Question on definition of a cover and results with compactness

Question

My professor defines a cover of $A$ to be a collection of sets whose union is equal to $A$. I am used to this being instead a superset of $A$.

Doesn't this lead to contradictions?

Then, there can be no open cover equal to a set like $[0,\infty)$.

Therefore, vacuously, $[0,\infty)$ is compact as it has no open cover.

This is a contradiction.

Am I missing something?

Answer 1

No it does not lead to contradictions, see this page: covering for information. Notice the alternative definition section

When you say a set is compact, you are imbuing it with the subspace topology which will "cut off" any overlap anyway.

You can see the proof with this lemma

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Just to clarify the important part of the proof is:

$(Y,\mathcal{J}_\text{subspace})$ is compact $\implies$ every covering of Y by sets open in X contains a finite subcovering