I have a non homework related question from a text and require a nice clear proof/disproof please

Is it true that a subset that is closed in a closed subspace of a topological space is closed in the whole space?

my ideas:

if $H$ is the subset of the topological space $X$

if the subset is closed in the closed subspace, the complement is open in the subspace, which means the complement is of form $U\cap H$ for some $U$ open in $X$

if the subspace is closed the complement is open which means complement of $H=U$ for some open $U$ in $X$

kind thanks

  • $\begingroup$ Look at closed subsets of $(0, 1)\subset \Bbb R$ $\endgroup$ – Arthur Feb 25 '13 at 15:15
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    $\begingroup$ @Arthur. But $(0,1)$ is not a closed subset of $\mathbb{R}$. $\endgroup$ – T. Eskin Feb 25 '13 at 15:19
  • $\begingroup$ Ahh, I didn't see that the mid-set had to be closed in the super-set. I stand corrected. $\endgroup$ – Arthur Feb 25 '13 at 15:21
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    $\begingroup$ This is useful i connection with your question: Closed Set in Topological Subspace $\endgroup$ – Martin Sleziak Mar 21 '13 at 6:39

Suppose that $H$ is a closed subspace of $X$, and $F$ is closed in the subspace $H$. By definition of the relative topology there is a closed set $C$ in $X$ such that $F=C\cap H$. But then $F$ is the intersection of two closed subsets of $X$, so $F$ is closed in $X$.

If you don’t already know this characterization of closed sets in the relative topology, it’s worth proving as a separate

Proposition. Let $Y$ be a subspace of a space $X$. Then a set $H\subseteq Y$ is closed in $Y$ if and only if $H=F\cap Y$ for some closed set $F$ in $X$.

Of course for this you do need to look at complements, but it’s very easy. $Y\setminus H$ is open in $Y$, so there is an open $U\subseteq X$ such that $Y\setminus H=U\cap Y$. Let $F=X\setminus U$; then $F$ is closed in $X$, and $F\cap Y=(X\setminus U)\cap Y=Y\setminus U=H$.

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    $\begingroup$ The definition of the subspace topology with which I am familiar says that open sets in the subspace topology have the form $G\cap H$ where $H$ is the subspace and $G$ is open in $X$. Then the statement you made in your third sentence is a theorem, and although it's simple, it seems to require exactly the sort of consideration of complements that you claim can be avoided. Am I missing something? $\endgroup$ – MJD Feb 25 '13 at 15:35
  • $\begingroup$ @MJD: No. I’ve internalized the concept to such an extent that I simply didn’t think of that as a separate result. I should probably rephrase. (Of course, if one defines a topology in terms of closed sets, one will presumably define the relative topology similarly, but that’s hardly likely to be relevant here!) $\endgroup$ – Brian M. Scott Feb 25 '13 at 15:41
  • $\begingroup$ Why $Y=(X\setminus U)\cap Y=Y\setminus U=H$ ? $\endgroup$ – Vinod Nov 5 '18 at 9:47

Let $X'$ be a closed subset of $X$, and say that $H$ is closed in $X'$ in the subspace topology. Then $X'\setminus H$ is open in $X'$, and is therefore of the form $G\cap X'$ for some set $G$ that is open in $X$. Then $X\setminus G$ is closed in $X$, and $(X\setminus G)\cap X' = H$ is an intersection of sets closed in $X$ and is thus closed in $X$.


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