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A space $X$ is countably compact if every countable open of $X$ has a finit subcover.

can we say:

A space $X$ is countably compact iff for every ascending sequence of open sets of $X$ like $U_1 ‎\subseteq‎ U_2 ‎\subseteq‎.....$that covers $X$, there is natural number like $n$ such that $X = U_n$.

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Yes. It’s trivial that if $X$ is countably compact, then every strictly increasing sequence of open sets that covers $X$ is finite. For the other direction, assume that $X$ is not countably compact, let $\mathscr{U}=\{U_n:n\in\omega\}$ be a countable open cover of $X$ with no finite subcover, and let $\mathscr{V}=\{V_n:n\in\omega\}$, where $V_n=\bigcup_{k\le n}U_k$. Then $V_n\subseteq V_{n+1}$ for each $n\in\omega$, and $\mathscr{V}$ covers $X$. In the second paragraph of my answer to your earlier question here I give a detailed argument that there is no $n\in\omega$ such that $V_n=X$.

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