# Showing that the Closure of a Connected set

1. Jun 10, 2012

### trap101

Showing that the Closure of a Connected set......

Show that the Closure of a Connected set is connected.

Attempt: Assume that the closure of a conncted set S is disconnected.

==> S = U $\cup$ V is a disconnection of S. (bold for closure)

==> (S$\cap$U) $\cup$ (S$\cap$V) is a disconnection of S.

This is where I'm stuck, I know some how I'm suppose to get a contradiction in that the closure of this set is actually connected. But I can't see how to form it.

In the solutions they make use of (S$\cap$U) or (S$\cap$V) being empty, but I still wasn't able to follow that either.

2. Jun 10, 2012

### HallsofIvy

Staff Emeritus
Re: Showing that the Closure of a Connected set......

A little more detail please- what do you mean by a "disconnection of S"?

3. Jun 10, 2012

### trap101

Re: Showing that the Closure of a Connected set......

oh, by the definition of connectedness or disconnectedness I guess since we don't define connectedness per say. So I mean the union of two non-empty subsets in which neither intersects the closure of the other one.

4. Jun 10, 2012

### christoff

Re: Showing that the Closure of a Connected set......

Ok, so you have that $\overline{S}=U\cup V$, which is good. But since $S\subset\overline{S}$ and $S$ is connected, you must have that either $S\subset U$ or $S\subset V$ since otherwise $S$ would be disconnected. Let's assume without loss of generality that $S\subset U$.

Then in fact, $S\cap V=∅$ since $U$ and $V$ are disjoint (where your textbook's hint comes in). Moreover, we must have $S\subset V^C$. Since the closure operator respects subsets and $V^C$ is closed (as $V$ is open), we can say some pretty interesting things about $\overline{S}$... Remember that by assumption, $\overline{S}=U\cup V$.

Hope I haven't given too much away :)