5

Given: $X$ is locally connected, $C \subseteq X$ is closed and connected, $U$ is a component of $X \backslash C$.

I want to show that $X \backslash U$ is connected. Here's what I know:

$C$ is closed $\implies X \backslash C$ is open $\implies U$ is open since $X$ is locally connected $\implies X \backslash U$ is closed.

If I suppose that $X \backslash U$ is disconnected, then there are disjoint closed sets $V,W$ so that $X \backslash U \subseteq V \cup W$.

Since $U$ is merely one component of $X \backslash C$, we have $X \backslash C = U \cup \bigcup U_\alpha$ where each $U_\alpha$ is a component of $X \backslash C$. Then $X \backslash U = C \cup \bigcup U_\alpha$.

I would just like some direction on where to go from here to arrive at a contradiction.

Julien
  • 44,791
Polly Nomial
  • 153
  • 7
  • Typo: since $X$ is locally connected. +1 for your nickname. – Julien Apr 16 '13 at 20:34
  • If $X$ has three connected components $U,C,V$, then $U$ is a component of $X\setminus C$ but $X\setminus U$ has components $C$ and $V$. Maybe $C$ should be non-open? – Stefan Hamcke Apr 16 '13 at 20:35
  • You must have missed some hypotheses like, for example, $U$ is only a component of $X-C$ but not of $X$, or something like that. – Stefan Hamcke Apr 16 '13 at 20:49

2 Answers2

3

This is false. Consider $$X=\{0\}\cup\{1\}\cup\{2\}\qquad C=\{0\}\qquad U=\{1\}\qquad X\setminus U=\{0\}\cup\{2\}$$ with the discrete topology.

In view of Stefan H.'s comment, this is still false if $C$ is not open and $U$ is not a component of $X$. Take $$ X=[0,2]\cup\{3\}\qquad C=[0,1]\qquad U=(1,2]\qquad X\setminus U=[0,1]\cup\{3\} $$ with the topology induced by the standard topology on $\mathbb{R}$.

Julien
  • 44,791
3

It would be right if $X$ were connected. So let us assume $X$ is connected.

So $C$ is closed and connected, and $U$ is a component of $X-C$. Since $X$ is locally connected, $U$ is open being a component of an open set in a locally connected space. $U$ is also closed in $X-C$, so there is a closed set $D$ such that $D-C=U$. Without loss of generality we can take $D=U\cup C$. If $X-U=A\sqcup B$ where both $A$ and $B$ are closed, then $C\subseteq A$ since $C$ is connected. Therefore $A\cup D=A\cup U=X-B$ is closed. Thus $B$ is closed and open and must be empty. This implies that $X-U$ is connected.

Edit: I found this post where they show that one can drop the condition that $C$ is closed and $X$ is locally connected. You merely need that $X$ and $C$ are connected and $U$ is a component of $X-C$.

Community
  • 1
Stefan Hamcke
  • 27,733