MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is a question which seems true to me but I can't rigorously show. Suppose $K$ is a compact subset of $\mathbb{R}^n$ such that $\mathbb{R}^n\setminus K$ is connected, does it follow that for any connected open set $U\subset \mathbb{R}^n$ such that $U\supset K$, $U\setminus K$ is also connected?

share|cite|improve this question

closed as too localized by Denis Serre, Ryan Budney, Will Jagy, Matthew Daws, Joel David Hamkins Nov 7 '11 at 9:45

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

of course not ! – Denis Serre Nov 7 '11 at 7:07
No. (Do you want a specific counterexample? This feels a bit like a homework question...) – Tom Smith Nov 7 '11 at 7:11
I am sorry to have omitted the condition $U\supset K$, this makes the question looks rather stupid. But I must say that this is not a homework question, this is a claim (without proof) in a proof of a paper I am reading. I don't think it is completely trivial, as this is false if we replace Rn with some other connected spaces, such as the torus. Therefore somehow we must use the property of $\mathbb{R}^n$ (e.g. Jordan's theorem), but I have no idea how to. Perhaps it's also interesting to see if we can replace $\mathbb{R}^n$ with other spaces, e.g. spheres. – Kwong Nov 7 '11 at 11:27
up vote 3 down vote accepted

Yes. Let $C$ be the closed complement of $U$, then by excision of $C$ we have $H_1(\mathbb{R}, \mathbb{R} - K) = H_1(U, U - K)$; since $H_1(\mathbb{R})=0$, you also have in fact $H_1(\mathbb{R}, \mathbb{R}-K)= H_1(U, U-K)= 0$ when $\mathbb{R}-K$ is connected.

So $H_0(U-K)$ injects into $H_0(U)$ and $U-K$ must be connected.

share|cite|improve this answer
I am not very familiar with relative homology, but I think we need more than that. I can only see that if $H_0(U, U-K)=0$, the inclusion induces a surjective map from $H_0(U-K)$ to $H_0(U)$, by the long exact sequence of relative homology. (Correct me if I am wrong. ) Also, if your argument is correct, doesn't it imply that if $K$ is the equator of the torus $T^2$, and $U$ is a tubular neighborhood of $K$, then $U-K$ is connected? – Kwong Nov 7 '11 at 10:24
you also have $H_1(\mathbb{R}, \mathbb{R} -K)=0$ using that $H_1(\mathbb{R})=0$ (crucially!), so by excision $H_1(U, U-K)=0$, too. The last part of my answer was incorrect, let me edit (sorry i answered too quickly). – Pierre Nov 7 '11 at 11:43
This answer has the advantage of answering his extension question (that is, on what manifolds this is still true), so bravo – Richard Rast Nov 7 '11 at 14:37
Thanks for the neat answer. So we can replace $\mathbb{R}^n$ by a connected space with vanishing $H_1$, and $K$ to be closed subset. Nice. – Kwong Nov 8 '11 at 5:19

No. $\;\;$ Let $n=2$, $\; K = [0,1]^2 \;$, $\;$ and $\; U = (0,1)\times (-\infty,\scriptsize+\normalsize\infty) \;$.

share|cite|improve this answer
$K$ is not a subset of $U$ (as e.g. $(0,0) \in K \setminus U$). – Henno Brandsma Nov 7 '11 at 18:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.