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

trap101

Show that the Closure of a Connected set is connected.


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

==> S = U [itex]\cup[/itex] V is a disconnection of S. (bold for closure)

==> (S[itex]\cap[/itex]U) [itex]\cup[/itex] (S[itex]\cap[/itex]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[itex]\cap[/itex]U) or (S[itex]\cap[/itex]V) 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

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

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

Hope I haven't given too much away :)