So, I'm going through a proposition, which states that if (X, d) is a metric space, then any set {x}, where x e X, is a closed subset of X.(adsbygoogle = window.adsbygoogle || []).push({});

First of all, could we do this proof to assume the contrary? Since then obviously for the point x from {x} there doesn't exist any real number r > 0 such that the open ball K(x, r) is contained in {x}?

The proof in the notes I'm going through relies on the fact that we have to prove that the complement of {x}, i.e. X\{x} is open. The proof is very simple too, although I'm not quite sure about one thing. Let x' be an element of X\{x}. Then d(x', x) = r > 0, so the open ball K(x', r/2) is contained in X\{x}, and if we take the union for all x' e X\{x} of all such open balls, we get X\{x}, and hence X\{x} is open.

Now, why is it r/2 ? Wouldn't open balls of type K(x', r) be contained in X\{x} too, since K(x', r) = {x'' in X : d(x''-x') < r}, and this set can't contain x, since d(x', x) = r? Perhaps I'm missing something trivially obvious here?

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# One point set closed proof

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