Why $\mathbb R$ with the metric $d(x,y)=\min\{|x-y|,1\}$ is not compact?

Question

Refer to Heine-Borel theorem, since $\mathbb R$ is a vector space of finite dimension, all closed and bounded of $\mathbb R$ should be compact. Here $\mathbb R$ is closed and bounded, therefore it should be compact, no ? But it's not and I don't understand why.

Answer 1

The Heine-Borel theorem says that $\mathbb{R}^n$, equipped with the Euclidean distance, has the given property. The statement remains true if the Euclidean distance is replaced with an equivalent distance. Unfortunately, your distance is not equivalent to the Euclidean one.

Answer 2

The teorem you are refering to apply to finite dimension Banach space (which turns out they can only be $\mathbb{R}^n$ or $\mathbb{C}^n$ with any norm - since they are all equivalent)

In this case the metric you are using does not induce a norm, simply because for any $x \in \mathbb{R}$, $d(x,0) \leq 1$. (for a detailed proof of this just ask me in the comment)

Now, in your case simply $$ \mathbb{R} = \bigcup_{x \in \frac{1}{2} \mathbb{Z}} B \left(x, \frac{3}{4} \right) $$ which prove that your space is not compact