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!


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.

  • $\begingroup$ 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 $\endgroup$ – Henno Brandsma Jan 4 at 23:40
  • $\begingroup$ @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 $A$? $\endgroup$ – Steven Wagter Jan 5 at 16:32
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Jan 5 at 17:19

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