Existence of compact/complete metric on countable set

Question

Let, $X$ be a countable set. Which of the following are true?

1. There exists metric $d$ on $X$ so that $(X,d)$ is complete.
2. There exists metric $d$ on $X$ so that $(X,d)$ is not complete.
3. There exists metric $d$ on $X$ so that $(X,d)$ is compact.
4. There exists metric $d$ on $X$ so that $(X,d)$ is not compact.

4 is true as we can take discrete metric. Then $\{\{x\}:x \in X\}$ is an open cover containing no finite subcover.

If there exists a compact metric then it would automatically become complete also. Hence if 3 is true, so is 1.

What about 2? Any other comment?

Answer 1

If by "countable" you mean "countably infinite", then all four are true. For 1 and 3, fix a bijection $X\to\{0\}\cup\{\frac{1}{n}:n\in\mathbb N\}$ and use the standard metric on $\mathbb R$ to induce a compact (hence complete) metric on $X$. For 2 and 4, fix a bijection $X\to\{\frac1n:n\in\mathbb N\}$ and use the induced metric.

If by "countable" you mean "finite or countably infinite", then only 1 and 3 are true, as every metric on a finite set is complete, and the induced topology will be compact.