Let $X\subset F_1\cup F_2$, where $F_1$ and $F_2$ are closed sets. If $f\colon X\to \mathbb{R}$ is such that its restrictions $f\restriction_{(X\cap F_1)}$, $f\restriction_{(X\cap F_2)}$ are continuous, then $f$ is continuous.

There is a specific part of the proof where I'm not understanding the implications of $F_2$ being closed, and why both sets need to be closed in order to $f$ be continuous.

Proof. Let $a\in X$. To show $f$ is continuous at $a$, suppose a given $\varepsilon>0$. There is three possibilities.

First: $a\in F_1\cap F_2$; as $f\restriction_{X\cap F_1}$ is continuous at $a$, there is a $\delta_1>0$ such that $x\in X\cap F_1,\ \left|x-a\right|<\delta_1 \implies \left|f(x)-f(a)\right|<\varepsilon$. Similarly, there is a $\delta_2>0$ such that $x\in X\cap F_2,\ \left|x-a\right|<\delta_2 \implies \left|f(x)-f(a)\right|<\varepsilon$. We take $\delta=\min{(\delta_1,\delta_2)}$. If $x\in X$ and $\left|x-a\right|<\delta$, then $\left|x-a\right|<\delta_1$ and $\left|x-a\right|<\delta_2$. Therefore, whether $x\in F_1$ or $x\in F_2$, we have $\left|f(x)-f(a)\right|<\varepsilon$.

The second possibility WLOG is $a\in F_1$ and $a\not\in F_2$.

And here is where I'm not getting the proof.

We choose $\delta_1>0$ so $x\in X\cap F_1,\ \left|x-a\right|<\delta_1\implies \left|f(x)-f(a)\right|<\varepsilon$.


As $F_2$ is closed, we can obtain a $\delta_2>0$ such that doesn't exist $x\in F_2$ with $\left|x-a\right|<\delta_2$, remember that $a\not\in F_2$.

How can we say that? Why the fact that $F_2$ is closed matters here?

Let $\delta=\min{(\delta_1,\delta_2)}$; if $x\in X$ and $\left|x-a\right|<\delta$, then $x\in F_1$ and $\left|x-a\right|<\delta_1$ and therefore $\left|f(x)-f(a)\right|<\varepsilon$.

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    $\begingroup$ The complement of a closed set is open. We use that some open $\delta$-ball around $a$ is contained in the complement (i..e, is disjoint from $F_2$) $\endgroup$ – Hagen von Eitzen Nov 18 '20 at 17:01
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    $\begingroup$ If $F_2$ is not closed then it is possible that for any $\delta >0$ there exists $x_{\delta}\in F_2$ such that $|x_{\delta}-a|<\delta.$ But we have no information about $|f(x_{\delta})-f(a)|$ since $x_{\delta}\in F_2, a\in F_1.$ $\endgroup$ – mfl Nov 18 '20 at 17:02

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