Open sets and limit points

  • A set A is open iff for every point a\in A, there exists a \delta such that the neighborhood \left(a-\delta,a+\delta\right) surrounding a is completely contained within (is a subset of) A.

  • A point x is a limit point of a set A iff given any \varepsilon, all neighborhoods \left(x-\varepsilon,x+\varepsilon\right) intersect the set A in some point other than x. Note that x need not be a point in A.

Let a be a point in A. Since A is open, there is a number \delta > 0 such that (a - \delta, a + \delta) \subset A.

We now show a is a limit point of A. To that end, let \epsilon > 0 be given. We may assume, without loss of generality, that \epsilon < \delta. Now, (a - \epsilon, a + \epsilon) \subset (a - \delta, a + \delta) \subset A, so (a - \epsilon, a + \epsilon) \cap A = (a - \epsilon, a + \epsilon). Since (a - \epsilon, a + \epsilon) intersects A in a point other than a, we conclude a is a limit point.

If \delta < \epsilon, then (a - \delta, a + \delta) \subset (a - \epsilon, a + \epsilon), so (a - \delta, a + \delta) \subset (a - \epsilon, a + \epsilon) \cap A). Openness already gives you an open \delta-interval around a, so you only need to worry about the case where \epsilon is smaller than that.

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