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I just started reading topology from "topology without tears" by SIDNEY A. MORRIS

Im stuck on the proof of "Let A be a subset of a topological space (X, τ ).A point x ∈ X is a limit point of A if and only if every neighbourhood of x contains a point of A different from x".

While the proof of second statement assuming first is straightforward.I'm confused about the converse.

Definition of neighborhood is specified as: N is said to be neighborhood of x if there exists a open set U such that U ⊆ N and x belongs to U.

So my question here is if we assume converse to be true, then for every neighborhood of x, there exists an open set U such that U ⊆ N and N∩A is not empty set.But to prove x is a limit point of A, we need to show U∩A is not empty. I'm unable to figure how.

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  • $\begingroup$ What is your definiton of limit point? $\endgroup$
    – mfl
    Aug 21 '18 at 19:15
  • $\begingroup$ x is said to be a limit point of A if for every open set containing x,lets say U, U∩A is not empty. $\endgroup$
    – mathgirl
    Aug 21 '18 at 19:15
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Assume that every neigbourhood of $x$ contains a point from $A$ different of $x.$ Since open sets containing $x$ are neigbourhoods of $x$ they contain a point of $A$ different from $x.$ This shows that $x$ is a limit point of $A.$

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  • $\begingroup$ oh yes! thank you! $\endgroup$
    – mathgirl
    Aug 21 '18 at 19:24

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