This is a topology question.

First of all, I am sorry this is a really dummy question. As a math student, it is a shame I haven't taken any courses in topology.

A closure of a set $A$ is usually defined as $\operatorname{int}(A)\cup \partial A $. I would like to know how to prove the equivalence of this statement to $A \cup A'$, where $A'$ is the derived set.

Thanks in advance.


$\operatorname{int}(A) = \bigcup\{O :\, O\text{ is open and }O\subseteq A \}$

$\partial A$ is defined as for any $x$, an open set containing it intersects both $A$ and $A^{c}$

$A'$ is a set containing all limit points that are defined as every open set containing a limit point also at least contains an element of $A$.

I hope it is clearer now. And It would be appreciated if a 'standard' definition of closure could be given.


Thanks Asaf Karagila for editing the text. I am sorry I only can accept one answer, so I go for the first and most voted one. But I really appreciate other's contibutions. Thank you all!


  • $\begingroup$ Well, you are using definitions that are different from what I consider the "standards", so just to make sure we are all on the same page, what are your definitions of $\mathrm{int}(A)$, $\partial A$, and $A'$? (As with "closure", there are many different-though-equivalent ways of defining them, so I just want to make sure we are all using the same definitions). $\endgroup$ – Arturo Magidin Jul 26 '11 at 17:18
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    $\begingroup$ In my experience the two most common definitions of $\mathrm{cl}A$ are $\bigcap\{H:A\subseteq H\text{ and }H\text{ is closed}\}$ and $\{x\in X:\forall V\in \mathcal{N}(x)(V\cap A\ne\varnothing\}$, where $\mathcal{N}(x)$ is the set of open nbhds of $x$. Munkres, Willard, Engelking, and Greever use the first of these, Dugundji the second. $\endgroup$ – Brian M. Scott Jul 26 '11 at 18:05
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    $\begingroup$ I have corrected a mistake in the definition of $\operatorname{int}(A)$, the $\bigcup$ should be outside the set defined. $\endgroup$ – Asaf Karagila Jul 26 '11 at 18:09
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    $\begingroup$ @newbie: I think your definition of $\partial A$ is not quite correct: you should not require $x$ to be in $A$; otherwise, $\mathrm{int}(A)\cup\partial A$ would be a subset of $A$, which means that it is not in general equal to the closure. $\endgroup$ – Arturo Magidin Jul 26 '11 at 18:10
  • $\begingroup$ It might be useful for you to learn not only definition but also some basic properties of closure in topological spaces. (In particular the characterization via neighborhoods mentioned by Brian is often useful.) You migh consult some standard textbook, e.g. Willard, p.25 books.google.com/… or you might have a look at wikipedia article en.wikipedia.org/wiki/Closure_(topology) $\endgroup$ – Martin Sleziak Jul 27 '11 at 7:43

The "standard" definition of closure that I usually encounter is the dual definition of interior. One defines interior as the largest open set that is contained in $A$; this is achieved with the definition you give (the union of all open sets contained in $A$) since the union of open sets is open, and the union of subsets of $A$ is a subset of $A$.

To achieve the dual definition, we define the closure of $A$ to be the smallest closed set that contains $A$, which is then easily seen to be equivalent to: $$\mathrm{cl}(A) = \bigcap\{ C\mid A\subseteq C\quad\text{and}\quad C\text{ closed}\}.$$ (Or, one can define the closure of $A$ to be the complement of the interior of the complement of $A$, $$\mathrm{cl}(A) = \left(\mathrm{int}(A^c)\right)^c,$$ since the smallest closed set that contains $A$ is the complement of the largest open set that is contained in $A^c$).

Now, you want to prove that $\mathrm{int}(A)\cup \partial A = A\cup A'$.

To prove $\mathrm{int}(A)\cup\partial A \subseteq A\cup A'$, note that $\mathrm{int}(A)\subseteq A \subseteq A\cup A'$ by definition, so it suffices to show that $\partial A \subseteq A\cup A'$. Take $x\in\partial A$. If $x\in A$, there is nothing to do. Now assume $x\notin A$; use the definition of $\partial A$ to show that you will necessarily have $x\in A'$.

The converse inclusion is perhaps a bit more delicate, since you don't have any obvious inclusions. I would proceed as follow: first show that $A\subseteq \mathrm{int}(A)\cup \partial A$, by showing that if $a\in A-\mathrm{int}(A)$, then $A\in\partial A$; to do this, show that any open set containing $a$ cannot be completely contained in $A$. Then we need to show that $A'\subseteq \mathrm{int}(A)\cup\partial A$. Show that if $x\in A'-A$, then $x\in \partial A$. Then simply note that if $x\in A'\cap A$, then you're done already, since you already know that $\mathrm{int}(A)\cup \partial A$ contains every point of $A$.

Finally, we probably want to check that the "standard" definition I gave agrees with the ones you were given.

If $x\notin \mathrm{int}(A)\cup\partial A$, then $x\notin\partial A$, so there is an open set that contains $x$ and is either completely contained in $A$, or completely contained in $A^c$; but if it were completely contained in $A$, then we would have $x\in\mathrm{int}(A)$, a contradiction to the choice of $x$; thus, there is an open set $U$ with $x\in U$ and $U\cap A=\emptyset$. That means that $C=U^c$ is closed and contains $A$, so $\mathrm{cl}(A)\subseteq C$. Since $c\notin C$, then $x\notin \mathrm{cl}(A)$. We have shown that $\mathrm{cl}(A)\subseteq \mathrm{int}(A)\cup\partial A$.

