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

Let $A$ be a closed set and let $B$ be a connected set such that $A \subset B$.

Does there always exist a closed connected subset $C$ of $B$ that contains $A$?

What if $B$ is path connected, is there always a path-connected $C$? A connected $C$?

share|cite|improve this question
I would mention in relation to this question, and the OP's answer to it. – David Roberts Nov 17 '10 at 1:33
Closed in what topology? I tend to view B as a closed subset of B ... – Yemon Choi Nov 17 '10 at 2:08
Yemon, in light of the question linked to by David, the OP is likely thinking of having A and B in some other background space, rather than the subspace topology on $B$. – Joel David Hamkins Nov 17 '10 at 2:26
Joel: that was my reading too, but I was hoping that the OP might come back and clarify (this tending to help in answering one's own questions, or getting other people to answer them). – Yemon Choi Nov 17 '10 at 4:13
up vote 4 down vote accepted

The answer to your first question is no. Let $A$ be a sequence on the $x$ axis converging to the origin in the real plane $\mathbb{R}^2$, which is the background topological space for these examples. Let $B$ be a cone over the sequence, connecting each point in the sequence to the point $(0,1)$, together with the origin, but not connecting the origin to $(0,1)$. Then $B$ is connected, but not closed, and there is no connected closed set $C$ contained in $B$ and containing $A$, since any connected set $C$ contained in $B$ and containing $A$ would have to contain all the lines and thus be all of $B$, which is not closed in the plane.

The example can be extended to an answer for your path-connected question as well. We simply add a roundabout path from the origin to $(0,1)$. That is, $A$ is the sequence $(\frac{1}{n},0)$ plus $(0,0)$ in the real plane. The set $B$ contains lines from every $(\frac{1}{n},0)$ to $(0,1)$, and $B$ also includes a circular curve connecting $(0,0)$ to $(1,0)$ through the left half-plane. So $A$ is closed, $B$ is path-connected, but any closed path-connected set $C$ with $A\subset C\subset B$ will have to contain all those lines and hence $C=B$, but this is not closed in the plane.

Similarly, for your last question, any closed connected $C$ with $A\subset C\subset B$ for the same example will have to contain all the lines from the points on the sequence to $(0,1)$, and hence not be closed (since it will miss the line joining the origin to $(0,1)$, which is not in $B$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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