# Confused about a step in a proof; For a closed set, $F$, and a point $x\in F$, there's an open interval around $x$ and contained in $F$?

I've underlined the step in the proof that isn't clear to me.

I know what I said in the title is wrong; but I'm not really sure what motivates the step that I underlined.

This is taken from Royden, 4th edition, page 66.

Any thoughts would be greatly appreciated.

Edit: To be more clear, I'm confused about how we get an open interval fully contained in $$F_i$$ which also contains $$x$$. Why must $$F_i$$ fully contain an open interval that also contains $$x$$?

• Your question does not match your title. Commented Jun 21, 2020 at 15:15
• The complement of a closed set is open. Commented Jun 21, 2020 at 15:16
• Well, the intersection of an open interval and $F$ is open relative to $F$. Commented Jun 21, 2020 at 15:17
• I've added an edit to my post to clarify what I'm confused about. Sorry if it was unclear. Commented Jun 21, 2020 at 15:20
• Royden never asserts that $F_i$ contains an open interval. Commented Jun 21, 2020 at 17:10

To say that $$g$$ is continuous on $$F$$ means that for any $$x \in F$$ and for any $$\epsilon>0$$ there is some $$\delta >0$$ such that if $$|y-x|<\delta$$ and $$y \in F$$ then $$|g(y)-g(x)| < \epsilon$$.

In the above case, we have $$x \in F_i$$, and since the $$F_k$$ are disjoint and closed, the set $$\cup_{k\ne i} F_k$$ is closed and disjoint from $$F_i$$. Hence $$x \notin \cup_{k\ne i} F_k$$ and so there is some interval $$(a,b)$$ containing $$x$$ such that $$(a,b) \cap (\cup_{k\ne i} F_k) = \emptyset$$.

Hence if $$y \in F \cap (a,b)$$ we must have $$y \in F_i$$ and hence $$g(y) = g(x)$$.

So, given $$x \in F$$ and $$\epsilon>0$$, we can find some $$i$$ and an interval $$(a,b)$$ as above such that $$x \in (a,b)$$ and $$(a,b) \cap F \subset F_i$$ (in particular we have $$g(x) = a_i$$).

Now choose $$\delta>0$$ such that $$B(x,\delta) \subset (a,b)$$, then if $$|y-x| < \delta$$ and $$y \in F$$ we have $$y \in (a,b) \cap F_i$$ and so $$g(y) = a_i$$ (and so $$|g(y)-g(x)| < \epsilon$$, of course).

• Well, I did show continuity above. Let me elaborate a little. Commented Jun 21, 2020 at 15:23
• Okay. I think I see. So the argument rests on the fact that $\cup_{k\neq i}F_k$ are closed, and thus the complement is open. Correct? Commented Jun 21, 2020 at 15:24
• @Bears: That is correct. Commented Jun 21, 2020 at 15:25