The first one is straightforward. If , thenxis in an open ballScontained inAor one contained inB; in either case, and so .

For the second part, proving is equally straightforward.xis in an open ballScontained in and and .

For the reverse inclusion, let .

such that . Similarly . such that .

Now take .

Then proving that .

For the example, take the closed intervals and in . Then but and .