Take the 2-minute tour ×
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.

If a set $X$ in a topological space $T$ has the property that for all sequences $x_n \in X, x_n \to x \implies x\in X$ must X be closed? I know this is true for metric spaces but is it true for a general topological space. Here I am defining $X$ is closed iff $T\setminus X$ is open.

share|improve this question
2  
$T$ being first countable should be sufficient. –  ronno Nov 12 '12 at 15:38
add comment

3 Answers

up vote 3 down vote accepted

We say $x_n\to x$ iff for every open set $U\ni x$, almost all $x_n\in U$. (Note that $x_n\to x$ and $x_n\to y$ need not imply $x=y$ in general!)

Let $X=\mathbb R$ with co-countable topology, i.e. $A$ is open iff $A=\emptyset$ or $\mathbb R\setminus A$ is at most countable. Then $T:=(-\infty,0]$ contains all its limit points because for any sequence $x_n$ in $T$ and point $y>0$, the set $\mathbb R\setminus\{x_n\mid n\in\mathbb N\}$ is an open neighbourhood of $y$ not containing any members of the sequence, i.e. $y>0$ is not limit of any sequence in $T$. But $T$ is not closed because it is not countable (and not all of $\mathbb R$).

The reason is (cf. ronno's comment) that $X$ is not first countable.

share|improve this answer
    
Thanks, I believe this space also has uniqueness of limits which is nice. –  Dave Nov 12 '12 at 16:08
add comment

A subset $F$ of a topological space $X$ is called sequentially closed if every limit point $x$ of some sequence $(x_n)_{n\in\mathbb N}$ in $F$ is an element of $F$. A subset $U$ of $X$ is called sequentially open if every sequence $(x_n)$ converging to a point of $U$ is eventually in $U$. It is easy to show that the complement of a sequentially open set is sequentially closed, and a bit harder to prove the other direction.

A space $X$ is called sequential if every sequentially closed subset of $X$ is closed. The other implication is always true: A closed subset is always sequentially closed. For an example of a non-sequential space see Hagen's answer.

A space with the property that for every adherence point $x$ of a subset $A$ there is a sequence $(x_n)\subseteq A$ converging to $x$ is sequential, i.e. every first-countable space is sequential.

share|improve this answer
add comment

In general topological space you can't even talk about sequences unless you have some structure on them.

share|improve this answer
5  
I think we can, a sequence is just a function from the naturals to the topological space. –  Dave Nov 12 '12 at 15:31
    
@Dave, that's correct. A sequence is literally that, a sequence of points. Maybe he's talking about convergence? I'm not quite sure. –  Michael Dyrud Nov 12 '12 at 15:32
add comment

Your Answer

 
discard

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.