[itex]Cl(S \cup T)= Cl(S) \cup Cl(T)[/itex]
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 [itex]Cl(S \cup T) \subset Cl(S) \cup Cl(T)[/itex]. 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 [itex]S \cup T[/itex] is a subset of the limit points of S union the limit points of T. So, if x is a limit point of [itex]S \cup T[/itex] then every open set U containing x intersects [itex]S \cup T[/itex] 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 [itex]S \cup T[/itex] 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?