ is a compact topological space, and $\{ f_n \}$ is a monotonically increasing sequence (meaning $f_n(x) \leq f_{n+1}(x)$ for all $n$ and $x$) of continuous real-valued functions on $X$ which converges pointwise to a continuous function $f$, then the convergence is uniform.

The same conclusion holds if $\{ f_n \}$ is monotonically decreasing instead of increasing.

(Note: I have proven both cases)

But, what if for every $n$ $\{f_n(x0)\}$ is monotonic but for some values of $n$ it's monotonically decreasing and for other it's monotonically decreasing. for example; for all even values it is increasing and for non-even values it is decreasing.

**How could I prove that Dini's theorem is effective in this case?
In other words, how to prove that the convergence is uniform**

Let $A=\{x: f_n(x) \leq f_{n+1}(x) \forall n\}$ and $B=\{x: f_n(x) \geq f_{n+1}(x) \forall n\}$. Note that $A$ and $B$ are closed sets and hence they are also compact. Also $A \cup B=X$. $f_n \to f$ uniformly on each of these sets. Given $\epsilon >0$ there exist $n_1, n_2$ such that $|f_n(x)-f(x)| <\epsilon$ for all $x \in A$ for all $n >n_1$ and $|f_n(x)-f(x)| <\epsilon$ for all $x \in B$ for all $n >n_1$. Let $n_0=\max {n_1,N_2\}. Then $|f_n(x)-f(x)| <\epsilon$ for all $x \in X$ for all $n >n_0$.

May 22, 2020 23:18 PM

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