Now let $x\notin \mathrm{cl}(A)$. Then there exists a closed set $C$ such that $A\subseteq C$ and $x\notin C$. Let $U=C^c$; then $U$ is open, $U\cap A=\emptyset$, and $x\in U$. In particular, $x\notin A$, and $x\notin A'$, so $x\notin A\cup A'$. Thus, we have shown that $A\cup A'\subseteq \mathrm{cl}(A)$.

Hence: $$A\cup A' \subseteq \mathrm{cl}(A) \subseteq \mathrm{int}(A)\cup\partial A = A\cup A',$$ proving the desired equality.

  • $\begingroup$ thanks for the answer.in your argument, it asserts $\partial A \subseteq A'$. If considering the following example: $[0,1]\cup{2}\cup[3,4]$ in $\mathbb{R}$ with standard topology. Is 2 a boundary point but not a limit point? $\endgroup$ – newbie Jul 27 '11 at 11:34
  • $\begingroup$ @newbie: Note that I said $\partial A\subseteq A'$, not $\partial A\subseteq A'$; it was only the elements of $\partial A$ that were not in $A$ that I asserted would lie in $A'$. That said, I would say $2$ is a limit point in your example, since every open neighborhood of $2$ contains points of $A$ (namely, $2$ itself). (Alternatively, the constant sequence $\{a_n\}$ with $a_n=2$ for all $n$ is a sequence of points of $A$, so its limit is in $A'$; but the limit is $2$). $\endgroup$ – Arturo Magidin Jul 27 '11 at 15:55
  • $\begingroup$ aha! yes, $\partial A \subseteq A\cup A'$...sorry, I overlooked it. $\endgroup$ – newbie Jul 27 '11 at 17:02
  • $\begingroup$ As I realized, the definition of limit point states that $x$ is a limit point if every neighborhood of $x$ contains at least one point $y\in A$ and $y\neq x$. $\endgroup$ – newbie Jul 27 '11 at 17:07
  • $\begingroup$ @newbie: Okay, with that definition, $2$ is in $\partial A$ but not in $A'$; however, since the claim was that $\partial A\subseteq A\cup A'$, this is not a problem. As I noted, if $x\in \partial A-A$, then you should prove it lies in $A'$, and that should not be a problem. $\endgroup$ – Arturo Magidin Jul 27 '11 at 17:09

There are many definitions of 'closure', and one which is especially useful is: $$ \overline{A} = \bigcap \{ C \;|\; A \subseteq C, C \;\textrm{closed} \} $$

Anyway, I'd like to prove $int(A) \cup \partial A = A \cup A'$ directly. (It's sorta fun, really.)

I. Show $int(A) \cup \partial A \subseteq A \cup A'$. Since $int(A) \subseteq A$, all we must show is $\partial A \subseteq A \cup A'$. Let $x \in \partial A$. By definition, any open $U$ containing $x$ must intersect both $A$ and $A^c$ nontrivially. Suppose $x \notin A$ (if $x \in A$, we're already done). Let $U$ be any open set containing $x$. Then $U \cap A \neq \emptyset$, which shows there is some point of $A$, not equal to $x$ (since $x \notin A$) living inside $U$, thus proving $x \in A'$.

II. Show $A \cup A' \subseteq int(A) \cup \partial A$. Here, we must show both $A \subseteq int(A) \cup \partial A$ and $A' \subseteq int(A) \cup \partial A$.

IIa. $A \subseteq int(A) \cup \partial A$. Let $x \in A$ and suppose $x \notin \partial A$. Then (by def. of $\partial A$), there exists some open set $U$ containing $x$ such that either $U \cap A = \emptyset$ or $U \cap A^c = \emptyset$. But $x \in U \cap A$ (hence non-empty), so we must have $U \cap A^c = \emptyset$. This is equivalent to $U \subseteq A$. Now we have an open set entirely within $A$, containing $x$. By definition $x \in int(A)$.

IIb. $A' \subseteq int(A) \cup \partial A$. Let $x \in A'$ and suppose $x \notin \partial A$. As in IIa, this implies there exists an open $U$ containing $x$ such that either $U \cap A$ or $U \cap A^c$ is empty. But by definition of $A'$, every open set containing $x$ must intersect $A$ in an element not equal to $x$. So $U \cap A \neq \emptyset$, implying that $U \cap A^c = \emptyset$, which (as in IIa) implies $x \in int(A)$.

By the way, most proofs in elementary point-set topology can be churned out in this manner. Argue from definitions, use set theory, use logic to break down unions into "OR" and intersections into "AND", etc.

Hope this helps!


The definition of open set that you gave:

$\operatorname{int}(A) = \bigcup\{O :\, O\text{ is open and }O\subseteq A \}$

Basically describes the interior of $A$ as the largest open set contained in $A$. Can you think of a dual notion for closed sets? Think, though, that counter/dual to the notion of an open set, the closure of a set must contain the set.

  • $\begingroup$ @newbie: And that definition doesn’t say quite what you want: it should read $\mathrm{int}(A) = \bigcup\{O:O\text{ is open and }O\subseteq A \}$, with the union on the outside. $\endgroup$ – Brian M. Scott Jul 26 '11 at 18:11
  • $\begingroup$ I have corrected a mistake in the definition of $\operatorname{int}(A)$. Note that the current one in your answer is incorrect. $\endgroup$ – Asaf Karagila Jul 26 '11 at 18:12
  • $\begingroup$ @Brian: You are writing your comment in the wrong place :-) $\endgroup$ – Asaf Karagila Jul 26 '11 at 18:12
  • $\begingroup$ Yes, I noticed, thanks; I just changed it. $\endgroup$ – gary Jul 26 '11 at 18:26

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