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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.

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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