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in almost all proofs using the definition of compactness as the topological space in which every open cover have a finite subcover, there is a step that i can not understand and in my standpoint is not true

Let $\cal C$ be an open cover of $U_1\cup U_2$. Then $\cal C$ is an open cover of both $U_1$ and $U_2$.

This is for example the part that I am having trouble. I think $\cal C$ would not be an open cover of both what do you think? (this is part of the proof in proofwiki website and the same idea is used in a lot of other places)

My argument is that $\cal C$ may contain open sets that are not in the relative topology on $U_1$ or in $U_2$, so $\cal C$ would not be made only of open sets and then would not be an open cover for $U_1$ or $U_2$. If is not this, then what is the topology of the set of union that is used?

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What you’re missing is that the sets don’t have to be relatively open sets, because we have the following proposition.

Proposition. Let $X$ be a topological space, and let $K\subseteq X$. The following are equivalent:

  • Every cover of $K$ by relatively open sets has a finite subcover.
  • Every cover of $K$ by sets open in $X$ has a finite subcover. (Here we say that a family $\mathscr{U}$ covers $K$ if $K\subseteq\bigcup\mathscr{U}$.)

Suppose that every cover of $K$ by sets open in $X$ has a finite subcover. If $\mathscr{U}$ is a cover of $K$ by relatively open sets, for each $U\in\mathscr{U}$ there is an open $V_U$ in $X$ such that $U=K\cap V_U$. Then $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$ is a cover of $K$ by sets open in $X$. By hypothesis $\mathscr{V}$ has a finite subcover of $K$, so there is a finite $\mathscr{F}\subseteq\mathscr{U}$ such that $\{V_U:U\in\mathscr{F}\}$ covers $K$, i.e., such that $K\subseteq\bigcup_{U\in\mathscr{F}}V_U$. Then $K=\bigcup\mathscr{F}$, so $\mathscr{F}$ is a finite subcover of $\mathscr{U}$.

I’ll leave the other direction for you; if anything, it’s the easier direction.

Thus, compactness of $K$ in terms of relatively open sets is the same as compactness of $K$ in terms of sets open in the containing space $X$.

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If $X$ is a topological space, we say $\mathcal{C}$ is an open cover of $Y\subseteq X$ if all $U\in \cal C$ are open in $X$ and $Y\subseteq\bigcup_{U\in \mathcal C}U$. It doesn't matter if each $U$ is an open set of the relative topology on $Y$.