Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a topological space and $X^*$ be its supspace. It is stated in my textbook that if $c(A)$ represents the closure of set $A$ in $X$, then $c(A) \bigcap X^*$ is closed in $X^*$.

A closed set is one which contains all its limit points, and a limit point of a set is a point such that every open set containing it contains a different point from the aforementioned set.

Let $l$ is an external limit point of set $A$. If there is an open set containing $l$, it has to contain a point in $A$- let's call it $p$. Let the open set containing $p$ and $l$ not contain any other point in $A$. I don't see why that should be a problem at all. Let the subspace $X^*$ contain $l$, but not $p$.

$c(A) \bigcap X^*$ will contain $l$, but $l$ will be not a limit point of $A$, as there is an open set containing $l$ and no point in $A \bigcap X^*$ ($p$ is not there in $X^*$). How is $A \bigcap X^*$ closed in $X^*$ then?

share|cite|improve this question
This statement stands justified if we look at it from this perspective- if $A$ is an open set in $X$, we declare an axiom that $A \bigcap X^*$ is also an open set in $X^*$. A closed set is the complement of an open set. Hence, $c(A) \bigcap X^*$ is closed in $X^*$. I don't understand why I'm facing a problem looking at it from the aforementioned perspective though. – Ayush Khaitan May 2 '13 at 11:49
up vote 2 down vote accepted

The fact that $l$ is not a limit point of $A\cap X^*$ has nothing to do with whether $c(A)\cap X^*$ is closed in $X^*$. Closure of $c(A)\cap X^*$ in $X^*$ requires that every limit point of $c(A)\cap X^*$ in $X^*$ belong to $c(A)\cap X^*$; it says nothing at all about points of $X^*$ that are not limit points of $c(A)\cap X^*$.

share|cite|improve this answer
Another part of the text says $c(E)=E \bigcup d(E)$, where $d(E)$ is the set of all limit points of $E$- the derived set. I don't think a closed set could contain any other points apart from those inside $E$ and $d(E)$. – Ayush Khaitan May 2 '13 at 11:55
I'm referring to "Foundations in Topology" by Pervin, for reference. Pg. 38. – Ayush Khaitan May 2 '13 at 11:57
@Ayush: This is correct, but it has nothing to do with whether $c(A)\cap X^*$ is closed in $X^*$. I think that you may be confusing ‘$c(A)\cap X^*$ is closed in $X^*$’ with ‘the closure of $A\cap X^*$ in $X^*$ is $c(A)\cap X^*$’; the first statement is true, and the second need not be. – Brian M. Scott May 2 '13 at 11:57
That was mind-bending! Thanks!! So it's possible to say that a set is closed without specifying whose closure it is. – Ayush Khaitan May 2 '13 at 12:09
@Ayush: Absolutely! A closed set is simply the complement of an open set. However, I’ll point out that it is always possible to identify a set whose closure it is: if $A$ is closed, then $A=c(A)$! – Brian M. Scott May 2 '13 at 12:10

A specific example may be helpful for you.

Let us consider the closed interval $[0,1]$ with usual topology and its subspace $[0,\frac12)$.

It is easy to see that $[0,\frac12)$ is closed in $[0,\frac12)$ as the subsapce of $[0,1]$, since $ $ $[0,\frac12)=[0,\frac12]\cap [0,\frac12)$. However $\frac12$ is the limit point of $[0,\frac12)$ in the whole space $[0,1]$, but not the limit point of $[0,\frac12)$ in the subspace $[0,\frac12)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.