Does there exist a compact topological space that doesn't possess the Bolzano-Weierstrass property?

Limit point $$a$$ of $$A$$: each neighbourhood of $$a$$ contains a point of $$A$$ unequal to $$a$$ itself.

Point of accumulation $$a$$ of a set $$A$$: each neighbourhood of $$a$$ contains infinitely many distinct points of $$A$$.

Topological space $$X$$ with the Bolzano Weierstrass property: each infinite subset of $$X$$ has at least one point of accumulation.

In my book it was proven that in Hausdorff spaces, a point $$a$$ is a limit point of the set $$A$$ if and only if it is an accumulation point of $$A$$.

Another relevant theorem is that in compact spaces, every infinite subset of a topological space has at least one limit point in the topological space.

I was wondering if there is an example of a topological space which is compact but doesn't possess the Bolzano-Weierstrass property. I couldn't come up with any, but we know that such a space must be non-Hausdorff by the aforementioned theorems and definitions.

Thanks in advance!

1 Answer

Every compact topological space has the Bolzano-Weierstrass property.

Let $$X$$ be a compact space, and suppose that $$A \subseteq X$$ has no accumulation points. That is, every $$x \in X$$ has an open neighbourhood $$U_x$$ such that $$U_x \cap A$$ is finite. Then $$\{ U_x : x \in X \}$$ is an open cover of $$X$$, and so by compactness there are $$x_1 , \ldots , x_n \in X$$ such that $$U_{x_1} \cup \cdots \cup U_{x_n} = X$$. But then $$A = A \cap ( U_{x_1} \cup \cdots \cup U_{x_n} ) = ( U_{x_1} \cap A ) \cup \cdots \cup ( U_{x_n} \cap A )$$ is a finite union of finite sets, and so is finite.

• In fact $X$ is compact iff for every infinite subset $A$ there is a point $x \in X$ such that for all open neighbourhoods $O$ of $x$ we have that $|O \cap A| = |A|$. This is the countable case of the right hand property. @StevenWagter – Henno Brandsma Jan 4 at 23:40
• @HennoBrandsma so if I understand you correctly, if $A$ is uncountable, then it doesn't suffice to show that there is an $x$ such that each nbhd contains infinite points of $A$; it would need to contain uncountably infinite points of $A$? – Steven Wagter Jan 5 at 16:32
• @StevenWagter as many points as $A$ has, yes, to show compactness. And for all cardinalities. It shows how much stronger compactness is than countable compactness. – Henno Brandsma Jan 5 at 17:19