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I'm trying to prove the following:

Show that a subset of a topological space is closed if and only if it contains all of its limit points.

Is my proof valid?

Definition of limit point:

$p$ is a limit point of a subset, if every neighborhood of $p$ contains a point in the subset other than $p$ (aka accumulation point).

Lets call the subset, $A$.

In this case we will take $A$ is a closed subset as a given. Lets assume that $p$ is a limit point of $A$, and $p \notin A$. Thus $p \in \partial A $ because only at the boundary can a point, not in the set, have every neighborhood with points that ARE in the set.(More specifically, because EVERY neighborhood of $p$ intersects $A$.) However, we are given that $A$ is closed and closed sets contain all their boundary points. Thus, $p$ cannot exist (RAA). So if a subset is closed it must contain all of its limit points.

Now the converse. In this case we will take $A$ contains all of its limit points as a given. Lets assume that $A$ is not closed, and thus does not contain all of its boundary points. Let $b$ be a point such that, $b \in \partial A$ and $b \notin A$. However $b$ must be a limit point, because every neighborhood of a boundary point contains a point in A. Thus $A$ does not contain all of it limit points (RAA). So, if a subset contains all of its limits points it must be closed.

QED

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  • $\begingroup$ Clarify please: RAA? $\endgroup$ – Mathemagician1234 Jun 14 '16 at 1:59
  • $\begingroup$ RAA, stands for reductio ad absurdum, which is proof by contradiction. It is a very common way to prove things. en.wikipedia.org/wiki/Reductio_ad_absurdum $\endgroup$ – Michael Maliszesky Jun 14 '16 at 2:17
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    $\begingroup$ Careful with your wording. In the 1st part of your proof , if $p$ is a limit point of $A$ and $p\not \in A$ then $p\in \partial A $ but NOT because if has a nbhd intersecting $A$, but rather because EVERY nbhd of $p$ intersects $A$.... BTW, which def'n of "closed" are you using? $\endgroup$ – DanielWainfleet Jun 14 '16 at 2:31
  • $\begingroup$ You say in the next to last sentence "However b must be a limit point, because points on the boundary have a neighborhood with points in A." What you mean is that every neighborhood of a boundary point contains a point in A. $\endgroup$ – John Wayland Bales Jun 14 '16 at 2:38
  • $\begingroup$ @ user254665, I just did an edit and added "EVERY". My book defines a closed set a one who complement is open. But right before this proof, it give 3 other "equivalent" definitions. One being "closed sets contain all their boundary points" $\endgroup$ – Michael Maliszesky Jun 14 '16 at 2:48
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Your corrected proof is correct. Here's a shorter and simpler way to prove the second direction: Let X be the topological space where $A\subseteq X$.Recall that the closure of a set $\bar A$ =$A\cup A'$ where A'= { x | x is an accumulation point of $A\subseteq X$.Recall also that $\bar A$ is a closed set because it is the intersection of all the closed subsets of X that contain A as a subset and the intersection of any number of closed sets is closed. So assume $A\subseteq X$ contains all it's accumulation points. Then $A'\subseteq A$.Then $A\cup A'\subseteq A$.But $A\subseteq A\cup A'$. Therefore, $A = A\cup A'$=$\bar A$ and therefore A is closed! Q.E.D.

In closing, here's one for you to mull your mind over: Consider $A\subseteq X$ where X is a topological space. Assume A has no limit points. Then is it closed? Why or why not?

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    $\begingroup$ An examples of a set with no limit points would be a discreet set, and thus BOTH open and closed. $\endgroup$ – Michael Maliszesky Jun 14 '16 at 11:56
  • $\begingroup$ @MichaelMaliszesky My question was somewhat misleading,I'm sorry. I wanted to ask merely if the set was closed-it may ALSO be open,but that wasn't the question's intent. I fixed it-it should be clearer now. Your comment above is quite correct. But my question was really why is a set with no limit points closed?The answer's very simple when you think about it,try it. $\endgroup$ – Mathemagician1234 Jun 14 '16 at 18:15
  • $\begingroup$ Boy, was that a beautiful, easy-to-follow proof! Sometimes textbooks make it so hard, I tell you... $\endgroup$ – étale-cohomology Aug 8 '17 at 23:18

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