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.

Find a metric space $X$ and a subset $K$ of $X$ which is closed and bounded but not compact.

I can find a metric space $X$ like the below.

Let $X$ be an infinite set. For $p,q\in X$, define $d(p,q)=\begin{cases}1,&\text{if $p\ne q$}\\0,&\text{if $p=q$}\end{cases}$

Then, with the metric space above, I can find a subset $K$ of $X$, which is a ball which centre is $x$ and radius is $1$. I know this is closed(since it has no limit points) and bounded.

enter image description here

(I'm confused again... I think $X$ is not an infinite set. Isn't the above triangle the metric space $X$?)

Some helps will be really appreciated!

Thank you!

share|improve this question
    
Consider the balls $B_{0.5}(p)$ as an open cover. Does a finite sub cover exist? –  Calvin Lin Jul 16 '13 at 1:44
1  
Under your metric, balls of radius $1$ are just singletons which are compact. –  Vectk Jul 16 '13 at 1:45
1  
What on earth does your diagram have to do with your example? You definitely can't draw $X$ in the plane. Also what on earth does your title have to do with your question? –  Sharkos Jul 16 '13 at 1:48
add comment

5 Answers

up vote 7 down vote accepted

HINT: Let $\langle X,d\rangle$ be the metric space that you described in the problem. For any $x\in X$, the open ball $B(x,1)=\{y\in X:d(x,y)<1\}=\{x\}$ is compact: it’s just the singleton set $\{x\}$. However, the closed ball $\overline{B}(x,1)=\{y\in X:d(x,y)\le 1\}$ is very different: $\overline{B}(x,1)=X$ for each each $x\in X$. (Why?) This shows that $X$ itself is a closed, bounded set in $X$

Consider $\{B(x,1):x\in X\}$. This is an open cover of $X$. Does it have a finite subcover? Is $X$ compact?

share|improve this answer
    
I draw the picture of metric space X. I think there exist only 3 points(which consist equilateral triangle when they are connected) in this space. I think B(x,1) is an open cover and has a infinite subcover since there exists only 3 distinct x in X. What's wrong here.... Omg.. It's so confusing... –  InfimumMaximum Jul 16 '13 at 2:04
1  
@InfimumMaximum: You’re forgetting that you started with an infinite set $X$, so you know that $X$ has more than $3$ points. For instance, $X$ could be $\Bbb N$, the set of natural numbers. Then $B(n,1)=\{n\}$ for each $n\in\Bbb N$, but $\overline{B}(n,1)=\Bbb N$ for each $n\in\Bbb N$, because $d(m,n)\le 1$ for every $m\in\Bbb N$. –  Brian M. Scott Jul 16 '13 at 2:12
add comment

Hint: Just let $X = K$. Argue why $X$ is not compact.

share|improve this answer
    
but then X is not bounded? –  InfimumMaximum Jul 16 '13 at 1:46
3  
@InfimumMaximum The entire space $X$ is bounded, since $X$ is contained in a ball of radius $2$. –  Vectk Jul 16 '13 at 1:48
    
oh............................ My bad.... Thank you! –  InfimumMaximum Jul 16 '13 at 1:50
add comment

If $X$ is an infinite-dimensional Banach space (over $\mathbb{R}$ or $\mathbb{C}$, either way) then the (closed) unit ball $B_1(X)$ is closed and bounded but not compact in the norm topology. See for instance http://planetmath.org/compactnessofclosedunitballinnormedspaces for a proof of this fact.

share|improve this answer
    
The OP is trying to understand the discrete metric. Do you expect him to understand this? –  Pedro Tamaroff Jul 16 '13 at 2:08
    
@PeterTamaroff, I thought it might be a little too advanced, but who knows? And it might be useful to others who look at the question. –  MTS Jul 16 '13 at 11:38
    
I agree with that. –  Pedro Tamaroff Jul 21 '13 at 6:42
add comment

What you're being given is called a discrete space $(X,d)$; where $d(x,y)=[x=y]$ ($[P]=1$ if $P$ is true, $0$ otherwise) is called the discrete metric. Note the set $X$ is assumed to be infinite. For instance, it can be $\Bbb N$, $\Bbb R$.

One can see that every singleton is an open set, since $\{x\}=B(x,1/2)$, say. What can you say about the compact sets in $X$, then? Is $X$ compact? Hint Note $X$ can be covered by singletons.

Spoilers

$(1)$ $X$ is closed, being the ambient space. It is bounded: $X\subset B(x,r)$ for any $x\in X$ whenever $r\geq 1$.

$(2)$ Suppose $F\subseteq X$ is compact. Then $F$ is finite. Reason: cover $F$ by singletons. The existence of a finite cover implies $F$ is itself finite.

$(3)$ Suppose $F\subseteq X$ is finite. Then $F$ is compact. Reason: Suppose $\mathscr C=\{C_\alpha\}_{\alpha\in A}$ covers $A$. Write $F=\{x_1,\ldots,x_m\}$. Since $F\subseteq \bigcup \mathscr C$ there must exist for each $i=1,2,\ldots,m$ an index $\alpha_{i}$ such that $x_i\in C_{\alpha_i}$, so $\mathscr C_0=\{C_{\alpha_i}:i=1,\ldots,m\}$ is a finite subcover.

Conclusion: $F\subseteq X$ is compact if and only if it is finite. Note that $(3)$ always holds in any (topolgical,metric) space, while $(2)$ certainly doesn't: $[a,b]$ is compact in $\Bbb R$ with the usual metric.

share|improve this answer
add comment

Take the reals, change the metric to $d(x,y)=\min(|x-y|,1)$. The topology is unchanged, so your favourite closed but not compact stays closed, not compact, but is now bounded.

share|improve this answer
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.