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.