From the definition of an(adsbygoogle = window.adsbygoogle || []).push({}); open setas a set containing at least one neighborhood of each of its points, and aclosed setbeing a set containing all its limit points, how can we show that the complement of an open set is a closed set (and vice versa)? Usually this is taken as a definition, but I'm sure it can be shown under these assumptions.

An open set is usually defined as a set S ⊆ U for which, for all x ∈ S, there exists a real number ϵ > 0 such that {y ∈ U : d(x, y) < ϵ} ⊆ S, i.e. there is a neighborhood of x entirely contained in S. Of course d: U^{2}→ ℝ is a metric on U. In some books, I've read the complement of an open set is called a closed set, simply enough. Elsewhere, a closed set is defined as a set containing all its limit points: if p ∈ U is such that the set {t ∈ U : d(p, t) < ϵ} contains infinitely many points of S, then p ∈ S. I want to use this second definition of a closed set and the stated definition of an open set to show that if S is open, (U / S) is closed.

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# Open and closed sets in metric spaces

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