# Does connectedness of set pass to connectedness in subspace topologies?

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?

• 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. Nov 30 '19 at 2:07
• @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 $\mathbb{R}^2\setminus\text{compact subset}$ Nov 30 '19 at 2:10
• Note that connectedness is an intrinsic property; it is redundant to ask whether a space is connected "in" another space. Nov 30 '19 at 2:47

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

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

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).

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