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I'd like to prove the following:

If $f_j$ are continuous functions on a compact set $K$, and $f_{1}(x) \leq f_{2}(x) \leq \dots$ for all $x \in K$, and the $f_j$ converge pointwise to a continuous function $f$ on $K$ then in fact the $f_j$ converge uniformly to $f$ on $K$.

Attempt:

Let $g_{j}(x) = f(x) - f_{j}(x)$ for all $j$. Then, since $f_j \rightarrow f$ pointwise, we see $g_j \rightarrow 0$ pointwise.

Now, let $\varepsilon > 0$ . And examine { $x \in K : g_{j}(x) < \varepsilon$ }.

I've been told that the next step should be to show that { $x \in K : g_{j}(x) < \varepsilon$ } is equal to the intersection of $K$ with some open set $U_j$. But I'm not certain why this is true?

Advice? Insight?

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2 Answers 2