The Heine-borel theorem states: A set $B$$\subset$$\mathbb{R}^n$ is compact $\Longleftrightarrow$ $B$ is closed and bounded.
In an exercise i need to prove that the $''\Longleftarrow''$ statement is not valid in some metric or topological spaces.
My counterexample is this:
Take the metric space $(\mathbb{N},d)$ where $d$ is the discrete metric.
The open sets in this metric space are the sets of the form $B(x,\epsilon)$$ =
\left\{\begin{matrix}
\{x\} & \epsilon < 1\\
X & \epsilon \geqslant 1
\end{matrix}\right.
$
Then $\mathbb{N}$ is bounded because $\mathbb{N}$$\subset$ $B(m,2)$
$\forall m\in \mathbb{N}$ and it also closed in the discrete metric space
Now suppose that $\mathbb{N}$ is compact.
The collection $C=\{B(m,1/2):m\in\mathbb{N}\}$ is an open cover of $\mathbb{N}$ so by our assumption there must be a finite subcover of $\mathbb{N}$
say $M=$$\{B(m_1,1/2),,,,B(m_j,1/2)\}$ where $\mathbb{N}$$\subseteq$$\bigcup$$M$.
But from this argument and the fact that the elements of $M$ are singletons( from the definition of the open ball in the discrete metric space) we deduce that $\mathbb{N}$ is a finite set which is a contradiction.
Hence $\mathbb{N}$ is not compact.
Is this counterxample valid or am i missing something?
If it valid i would appreciate a little help to construct another counterexample.
Thank you in advance!
Best Answer
Your counterexample works perfectly.
You can construct a huge family of counterexamples in the following way:
Pick any non-compact metric space $(X,d)$, e.g., any unbounded metric space like $\Bbb R,\Bbb Q$ and so on. Then, define a new metric $\bar{d}$ on $X$ by $\bar{d}(x,y)=d(x,y)$ if $d(x,y)<1$, $\bar{d}(x,y)=1$ otherwise.
It is a good exercise to show that the topology induced by $\bar{d}$ is the same as that induced by $d$, so that $(X,\bar{d})$ isn't compact.
But now $(X,\bar{d})$ is bounded and closed, which gives a counterexample to Heine-Borel.