Compactness of $\mathbb{R}$ with the trivial and the discrete topology Let $\mathbb{R}$ with the discrete topology, is $\mathbb{R}$ compact?
Answer :no, it doesn't exist an open  finite cover

Let $\mathbb{R}$ with the  trivial topology, is $\mathbb{R}$ compact?
Answer :no, there is only one possible open cover $\mathbb{R}$ itself, and it can't be finite
Is my answers correct?
 A: There is a finite open cover of $\mathbb R$ with the discrete topology, namely the cover whose only member is $\mathbb R$ itself. But some open covers, in particular $\{ \{x\} : x\in\mathbb R \},$ have no finite subcover, so this space is not compact.
You are right that there is only one open cover of $\mathbb R$ with the trivial topolgy, namely the open cover whose only member is $\mathbb R$ itself. But it is finite, since it contains only one member, namely $\mathbb R$ itself.
A: But $\mathbb R$ with the discrete topology does have a finite cover: $\{\mathbb R\}$ is a finite cover.
And in $\mathbb R$ with the trivial topology, $\{\mathbb R\}$ itself is also a finite cover.
You might wish to ponder the fact that a cover of a set $X$ is a set $\{C_i\}$ of subsets $C_i \subset X$. When talking about whether a cover is finite or infinite, what matters is not whether the subsets $C_i$ themselves contain finitely many or infinitely many elements. What matters is whether the set $\{C_i\}$ contains finitely many or infinitely many elements.
A: You are misreading the definition of compact. You have to know whether every cover by open sets has a finite subcover.
That does not mean the open sets in the cover have only finitely many elements, it means you need only finitely many of the open sets.
To show a space is not compact you can look for some cover by infinitely many open sets for which no finite subset covers. To show a space is compact you must reason about all possible open covers.
