# The Union of Two Open Sets is Open

#### G-X

##### New member
Let $$\displaystyle x ∈ A1 ∪ A2$$ then $$\displaystyle x ∈ A1$$ or $$\displaystyle x ∈ A2$$

If $$\displaystyle x ∈ A1$$, as A1 is open, there exists an r > 0 such that $$\displaystyle B(x,r) ⊂ A1⊂ A1 ∪ A2$$ and thus B(x,r) is an open set.

Therefore $$\displaystyle A1 ∪ A2$$ is an open set.

How does this prove that $$\displaystyle A1 ∪ A2$$ is an open set. It just proved that $$\displaystyle A1 ∪ A2$$ contains an open set; not that the entire set will be open? This is very similar to the statement: An open subset of R is a subset E of R such that for every x in E there exists ϵ > 0 such that Bϵ(x) is contained in E.

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#### castor28

##### Well-known member
MHB Math Scholar
The point is that the argument is valid for every $x\in A_1\cup A_2$.

If $C = A_1\cup A_2$, we have proved that, for every $x\in C$, there is an open ball $B(x,r)\subset C$ (where $r>0$ depends on $x$). That is precisely the definition of an open set.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
If $$\displaystyle x ∈ A1$$, as A1 is open, there exists an r > 0 such that $$\displaystyle B(x,r) ⊂ A1⊂ A1 ∪ A2$$ and thus B(x,r) is an open set.

Therefore $$\displaystyle A1 ∪ A2$$ is an open set.
Hi G-X, welcome to MHB!

As castor28 's pointed out, it's about the definition of an open set, which he effectively quoted.

Additionally that proof is not entirely correct and it is incomplete.
It should be for instance:

If $$\displaystyle x ∈ A1$$, as $A1$ is open, there exists an $r > 0$ such that $$\displaystyle B(x,r) ⊂ A1$$ (from the definition of an open set), which implies that $$\displaystyle B(x,r)⊂ A1 ∪ A2$$.
If $$\displaystyle x ∈ A2$$, as $A2$ is open, there exists an $r > 0$ such that $$\displaystyle B(x,r) ⊂ A2⊂ A1 ∪ A2$$.
Therefore for all $$\displaystyle x ∈ A1∪ A2$$, there exists an $r > 0$ such that $$\displaystyle B(x,r) ⊂ A1 ∪ A2$$.

Thus $$\displaystyle A1 ∪ A2$$ is an open set.

#### G-X

##### New member
I see, I think I had the misunderstanding that something from A2 might close A1.

But I don't think that is an issue you technically need to wrap your head around.

Because the definition states: We define a set U to be open if for each point x in U there exists an open ball B centered at x contained in U.

So, essentially looping over A1, A2 - making the reference that open balls exist at each of these points then all points in A1 ∪ A2 have open balls contained in the union thus by the definition it must be open.

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