# Union of intersections

• tomboi03
In summary, if x is in the neighborhood of a set A, then x is in the closure of A (cl(A)) provided that every neighborhood around x intersects A. However, this doesn't necessarily hold true, and x may not be in the closure of any of the A^a.

#### 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?

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.

shaggymoods said:
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.

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... :)

## What is the "Union of Intersections"?

The union of intersections is a mathematical concept that refers to the set of elements that are shared by two or more sets. It is written as A ∪ B, where A and B are sets.

## How is the union of intersections different from the intersection of unions?

The union of intersections and the intersection of unions are two different operations in set theory. The union of intersections focuses on finding the shared elements between sets, while the intersection of unions focuses on finding the unique elements in each set.

## What is an example of the union of intersections?

An example of the union of intersections would be if set A = {1, 2, 3} and set B = {2, 3, 4}, then the union of intersections, A ∪ B, would be {2, 3} because those are the elements shared by both sets.

## Can the union of intersections be performed on more than two sets?

Yes, the union of intersections can be performed on any number of sets. For example, if there are three sets A, B, and C, the union of intersections would be written as A ∪ B ∪ C and would include all elements that are shared by all three sets.

## What is the relationship between the union of intersections and the complement of the union?

The union of intersections and the complement of the union are related in that they are two different ways of looking at the same set of elements. The union of intersections focuses on the shared elements between sets, while the complement of the union focuses on the elements that are not shared by the sets.