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I already looked at about 3 problems on here (listed below), but they all seemed to use metric spaces. I was trying to understand this proof (from armstrong's basic topology), which just works in a general topological space. Basically, Armstrong shows for A contained in a topological space X, cl(A) is closed because X - cl(A) is open. He first shows if we take an element x in X - cl(A) , we can find an open neighborhood U around it that contains no elements of A (I think this is clear by definition of the complement). Then he says U can't contain any limit points since it's open. I think I understand everything expect that: Why would being a neighborhood of every point mean U couldn't contain any limit/accumulation points of A:

Armstrong's proof

These are articles I referenced, but I couldn't generalize it not in a metric space:

Closure of a set is closed proof

Prove the closure is closed and is contained in every closed set

[Proof Verification]: The closure of a set is closed.

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  • $\begingroup$ A point of clarification: the author says $U$ can't contain any limit points of $A$. He did not say $U$ can't contain any limit points period (which it may in general). $\endgroup$ – 0XLR Apr 21 '19 at 1:56
  • $\begingroup$ You might want to consider adding what Armstrong’s definition of the closure is. There are many equivalent definitions, some of which require some work to show that the closure is closed and some of which the result follows almost immediately. $\endgroup$ – Santana Afton Apr 21 '19 at 2:13
  • $\begingroup$ @ZeroXLR thanks that's a good point. I think I get my initial confusion. So U can't contain any limit points of A since we selected U to not have any points of A. Am I on the right track on understanding how we can guarantee such an open neighborhood U? I wasn't 100% solid on that either but was more caught up on the limit point stuff $\endgroup$ – Mariah Moschetti Apr 21 '19 at 2:18
  • $\begingroup$ @SantanaAfton his definition of the closure of a set A is the set containing all limit points (accumulation points) of A and all elements of A $\endgroup$ – Mariah Moschetti Apr 21 '19 at 2:19
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    $\begingroup$ @Chris Custer Ok so you're saying we guarantee the open set U because x∈ X-cl(A), which means x is NOT a limit point. And negating the definition of limit point means x has a nbhd st U∩A=∅ $\endgroup$ – Mariah Moschetti Apr 21 '19 at 21:30
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If $U$ contains a limit point $x$ of $A$, then $U$, being a neighboorhood of $x$, must contain elements of $A$ (by definition of limit points), contrary to the choice of $U$.

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One definition of the closure is as the intersection of all closed sets containing $A$. Or, the smallest closed set containing $A$. Hence closed.

As to the part you didn't understand: a limit point of $A$ can't have a neighborhood disjoint from $A$ (by definition of limit point).

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To show that $Cl(A)$ is closed, we can show that it’s complement $X-Cl(A)$ is open. Take any point $x$ in $X-Cl(A)$. Because $x$ is not in the closure of $A$, it is in particular not a limit point of $A$.

Therefore, there is some open neighborhood $U\ni x$ such that $U\cap A=\emptyset$. Since $x$ was arbitrary, this proves that $X-Cl(A)$ can be written as a union of open sets and is therefore open.

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