Prove that if $U$ is open and $\text{int}(S)\neq\emptyset$ then $\text{int}{\big(U\cap S\big)}\neq\emptyset$ when $S$ is path connected.

Question

Statement

If $U$ is open and $\text{int}(S)\neq\emptyset$ then $\text{int}(U\cap S)\neq\emptyset$ too when $U\cap S\neq\emptyset$ and when $S$ is path connected.

We know that $\text{int}(U\cap S)=\text{int}(U)\cap\text{int}(S)=U\cap\text{int}(S)$ so that if $\text{int}(U\cap S)=\emptyset$ and $U\cap S\neq\emptyset$ then $(U\cap S)\subseteq\text{Bd}(S)$ but unfortunately I don't see in this any contradiction. So I don't be able to prove the statement. Could be that it is generally false? Anyway it seems to me that in the case where $U$ and $S$ are path connencted subset of $\Bbb R^n$ then the statement holds so I think that although it is gerenally false it could be true if we consider some particular case. If it is more difficul understand when the statement is generally true I ask to prove it when $U$ and $S$ are subset of $\Bbb R^n$. Finally I point out that the statement is generally false when $U$ or $S$ are not path connected: e.g. if you take $U=(0,1)$ and $S=\{\frac 1 2\}\cup(2,3)$ or $U=\big[(0,1)\cup(2,3)\big]$ and $S=[1,2]$ then the statement is clearly false!

So could someone help me, please?

Answer 1

Connectedness problems often degenerate in $\Bbb R$, it's the plane where stuff gets interesting.

$U=\{x\mid \|x\|< 1\}$ which is open, path-connected.

$S= \{x \mid \|x\|\ge 1\} \cup \{(x,y):xy=0\}$, the exterior plus axes, clearly $U \cap S \neq \emptyset$ and the interior of $S$ is $\{x\mid \|x\| > 1\} \neq \emptyset$. $S$ is path-connected, and $S \cap U$ has empty interior (but is path-connected).

Answer 2

Ok here's a counter example. Below, set A is the interior of the blue ball. Set B is the red line and the interior of the red ball. Clearly, the interior of B is $R^2$ is the interior of the red ball. And the interior of the blue ball clearly doesn't intersect with that. Making this rigorous wouldn't be tricky. Just set two balls a certain distance apart, and pick a line to go from one into the other.

Now, your examples looked specifically at $R$. But if a set is path connected in R, then visually you can see that there will either be points in the interior either directly to the right of it, or directly to the left of it. That's becauuse, if we take another point y, and we take the continuous path mapping from x to y, by the intermediate value theorem all points inbetween them will be mapped to by our curve. That's why you seem to have trouble finding counter examples in R