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

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

• I think you misunderstood the problem. You are supposed to find a metric space in which the implication "closed and bounded $\implies$ compact" does not hold, not a metric space which is not compact (although this could provide a counterexample provided this metric space is itself bounded). – Stefan Hamcke May 2 '16 at 17:21
• One example, if you are willing to consider metric spaces that are not $\mathbb{R}^n$, is the metric space $(\mathbb{Q},d)$ where $d$ is the standard Euclidean distance function. Here we cannot conclude the Heine-Borel theorem because $\mathbb{Q}$ is not complete. – Geoff May 2 '16 at 17:24
• The question says "give an example of a metric space in which this criterion does not characterize compactness" and I was not sure how to interpret it. So I need to find a metric space which is closed and bounded, but cannot be such that any open cover has finitely many subcovers – thinker May 2 '16 at 17:25
• @Geoff Is $\mathbb{R}$ non complete also? It seemed so because it is infinite, but I am guessing that is incorrect – thinker May 2 '16 at 17:27
• Also your description of compactness is a bit unfortunate: A space is compact when any open cover has a finite subcover. Having finitely many subcovers is trivially satisfied. – Stefan Hamcke May 2 '16 at 17:27

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.

• When you say $\mathbb{R^2}$ is closed and bounded, do you mean the set of points in $\mathbb{R^2}$ which satisfy this metric is closed and bounded? – thinker May 2 '16 at 17:32
• I'm saying that the metric space $(\Bbb R^2,\bar{d})$ is closed and bounded. It is obviously closed (its complement is empty) and bounded because it is equal to $B_{\bar{d}}(0,2)$. – Nitrogen May 2 '16 at 17:57
• Oh ok I see what you mean. Thank you for that example. I am not quite sure how you got the closed ball though.. why is its radius equal to 2? Thanks again – thinker May 2 '16 at 18:02
• The distance between 2 points (with the bounded distance) is at most 1. So I took radius=2 to have some room left. I could have taken 1.5 or 1.01 or a closed ball of radius 1. – Nitrogen May 2 '16 at 18:13

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.