Is the Proof for Cl(S ∪ T) ⊆ Cl(S) ∪ Cl(T) Correct in Topology?

[arty21]

Homework Statement

$Cl(S \cup T)= Cl(S) \cup Cl(T)$

Homework Equations

I'm using the fact that the closure of a set is equal to itself union its limit points.

The Attempt at a Solution

I am just having trouble with showing $Cl(S \cup T) \subset Cl(S) \cup Cl(T)$. I can prove this one way, but I figured out another way that I prefer and want to know if there is any flaw in it:

Basically it boils down to showing the limit points of $S \cup T$ is a subset of the limit points of S union the limit points of T. So, if x is a limit point of $S \cup T$ then every open set U containing x intersects $S \cup T$ in a point other than x. Hence, x is a limit point of S or T.

I keep going back and forth. Sometimes I feel like this is fine and other times I feel like I'm making an error because if x is a limit point of S this means that every neighborhood of x intersects x in a point other than x. But in the proof I just gave, we know every neighborhood of x intersects $S \cup T$ in a point other than x, so we don't know if it necessarily always in S or T. I don't know why I'm having so much trouble with this! I think I am just over thinking it or something. So is this proof fine then?

 

[Dick]

Well, I think you've figured out exactly what the problem is. And it is a problem. It's good to have a feeling for when a proof is incomplete. You haven't proved it this way.

 

[kru_]

I'm not certain this can be proved directly like this, but the contrapositive is pretty easy to show. If x is not in Cl(S)∪Cl(T), then show it can't be in Cl(S∪T).