Proving the Locally Finite Family Problem

[radou]

Homework Statement

Let X be a topological space, and A a locally finite family of sets in X (i.e. such a family of sets that every point in X has a neighborhood which intersects a finite number of sets in A). One needs to show that Cl(U A) = U (Cl(A)) (i.e. the closure of the union of sets in A equals the union of the closures of sets in A).

The Attempt at a Solution

Inclusion $$\subseteq$$. Let x be in Cl(U A). Then every neighborhood of x intersects U A. Since A is locally finite, there exists some neighborhood N of x which intersects A in a finite number of sets.

This is where I'm stuck, right at the beginning. Somehow, I need to show that this very x is contained in some set of the family A, since then it's contained in U (Cl(A)), too. Any push in the right direction is highly appreciated.

 

[Eynstone]

Let V be a neighbourhood of x intersecting A1,A2,...An. Assume the contrary, that x doesn't belong to the closure of any Ai.Then, we can choose a neighbourhood Vi of x which doesn't intersect Ai for each i.
The intersection W of V & Vi (1<=i<=n) is open & doesn't intersect any of A's & hence is not in U A. Therefore, x is not in Cl(U A), a contradiction.

 

[radou]

Let V be a neighbourhood of x intersecting A1,A2,...An. Assume the contrary, that x doesn't belong to the closure of any Ai.Then, we can choose a neighbourhood Vi of x which doesn't intersect Ai for each i.
The intersection W of V & Vi (1<=i<=n) is open & doesn't intersect any of A's & hence is not in U A. Therefore, x is not in Cl(U A), a contradiction.

Eynstone, thanks a lot. So, for the other inclusion, it's similar: let x be in U (Cl(A)), then every neighborhood of x intersects at least one set from A, and if we assume the contrary, i.e. that x is not in Cl(U A), then there exists some neighborhood of x which doesn't intersect any set from the family A, which is clearly a contradiction. (Although I didn't use the finite intersection property here, but perhaps I don't need to, if this is okay.)