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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?

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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 $c(A)\cap X^*$.

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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 $[0,\frac12)$.

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