Compact iff every family of nonempty subsets has an accumulation point

157 Views Asked by At

In this presentation (slide $14$) the author states that a topological space $(X,\tau)$ is compact if and only if every family of nonempty subsets of $X$ has an accumulation point.

What does he means by accumulation point of a family of sets?

It might be that we can assume in his statement that the union of the family is infinite.

I've considered the property (A): every infinite subset has an accumulation point. I've noted that every compact space satisfies (A) but the space $[0,\omega_1)$ with the order topology, where $\omega_1$ is the first uncountable ordinal, seems to be a counterexample to the converse. So compactness is stronger than (A).

If I interpret "accumulation point of a family of sets" to mean accumulation point of the union of said family then the author's statement cound simply mean (A) $\Leftrightarrow$ compact, which is not true.

If I interpret it to mean accumulation point of at least one set in the family then the statement makes no sense since in a $T_1$ space every infinite family of singletons would have no accumulation points.

I'm stuck here.

1

There are 1 best solutions below

0
On BEST ANSWER

Let $\mathcal{C}$ be a family of nonempty subsets of $X$ and $x\in X$.

We say that $x$ is an accumulation point of $\mathcal{C}$ if, for every neighbourhood $U$ of $x$, there exists $C\in \mathcal{C}$ such that, for every $B\in\mathcal{C}$, if $B\subseteq C$, then $U\cap B \neq \emptyset$.

This is equivalent to compactness.