The Union of Two Open Sets is Open

[G-X]

Let [Math]x ∈ A1 ∪ A2[/Math] then [Math]x ∈ A1[/Math] or [Math]x ∈ A2[/Math]
If [Math]x ∈ A1[/Math], as A1 is open, there exists an r > 0 such that [Math]B(x,r) ⊂ A1⊂ A1 ∪ A2[/Math] and thus B(x,r) is an open set.

Therefore [Math]A1 ∪ A2[/Math] is an open set.

How does this prove that [Math]A1 ∪ A2[/Math] is an open set. It just proved that [Math]A1 ∪ A2[/Math] contains an open set; not that the entire set will be open?

 

[G-X]

Addition: 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.

 

[G-X]

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.

 

[Dr.Peterson]

You haven't quite stated the corrected proof, but it looks like you may have the right idea. Your original didn't consider the case where x is not in A1, so you just have to include that case.

You also didn't state what you were proving: "Let A1 and A2 be open sets, ..."

 

[G-X]

Thanks for the help. I was able to finish the proof last night. I did not post my version of the proof, unless I need proof verification, because one was already given above - even though it was slightly incomplete. And several other hints were given below by others. I won't post the full proof to allow others to explore the answer for themselves.