Is metric space $(\left<0,\infty\right),d_E)$ local compact complete separable?

Question

Consider metric space $(X,d_E)$, where $X=\left<0,\infty\right)$ and $d_E(x,y)=|x-y|$. I would like to show that $(X,d_E)$ is local compact complete separable metric space.

I think that if I take $\mathbb{Q}^+_0$, which is the countable set, then $\overline{\mathbb{Q}^+_0}=\left<0,\infty\right)$. It means that $(X,d_E)$ is separable space.

Intuitively I think that $(X,d_E)$ is also complete space, because $(\mathbb{R},d_E)$ is complete space and my new set is $\left<0,\infty\right)$, so the only one problem can be with number $0$ but $0\in\left<0,\infty\right)$.

Finally I have troubles with local compactness. So any help will be appreciated. Thank you very much.

Answer 1

First note that your metric $d_E$ is the "usual" norm that we endow on $\mathbb{R}$, which is well-known to be complete. Thus, the topology of $(X,d_E)$ is the same as the subspace topology inherited from $(\mathbb{R},d_E)$.

• $(X,d_E)$ is locally compact, as a closed subspace of a locally compact space is locally compact.

• $(X,d_E)$ is complete, as a closed subspace of a complete space is complete.

• $(X,d_E)$ is separable, as indeed $\mathbb{Q}^+$ is a countable dense subset of $X$. Alternatively, a metric space is separable iff it is second countable, and subspaces of a second countable space are second countable.