Seeking Clarification on the Proof of the Connectedness of the Closure of a Connected Subspace

Question

I was reviewing the proof regarding the connectedness of the closure of a connected subspace, and I have a question about a particular step in the proof. The proof goes as follows:

Link to the answer

My question pertains to the following part of the proof: "But now since cl(A)∩V≠∅, take any point there, v. Then v∈V implies that V∩A≠∅. Then V∩U isn't empty."

I understand that if v is not a boundary point, this step works as described. However, what if v is a boundary point? How does the proof handle this situation, and can we still conclude that V∩U isn't empty?

I'd appreciate any clarification on this matter. Thank you!

score 0 · Answer 1 · answered Oct 28 '23 at 06:15

0

$v \in Cl (A)$ implies that every neighborhood of $v$ intersects $A$. Here $V$ is one such neighborhood, so $V$ interecsts $A$.

answered Oct 28 '23 at 06:15

geetha290krm

36,632

what if V only has the boundary points of A? Then all such points v, need not intersect A – SYT Oct 28 '23 at 06:21
Are you forgetting that $V$ is open? @SYT – geetha290krm Oct 28 '23 at 06:39

Seeking Clarification on the Proof of the Connectedness of the Closure of a Connected Subspace

1 Answers1