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.

Let $(X,d)$ be a metric space and $a_1,a_2,\ldots\in X$. Define $A=\{a_n:n\in\mathbf N\}$. Is $A$ closed in $(X,d)$?

And is $A$ closed when X is a topological space?

share|improve this question

5 Answers 5

Another example of a countable, non-closed set is $\mathbb{Q}$ inside of $\mathbb{R}$ with the usual topology, since $\mathbb{Q}$ is dense and there are irrational numbers in $\mathbb{R}$.

share|improve this answer

No. Consider $a_n = \frac{1}{n}$ and $X= \Bbb{R}$ equipped with the euclidean metric. The set $A$ is not closed because $0$ is a limit point of $A$ such that for no $n$ is $a_n$ ever equal to $0$.

share|improve this answer
1  
So the negative answer is due to possible cluster points of $(a_n)_n$ not in $A$. –  Seirios Jul 9 '12 at 12:55
    
@Seirios: Yes, but I would write it cluster points of $a_n$, with just one subscript. –  Ross Millikan Jul 9 '12 at 13:12
2  
That is what "not closed" means: there are cluster points not in the set. –  GEdgar Jul 9 '12 at 14:13
    
I disagree with your second example. Sequence $\{x^n\}$ is closed in $C[0,1]$ equipped with the sup metric. –  GEdgar Jul 9 '12 at 14:15
    
@GEdgar I forgot that the pointwise limit of $\{x^n\}$ lies outside of $C[0,1]$. –  user38268 Jul 10 '12 at 8:40

Let $X$ be an infinite topological space. Then:

1) If $X$ is discrete, then (all subsets are closed, so) every countably infinite subset is closed.

2) If $X$ is non-discrete, separated and first-countable -- in particular, if $X$ is non-discrete and metrizable -- then there is a non-isolated point $x$ and a sequence $\{x_n\}_{n=1}^{\infty}$ of distinct elements of $X \setminus \{x\}$ converging to $x$. Thus $A_1 = \{ x_n: n \in \mathbb{Z}^+ \}$ is a countably infinite subset which is not closed, and $A_2 = A_1 \cup \{x\}$ is a countably infinite subset which is closed.

3) Above the hypothesis of first countability can be somewhat weakened: it is enough to assume instead that $X$ is sequential. But it cannot be dropped entirely: let $X$ be an uncountable set endowed with the cofinite topology, in which a proper subset is closed if and only if it is finite. Then $X$ is a separated space which has no countably infinite closed subsets.

At the moment I am not seeing a clean necessary and sufficient condition on a general topological space for it to admit a countably infinite closed subset. Maybe someone else can do better?

share|improve this answer

Here's an example from functional analysis:

Consider $B[0,1] = \{f: [0,1] \to \mathbb R \mid f \text{ bounded }\}$ with the metric $d(f,g) = \int_{[0,1]} |f(x)-g(x)| dx$. Then we define the following countable subset: $S = \{ f_n (x) \mid n \in \mathbb N_{\geq 2}\} \subset C[0,1] \subset B[0,1] $ where

$$ f_n(x) = \begin{cases} 0 & x \in [0, \frac{1}{2}-\frac1n] \\ nx - \frac{n}{2} + 1 & x \in [\frac{1}{2}-\frac1n, \frac{1}{2}]\\ 1 & x \in [\frac12, 1] \end{cases}$$

Then $f_n$ is continuous. Its pointwise limit $f(x)$, which is also its limit with respect to $d(\cdot, \cdot)$, is the (discontinuous) function that is $0$ on $[0,\frac12)$ and $1$ on $[\frac12,1]$ so the limit of $f_n$ is not in $C[0,1]$ and hence also cannot be in $S$.

Note:

If you use a different metric, such as for example $d(x,y) = \sup_{x \in [0,1]} |f(x) - g(x)|$, the pointwise limit $f(x)$ will no longer be a limit of $f_n$ with respect to the metric because the sequence is no longer Cauchy.

share|improve this answer
    
@BenjaLim Did you miss a "not" in that comment? (we want a set that is not closed) –  Matt N. Jul 10 '12 at 10:52
    
Sorry, I meant to say how does this show that family $f_n$ is not closed? I think the family $f_n$ has no limit points (as you noted the pointwise limit is not in $C[0,1]$ so we can't count it as a limit point in the space you're considering). –  user38268 Jul 10 '12 at 10:55
    
@BenjaLim Oh dear. Well spotted. I can salvage the example though I think. –  Matt N. Jul 10 '12 at 11:02
    
@BenjaLim Look, I think I fixed it. : ) –  Matt N. Jul 10 '12 at 11:04
    
@BenjaLim I think it's clear as it is : ) –  Matt N. Jul 10 '12 at 11:41

The definition of countably compactness maybe helpful for your question:

$X$ is countably compact, i.e., for every infinite subset $A$ there exists a cluster $a$ of A.

(Another equivalent definition of countably compactness is this: for every countable open cover $U$ of $X$, there exists finite subcover $V \subset U$ which covers the space $X$.)

So, if $a \in A$, then $A$ is closed in $X$; and if your $A$ does not contain his cluster point, then it is not closed.

Remark: However, metric space or topological space does not imply countably compactness. it is another topological property. But we have countably compactness = compactness in the metric spaces.

share|improve this answer

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.