# Open Sets: Exploring Disjoint Union Properties

• ehrenfest
In summary: Therefore, every set in the union is open.Consider any non-open set X and it's complement \bar{X}.Their union is clearly open since it's the entire space, they're disjoint by construction, and X is non-open by hypothesis. Therefore, every set in the union is open.
ehrenfest

## Homework Statement

If you have a collection of disjoint open sets in a general topological space whose union is open, is it true that each of them individually must be open? Why?

EDIT: this makes absolutely no sense. here is what I meant to ask:
EDIT:If you have a collection of disjoint sets in a general topological space whose union is open, is it true that
EDIT:each of them individually must be open? Why?

## The Attempt at a Solution

Last edited:
As stated the answer is trivially yes because they are disjoint and OPEN. Did you really mean to ask if you had a collection of disjoint set's whose union in a topological space is open is each necessarily open. In that case the answer is false consider the trivial topology on {a,b,c}. Then the collection {{a}, {b}, {c}} is a collection of disjoint sets whose union is open but none of the ndividual sets is open.

d_leet said:
As stated the answer is trivially yes because they are disjoint and OPEN. Did you really mean to ask if you had a collection of disjoint set's whose union in a topological space is open is each necessarily open. In that case the answer is false consider the trivial topology on {a,b,c}. Then the collection {{a}, {b}, {c}} is a collection of disjoint sets whose union is open but none of the ndividual sets is open.

What if we are in a locally path connected and path connected space?

What are the original sets that comprise the union? They can't be open individually if you are being asked that very question.

Mathdope said:
What are the original sets that comprise the union? They can't be open individually if you are being asked that very question.

Sorry. Reread the EDIT. I am foolish.

ehrenfest said:
Sorry. Reread the EDIT. I am foolish.

Consider any non-open set $X$ and it's complement $\bar{X}$.
Their union is clearly open since it's the entire space, they're disjoint by construction, and $X$ is non-open by hypothesis.

## 1. What is the definition of an open set?

An open set is a subset of a topological space in which every point has a neighborhood that is also contained within the set.

## 2. How are open sets related to disjoint union properties?

Open sets are related to disjoint union properties because when two sets are disjoint, their union is considered an open set. This is because each point in the union has a neighborhood contained within the union.

## 3. Can open sets have empty intersections?

Yes, open sets can have empty intersections. If two open sets are disjoint, then their intersection will be empty. This is because no points are shared between the two sets.

## 4. What are some examples of open sets?

Examples of open sets include an open interval on the real number line, a disk in the plane, or a sphere in three-dimensional space.

## 5. How do open sets contribute to the concept of continuity?

In topology, a function is continuous if the inverse image of every open set is also open. This means that open sets play a crucial role in determining the continuity of a function.

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