• 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.

Anuj holds professional certifications in Google Cloud, AWS as well as certifications in Docker and App Performance Tools such as New Relic. He specializes in Cloud Security, Data Encryption and Container Technologies.

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