I'm trying to prove that the closure of the union is the union of the closures.
Where for topological space ,
denotes the closure of .
I'm trying to prove
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This is what I have so far:
First, proving :
We know are closed,
so is closed.
Furthermore, ,
so .
Since is a closed set containing , we have .
Now to prove :
... This is where I'm stuck ...
It's not super important for this case, but it's an intertesting alternative way to define a topological space. Namely, call a closure space to be an ordered pair where is a non-empty set and (here is the power set) such that for every
Clearly every topological space induces a closure operator, the normal one. What's more interesting is that if one defines on a closure space the set then is a topology on . What's really interesting about this whole things is that these two notions of topological spaces (the normal definition and the definition which makes closure space and topological space synonomous) are really equivalent. What I mean by this is if you start with a topological space and creat from this the usual closure operator (the intersection of all the closed sets containing the set in question) and then create from this the topology induced by the closure operator (as I previously mentioned) then one can see that . Conversely, if one starts with a closure space and create the topology induced by this closure operator and then defines the usual closure operator on the topological space (the intersection of all closed supersets) then . Kind of interesting, huh? What's weird is that while the formulation of topological spaces seen today is probably the most natural the concept of a closure space predates it, as can be seen in the works of Kuratowski.
Sorry for the rant, just thought I'd share.