The Union of Two Open Sets is Open

[Guest]

Let [tex]x ∈ A1 ∪ A2[/tex] then [tex]x ∈ A1[/tex] or [tex]x ∈ A2[/tex]

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

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

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

[HallsofIvy]

It doesn't! It stops in mid-proof!

The definition of open sets, for a metric topology, is that "set X is open if and only if, given any [tex]x\in X[/tex] there exist [tex]\delta> 0[/tex] such that [tex]\{ y| d(x,y)< \delta\}[/tex] is a subset of X" (x is an "interior point" of X).

What was proved here is that A is an open set. But we already knew that!

Instead:

Let x be any point in [tex]A\cup B[/tex]. Then by the definition of "union" either [tex]x\in A[/tex] or [tex]x\in B[/tex].

case 1: [tex]x\in A[/tex]. Since A is open then, there exist [tex]\delta> 0[/tex] so that the [tex]\delta[/tex] neightborhood, [tex]\{ y| d(x, y)< \delta\}[/tex], is a subset of A. But A itself is a subset of [tex]A\cup B[/tex] so this neighborhood is a subset of [tex]A\cup\B[/tex].

case 2: (Really the same with B instead of A) [tex]x\in B[/tex]. Since B is open then there exist [tex]\delta> 0[/tex] so that the [tex]\delta[/tex] neightborhood, [tex]\{ y| d(x, y)< \delta\}[/tex], is a subset of B. But B itself is a subset of [tex]A\cup B[/tex] so this neighborhood is a subset of [tex]A\cup\B[/tex].

In either case x is an "interior point" of [tex]A\cup B[/tex]. Since x could be any point in [tex]A\cup B[/tex], every point in [tex]A\cup B[/tex] is an interior point. Therefore [tex]A\cup B[/tex] is an open set.