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[SOLVED] Perfectly Compact and countable basis implies compactness


Active member
Mar 21, 2017
The point set $M$ is said to be perfectly compact if and only if it is true that if $G$ is a monotonic collection of non-empty subsets of $M$ then there is a point $p$ that is a point or a limit point of every every element of $G$.

Problem: If $X$ is perfectly compact and has a countable basis, then $X$ is compact.

We are allowed to assume that X is Hausdorff and at most (if we need it) metric.

This is my first time seeing "perfectly compact"... do you approach this using contradiction? Assume that X is not compact (i.e. there is a cover with no finite sub cover) to give a contradiction?
Last edited:


Active member
Mar 21, 2017

X has a countable basis $\implies$ every open cover of X has a countable sub-cover (I won't write up the proof for this, but it can be easily verified)

Let $\mathcal{U}$ be an open cover of X. $\implies$ There is a countable sub-cover of $\mathcal{U}$, say $\mathcal{U'}=\left\{U_0, U_1,...\right\}$.

For each n, define
$V_n = \cup_{i\leq n}U_i$
$G_n = X - V_n$

Then, $\forall n, G_n$ is closed.
Also, as $n$ increases $V_n$ becomes a larger set $\implies G_0 \supset G_1 \supset...$

Case I: $\exists n$ such that $X = V_n$.
If this is the case, then we are done. Because then $V_n = \cup_{i\leq n}U_i$ would be a finite sub-cover of $\mathcal{U}$.

Case II: There does not exist such an $n$ as in Case I.
$\forall n, G_n \neq \emptyset$ and $\cap G_n \neq \emptyset$, by perfect compactness of X.
$\implies \exists x\in X, x\in \cap_{n\in \Bbb{N}} G_n \implies x\in \cap_{n\in \Bbb{N}}\left(X - V_n\right) \implies x\notin \cup_{n\in \Bbb{N}} V_n$
A contradiction with our construction of the $V_n$'s.
$\implies$ Case I is the only possible case. $\implies$ X is compact.