Negation Of A Definition Of A Compact Space

Question

I was given the following definition:

Let $M$ be a metric space, $M$ is compact if: for all open cover of $M$ there is a finite sub cover.

I know that the negation of for all is exists and vice versa, and in the case that depend quintifires we negate just the first one, for example in the case of the definition of a limit $\exists \varepsilon>0\forall \delta>0: |x-x_0|\leq \delta \rightarrow |f(x)-L|\leq \varepsilon $

So the negation will be: $M$ is no compact $\iff$there is an open cover of $M$ there is not finite sub cover?

Answer 1

Yes, you are right. If we can find at least one open cover of the metric space which does not have any finite sub-cover, then the space is not compact. The other way is also true: if the metric space is noncompact, then there exists an open cover of the metric space which does not have a finite subcover.