In the case of $X=\mathbb{R}^1$ are intersections of connected sets always connected?

Question

Intuitively yes seems to make the most sense: after all $E\subseteq \mathbb{R}$ is connected if and only if $x\in E, y\in E,$ and some $x<z<y,$ then $z\in E$. But I am struggling writing out a formal proof.

Answer 1

Your intuition is right. Here's an outline for how to prove it:

• First, you want to prove the fact you mention: that if $A\subseteq\mathbb{R}$ is connected, then it is convex - if $a, b\in A$ and $a<z<b$, then $z\in A$. HINT: suppose this fails. What can you say about $\{(-\infty, z), (z, \infty)\}$ and the connectedness of $A$?

• Now, you need to prove the converse of that fact: that if $A$ is convex, then $A$ is connected. HINT: show that a convex $A$ is of the form $(c, d), [c, d], (c, d]$, or $[c, d)$ for $c,d\in\mathbb{R}\cup\{-\infty,\infty\}$. (I'm being informal here - we need to ignore intervals with "$\infty$" or "$-\infty$" on the closed side, e.g. "$[-\infty, 8)$".)

• Finally, suppose $A$ and $B$ are connected. You want to argue that $C:=A\cap B$ is connected. By the second fact above, it's enough to show that $C$ is convex. So suppose $c, d\in C$, and $c<w<d$ - you want to show $w\in C$. To do this, you need to show $w\in A$ and $w\in B$. HINT: use the first fact above - that connectedness implies convexity . . .