Metric space on $\mathbb{R^n}$ where Heine-Borel criterion does not hold

Question

Heine-Borel criterion of $\mathbb{R^n}$ : closed and bounded $\implies$ compactness

Give an example of a metric space in $\mathbb{R^n}$ where this criterion does not characterize compactness

So I need a closed bounded metric space of $\mathbb{R^n}$ which is not compact

So I think I need to consider the definition of compactness where a space is compact if any open cover has a finite subcover

I am having trouble finding such a space

Answer 1

I don't understand your example, with this metric $\Bbb R^2$ is not bounded so what?

For an example where Heine-Borel does not hold, take the bounded distance on $\Bbb R^2$, i.e., $\bar{d}(x,y)=d(x,y)$ if $d(x,y)<1$ and $\bar{d}(x,y)=1$ if $d(x,y)\geq 1$. This is a metric that induces the usual topology on $\Bbb R^2$ (prove it).

With this metric, $\Bbb R^2$ is closed and bounded, but not compact.

Answer 2

First, $a^2 + b^2$ is not a metric. But you probably mean the Euclidean metric $d(a,b) = \left|a - b\right|$ on $\mathbb{R}$.

This is indeed not compact, but it is true that a closed and bounded subset of it is compact.

So it is not a counterexample to the question.

Now, it is true that $d'(a,b) = \min(\left|a-b\right|,1)$, which is the truncated Euclidean metric, which induces the same topology on the reals, but which does not obey the Heine-Borel criterion. This is because all subsets of $\mathbb{R}$ are bounded (by $1$) and so $\mathbb{R}$ itself, which is not compact, is closed and bounded.

Another example is the discrete metric ($d(x,y) = 1$ for $x \neq y$) on any infinite set. There all subsets are closed and bounded but only the finite ones are compact.