Heine–Borel theorem: why is the condition closed and bounded?

The Heine–Borel theorem says:

A subset of $\mathbb R^k$ is compact if it is closed and bounded.

At the same when I think that any either closed or bounded is enough: if it is bounded it must be a subset of a $k$-cell and thus compact. If it is closed, every subsequence converges to the limit that lies in it and hence compact.

So, why don't they say a set is compact if either it is closed or bounded?

• "Because if it is bounded it must be a subset of a k-cell and compact." No, not at all. This is not true. The bounded interval $(-1,1)$ is NOT compact. – Jack Nov 30 '17 at 14:10

There are closed sets that are not bounded, for example $\{ x : x \ge 0 \}$ as a subset of $\mathbb{R}$, and there are bounded subsets that are not closed, for example $\{ x : -1 < x < 1 \}$, again as a subset of $\mathbb{R}$.