# $f$ is continuous if its restriction to closed set is continuous

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

Ok.

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 $$\left|f(x)-f(a)\right|<\varepsilon$$.

• 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 $F_2$) – Hagen von Eitzen Nov 18 '20 at 17:01
• 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 $x_{\delta}\in F_2, a\in F_1.$ – mfl Nov 18 '20 at 17:02