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

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.

(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!

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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
Under your metric, balls of radius $1$ are just singletons which are compact. –  Vectk Jul 16 '13 at 1:45
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

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?

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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
@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

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

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but then X is not bounded? –  InfimumMaximum Jul 16 '13 at 1:46
@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

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.

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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

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.

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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.

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