examples-counterexamplesgeneral-topologypath-connectedreal-analysis
Let $A, B \subseteq \mathbb{R}^{n}$ be two path-connected sets. Is it true that $A \cap B$ is also path-connected?
(A subset of $\mathbb{R}^{n}$ is path-connected if every pair of points in $A$ can be joined by a path in $A$).
Intuitively, I think that the answer is no. I don't know how to disprove this though. Can someone please help me with this exercise? I am having trouble finding a counterexample.
Others have mentioned the topologist's sine curve, that's the canonical example. But I like this one, though it is the same idea:
This is the image of the parametric curve $\gamma(t)=\langle (1+1/t)\cos t,(1+1/t)\sin t\rangle,$ along with the unit circle. There is no path joining a point on the curve with a point on the circle. Yet, the space is the closure of the connected set $\gamma((1,\infty))$, so it is connected.
It is unnecessary. Your informal proof is correct. See Rotman's Theorem 1.15.
Pick $c \in A \cap B$. Then each $a \in A$ satisfies $a \sim c$ and each $b \in B$ satisfies $b \sim c$, thus each $x \in X$ satisfies $x \sim c$. Since $\sim $ is transitive, any two points $x, y \in X$ satisfy $x \sim y$.
Best Answer
Take the two (closed) halves of a circle. Each of them is path-connected, but their intersection isn't even connected.