Showing that the Closure of a Connected set

[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\capU) \cup (S\capV) 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\capU) or (S\capV) being empty, but I still wasn't able to follow that either.

 

[HallsofIvy]

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

 

[trap101]

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

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.

 

[christoff]

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 :)