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Definitions:

$A'$ is the set of all accumulation or limit points.

$\bar{A} = A \cup A'$ - this is known as the closure of $A$.

Let $A$ be a subset of $\mathbb{R}$. A point $p\in\mathbb{R}$ is an accumulation point of $A$ if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$.

Prove: $\bar{A}$ is closed

proof: Suppose $p$ is not in $\bar{A}$. Then it has a neighborhood $N_{r}(p)$ that is included in $\bar{A}^{c}$. This neighborhood is open, so none of its points is in $\bar{A}$. This the compliment of the closure is open, so the closure is closed

I am not sure if I am right, any suggestions would be greatly appreciated

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    $\begingroup$ No, the point is to show that $p$ has an open set that doesn't intersect $\bar{A}$. You have just assumed it... $\endgroup$ – copper.hat Mar 6 '15 at 17:45
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    $\begingroup$ Can you add the definitions you are using for accumulation and limit points? They might differ from narrative to narrative. $\endgroup$ – Alp Uzman Mar 6 '15 at 17:50
  • $\begingroup$ This thread might be useful. $\endgroup$ – Alp Uzman Mar 6 '15 at 17:53
  • $\begingroup$ Added the accumulation point definition $\endgroup$ – Wolfy Mar 6 '15 at 17:54
  • $\begingroup$ Depending on the text, accumulation points and limit points can have different definitions, so it is really important to know what definition you are using in order to know how you are supposed to write the proof. $\endgroup$ – jdods Mar 6 '15 at 17:54
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Your proof is assuming that $\overline{A}^c$ is open in the assertion that $N_r(p)$ lies in it. Equivalently, you're kind of assuming that $\overline{A}$ is closed. You must do a bit more work to show that this lies in $\overline{A}^c$.

You are on the right track, though. Since $p$ is not in $\overline{A}$, it is not in $A$ nor is it a limit point of $A$. Therefore there must be some neighborhood $N$ of $p$ that does not intersect $A$ at all.

Can $N$ contain any limit points of $A$? No. If it contained one, $a$. Then by definition of limit point $N$ must contain another point of $A$. But $N$ contains no points of $A$, so this is ridiculous. Thus $N$ must be disjoint from both $A$ and its set of limit points, so $N \subseteq \overline{A}^c$, as desired.

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  • $\begingroup$ May be a silly question, but how does one then go from $N \subseteq \bar{A}^c$ to $\bar{A}$ being closed? $\endgroup$ – Vedvart1 Feb 24 '17 at 14:34
  • $\begingroup$ It shows that $\overline{A}^c$ is open (which is equivalent to $\overline{A}$ being closed. Given any point $p$ in it, there is an open neighborhood containing it, that lies inside $\overline{A}^c$. So this set is open. $\endgroup$ – BigMathTimes Mar 6 '17 at 22:03
  • $\begingroup$ Why does $N$ not contain any points of $A$? $\endgroup$ – Reznick Aug 28 '17 at 10:53
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I am new in mathematics analysis and also I was confused with the proof of this proposition, then I based on the proof of W. Rudin, Principles of mathematical analysis, 3rd ed., McGraw Hill, 1976. And I try to complement it.

My english is poor, so...

Let $X$ be a metric space, and $ E_{c} $ the closure of $ E $, and $ E_{c}^c $ the complement of $ E_{c} $.

If $p\in X$ and $p\in E_{c}^c$ then $p$ is neither a point of $E$ nor a limit point of $E$.

Let $ E_{a} $ be the set of limit points of $ E $ such that it contains no elements of $ E $. If there exist a point $ q\in E_{c}^c $ such that it is a limit point of $ E_{c} $, then it has to be a limit point of $ E_{a} $ and not a limit point of $ E $. So there exist a neighborhood of $ q $ such that $ N_{r}(q)\subset {E_{c}^c\cup E_{a}} $ and has some elements of $ E_{a} $, but every point of $ Nr(q), $especially the elementes of $ E_{a} $, have a neighborhood such that it is contained in $ Nr(q) $, wich implie that the elements of $ E_{a} $ inside $ Nr(q) $ aren’t limit points of $ E $. So there’s no limit points of $ E_{c} $ in $ E_{c}^c $, hence $ E_{c} $ is closed.

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  • $\begingroup$ Looks right to me but I had to do this proof without using metric spaces $\endgroup$ – Wolfy Jul 6 '15 at 1:58
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Here's an idea for proof by contradiction. Let $p \in \overline{A}^c$. If $p$ has no neighborhood disjoint from $\overline{A}$ that implies that $N_{\frac{1}{k}}(p) \cap A \neq \emptyset$ for all $k \in \mathbb{N}$. This impossible though, can you think of a reason why?

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  • $\begingroup$ What is $N_{\frac 1 k}(p)$? $\endgroup$ – Milos Mar 4 '18 at 0:55
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First suppose that we have a convergent sequence $x_n\in\overline{A}$. We need to show that $x_n\rightarrow x\in\overline{A}$.

If $x$ is in $\overline{A}^c$, then eventually all $x_n$ are in $\overline{A}\backslash A$. If there were infinitely many $x_n$ in $A$ (a convergent subsequence), then $x$ would be an accumulation point of $A$ and hence be in the closure of $A$. So what is left to be shown is that a sequence of accumulation points of $A$ always converges to an accumulation point of $A$.

Now suppose we have sequence $x_n\in\overline{A}\backslash A$ and suppose $x_n\rightarrow x\in\overline{A}^c$. For any $\epsilon>0$, $B_\epsilon(x_n)\cap A \neq \emptyset$ since each $x_n$ is an accumulation point of $A$.

Choose a sequence $\epsilon_n=1/n$ and choose $y_n\in B_{1/n}(x_n)\cap A$. Hence $y_n\rightarrow x$, but $y_n$ is a sequence in $A$, therefore $x$ is an accumulation point of $A$, and we have a contradiction. Therefore $x\in\overline{A}$ which is thus closed.

Is this argument ok? Probably not the best argument though.

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    $\begingroup$ I think there is a problem here, since we take the sequence from the closure, not the set itself. $\endgroup$ – Alp Uzman Mar 6 '15 at 17:56
  • $\begingroup$ Ya, I guess I overlooked that! I've assumed $\overline{A}$ is closed! whoops... $\endgroup$ – jdods Mar 6 '15 at 17:58
  • $\begingroup$ @Uzman, I fixed my answer. If you feel like it, please take a look. Thanks for spotting my horrendous error! $\endgroup$ – jdods Mar 6 '15 at 18:53

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