Real Analysis Proof: If U, V open, then U-union-V is open

[sparkles7922]

I am working on some Real Analysis and I have some troubles. I have to prove:
If U and V are open, then U union V is open.

I can only find stated that The union of two open sets is always open but I do not know how to prove this.
I need some help. Thank you.

 

[royhaas]

Re: Real Analysis Proof

Since open means the "distance" beween two points is small, looks like you need to consider one point in each set, then consider what "union" means.

 

[daon]

Assume \(\displaystyle U\) and \(\displaystyle V\) are open. If either are empty, then its trivially true. WLOG ("without loss of generality"), assume \(\displaystyle x \in U\) (and hence nonempty). Since \(\displaystyle U\) is open, there is an open set \(\displaystyle U_x \subset U\) s.t. \(\displaystyle x \in U_x\). Note that \(\displaystyle U_x\) is also contained in \(\displaystyle U \cup V\).

 

[PAULK]

I am working on some Real Analysis and I have some troubles. I have to prove:
If U and V are open, then U union V is open.

I can only find stated that The union of two open sets is always open but I do not know how to prove this.
I need some help. Thank you.

What style of definition of open sets do you have? [Many authors have their own style.]

Mine is: X is open if every point of X is an interior point. That would mean if x in X, there is a nbhd of x completely within X.

So if A, B both open, let x be a point of A+B. [using + for union]

Then assume x in A
Then x is an interior point of A.
Then there is a nbhd of x completely within A.
Then every point of that nbhd of x is in A, therefore in A+B.

Are we done yet?

HOW ABOUT:

If A, B both open, then A', B' both closed. [A' is complement.]
Then A'B' is closed. [AB or XY is intersection.]
Then (A'B')' is open.
But (A'B')' = (A')'+(B')'

Are we done yet?