If $X$ is a topological space and $A\subset X$ is a connected subset of $X$ and $Β$ is an open subset of $X$, is it true that $A\cap B$ is connected in the subspace $B$? I tried to prove this using the classic definition of connectedness but I can't get around it.

I actually need this for the special case where $X=\mathbb{R}^2$ and $A$ is a proper, open and connected subset of $\mathbb{R}^2$ and $B=\mathbb{R}^2\setminus K$, where $K$ is a compact subset of $A$, so maybe it holds in this case, maybe employing the equivalent path-connectedness of domains in $\mathbb{R}^2$.

Any ideas?

$\textbf{Edit:}$ As pointed out, the first part gets a negative answer. What about the 2nd part though? What if $B=\mathbb{R}^2\setminus K$, where $K\subset A$ is a compact set and $A$ is open?

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    $\begingroup$ When you say "I actually need this", it might be good to be more specific about the bigger thing you're trying to prove; as you've stated it, the answer is "no" because you could take $A$ to be U-shaped and $B$ to cut off the tops of the U without capturing the connecting segment - but maybe there's a related statement that would help your bigger goal. $\endgroup$ Nov 30 '19 at 2:07
  • $\begingroup$ @MiloBrandt Thanks; the bigger goal is pretty much this: I am simply trying to prove that a domain in $\mathbb{R}^2$ minus a compact subset is connected in the subspace $\mathbb{R}^2\setminus\text{compact subset}$ $\endgroup$ Nov 30 '19 at 2:10
  • $\begingroup$ Note that connectedness is an intrinsic property; it is redundant to ask whether a space is connected "in" another space. $\endgroup$ Nov 30 '19 at 2:47

The answer is no. Consider the following subsets of $\mathbb R^2$

$$A=\{(x,y)\in\mathbb R^2\mid 2<x^2+y^2<4,\;y>-1\}$$$$B=\{(x,y)\in\mathbb R^2\mid 2<x^2+y^2<4,\;y<1\}$$

Note that both sets are connected and open, but their intersection isn't (the sets form horseshoe shapes and intersect at the ends of their respective horseshoe).

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    $\begingroup$ @Downvoter any problem with the answer? $\endgroup$ Nov 30 '19 at 2:13
  • $\begingroup$ +1 since you are answering the first part. I will edit my question for the most specific part, therefore I'm not accepting yet $\endgroup$ Nov 30 '19 at 2:22
  • $\begingroup$ I guess the downvote is because you are saying something that has already been mentioned as a comment $\endgroup$ Nov 30 '19 at 2:27
  • $\begingroup$ Please see the update. $\endgroup$ Nov 30 '19 at 2:32
  • $\begingroup$ Your rectangle isn't a compact subset of $A$? $\endgroup$ Nov 30 '19 at 2:39

You're asking if an open connected proper subset of $\mathbb R^2$ remains connected after deleting a compact subset.

Hint: the continuous image of a circle is compact.


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