6
$\begingroup$

Let X be a topological space.

Given a decreasing sequence of closed and connected subsets $J_1\supset J_2 \supset J_3\supset\cdots$, consider $J:=\bigcap_{i=1}^\infty J_i$.

Is it always true that $J$ is connected when $X=\mathbb{R}^2$ ?

and when $X$ is compact but not Hausdorff ? (if $X$ is compact and Hausdorff $J$ is connected)

$\endgroup$
2
  • $\begingroup$ Without further conditions on the J_i it's not even true in [0,1] x [0,1]. $\endgroup$ Commented Jan 29, 2017 at 14:33
  • $\begingroup$ @DavidHartley sorry $J_i$ are closed sets $\endgroup$ Commented Jan 29, 2017 at 14:42

2 Answers 2

10
$\begingroup$

Counterexample:

Consider $J_n=(\mathbb R^\times\times \mathbb R)\cup (\{0\}\times (n,\infty))$. Each $J_n$ is connected, but the intersection is $\mathbb R^\times\times\mathbb R$ which is disconnected. Basically, each $J_n$ eats away a bit more of the connection, and eventually every part of the connection gets consumed.

Edit: An edit to the question now added the condition of closed sets, which is violated by the example above. But the basic idea can still be used.

Now consider $J_n = (((-\infty,-1]\cup[1,\infty))\times\mathbb R)\cup ([-1,1]\times [n,\infty))$. It is not hard to check that those sets are closed and connected, but the intersection is $((-\infty,-1]\cup[1,\infty))\times\mathbb R$ which is disconnected.

$\endgroup$
3
  • 1
    $\begingroup$ what is $\mathbb{R}^\times$? $\endgroup$ Commented Jan 29, 2017 at 14:40
  • $\begingroup$ @TrueTopologist: The set of invertible real numbers, that is, $\mathbb R\setminus\{0\}$. $\endgroup$
    – celtschk
    Commented Jan 29, 2017 at 14:43
  • $\begingroup$ wow! the idea of the consuming of the connection is amazing. $\endgroup$ Commented Jan 29, 2017 at 14:59
5
$\begingroup$

Here is a counterexample when you remove the Hausdorff assumption. Let $X=[-1,1]\cup\{0'\}$ be topologized like the line with two origins, so neighborhoods of $0'$ are neighborhoods of $0$ but with $0$ replaced by $0'$. Note that $X$ is compact but is not Hausdorff, since $0$ and $0'$ do not have disjoint neighborhoods. Let $J_n=[-1/n,1/n]\cup\{0'\}$. Then each $J_n$ is closed and connected, but their intersection is just $\{0,0'\}$, which has the discrete topology and thus is disconnected.

$\endgroup$
1
  • $\begingroup$ great example! It is a common example of a non Hausdorff space, I should have thought of it. $\endgroup$ Commented Jan 29, 2017 at 18:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .