Is the Closure of a Subset Always in the Union of Neighborhoods?

[tomboi03]

Let A, B, and A_$$\alpha$$ denote subsets of a space X.
neighborhood of $$\bigcup$$A_$$\alpha$$ $$\supset$$ $$\bigcup$$ neighborhood of A_$$\alpha$$; give an example where equality fails.Criticize the following "proof" of the above statement: if {A_$$\alpha$$} is a collection of sets in X and if x $$\in$$ neighborhood of $$\bigcup$$A_$$\alpha$$, then every neighborhood U of x intersects $$\bigcup$$ A_$$\alpha$$. Thus U must intersect some A_$$\alpha$$, so that x must belong to the closure of some A_$$\alpha$$. Therfore, x $$\in$$ $$\bigcup$$ neighborhood of A_$$\alpha$$.

I know that the error is in the statement "x must belong to some A_$$\alpha$$ closure over the whole thing." it is false...

but I don't know how to explain it...

Can you help me out guys?

 

[shaggymoods]

Well, first off, the proof "proves" a reversed inclusion from the one you have listed. Also, I'm not sure what this is looking for: "give an example where equality fails." as you don't have any equality in your problem.

A few things about the proof though:
1.) Just because x is in a neighborhood around a point (or a set), this doesn't mean that x is in the closure of that set (i.e., not every neighborhood around x will necessarily intersect the given set). Thus the first statement is incorrect.

2.) Assuming every neighborhood of x *does* intersect the union, for x to be in the closure of A^a for some a, every neighborhood around x would have to intersect the *same* A^a and it isn't clear that this happens; thus x is not necessarily in the closure of any of the A^a.

Obviously, the first statement alone is enough to invalidate the entire proof.

 

[Symmetryholic]

A few things about the proof though:
1.) Just because x is in a neighborhood around a point (or a set), this doesn't mean that x is in the closure of that set (i.e., not every neighborhood around x will necessarily intersect the given set). Thus the first statement is incorrect.
Obviously, the first statement alone is enough to invalidate the entire proof.

I don't think the following is false.

"If x is in the neighborhood of a set A, then x is in the closure of A (cl(A))".

Proof.
Let $$O \subset A$$ be a neighborhood of x. Then O belongs to int(A). Since int(A) belongs to cl(A), O belongs to cl(A). Thus, x is an element of cl(A).

Please correct me if I am wrong.

 

[shaggymoods]

Ah, but for your proof you have assumed something that we aren't given, namely that x is an element of A. If x is not an element of A, then a neighborhood of x is not a subset of A.

Also, consider this example: let (X, T) be the discrete topology. Consider any proper subset A of X. Then any point in X\A is also in *a* neighborhood of A, namely X. From this we cannot conclude that every open neighborhood around x intersects A; consider the singleton set {x}. This is an open set in the discrete topology and is thus an open neighborhood around x. But we chose x such that it is not in A. Consequently {x} intersected with A is empty. Thus being in just one neighborhood of a set is not sufficient to be in the closure of that set.

More generally,

$$x \in \bar{A} \Leftrightarrow \forall \text{neighborhoods U of x}, U \bigcap A \neq \emptyset$$

Note that we need every neighborhood of x to intersect A, not just one. This was my original point. Sorry if I rambled... :)