# Bolzano-Weierstrass Property Implies Countably Compact?

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## Solution 1

The problem with your argument arises from your understanding of the term accumulation point. The Bolzano-Weierstrass property as usually stated requires that every infinite set have a limit point, and a point $x$ is a limit point of a set $S$ if every open nbhd of $x$ contains a point of $S\setminus\{x\}$, i.e., a point of $S$ different from $x$ itself. If the space is $T_1$, this implies that every open nbhd of $x$ contains infinitely many points of $S$, but if the space is not $T_1$, this need not be the case. For example, let $=\Bbb N$, for each $n\in\Bbb N$ let $V_n=\{k\in\Bbb N:k<n\}$, and let $\tau=\{\Bbb N\}\cup\{V_n:n\in\Bbb N\}$. Then $\tau$ is a $T_0$ (but not $T_1$) topology on $X$, and every $k>0$ is a limit point of the set $\{0\}$: if $0<k\in V_n$, then $n>k$, so $n>0$, and hence $0\in V_n$, i.e., every open nbhd of $k$ contains $0$, a point of $\{0\}$ different from $k$. And this is true even though the set $\{0\}$ is finite: $\operatorname{cl}\{0\}=X$, so the singleton $\{0\}$ is actually dense in $X$.

Thus, the final step of your argument works only if $X$ is $T_1$: otherwise it’s entirely possible for $x$ to be a limit point of $S=\{x_n:n\in\Bbb Z^+\}$ even though $C_i$ contains only finitely many points of $S$. In my example above, for instance, $\{V_n:n\in\Bbb Z^+\}$ is an open cover of $X$ with no finite subcover. For $n\in\Bbb Z^+$ let $x_n=n-1$; then $x_n\in V_n\setminus\bigcup_{k<n}V_k$. Then $\{x_n:n\in\Bbb Z^+\}=\Bbb N=X$, and we saw above that every open nbhd of $x_2=1$ contains $0=x_1$, so $x_2$ is a limit point of $\{x_n:n\in\Bbb Z^+\}$ even though $V_2$ contains only the two points $0$ and $1$.

## Solution 2

Your proof is correct and it shows that one of the implications. There is however the following issue: how exactly do you define Bolzano–Weierstrass property? There are several conditions:

1. Every sequence has an accumulation point.
2. Every injective sequence has an accumulation point.
3. Every (countably) infinite set has an $ω$-accumulation point.
4. Every (countably) infinite set as an accumulation point.

It doesn't matter when you restrict yourself to injective sequences and countably infinite sets. The point is that an countably infinite set can be interpreted as an injective sequence and vice versa, but an accumulation points of the sequence correspond to $ω$-accumulation points of the set. But accumulation point of a set is weaker notion than $ω$-accumulation point unless you are in a $T_1$ space.

To conclude, 1, 2, 3 are equivalent, 4 is equivalent with them on $T_1$ spaces. Countable compactness is equivalent to Bolzano–Weierstrass property if its definition is based on 1, 2, 3. Note that the definition forbidding infinite closed discrete subset is equivalent to 4.

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### Tom Collinge

Updated on June 06, 2020

• I thought that I could prove this, but it seems that it isn't generally true (for example, https://math.stackexchange.com/q/438474). So, I'd appreciate it if anyone could point out my error. Even better would be to show where an additional property (e.g. Hausdorff or first-countable) would complete the proof.

Assume $S$ has some countable open cover $\mathscr C = \{{C_i\}}$ which has no finite subcover.

Then for all $n > 0, \; ⋃_{i = 1, n} C_i$ does not cover S, and so there is $x_n \in S$ \ $⋃_{i = 1, n} C_i$

Choose one such $x_n$ for each n, forming the sequence $(x_n)$.

So, by the Bolzano-Weierstrass Property the sequence $(x_n)$ has an accumulation point $x \in S$.

Since $\mathscr C$ is countable and covers $S$, then for some $i$, $x \in C_i$.

By the construction of $x_n$ , for $n > i$, then $x_n \not \in C_i$, so there are a finite number of $x_n \in C_i$

But this contradicts $x$ being an accumulation point, which requires an infinite number of $x_n$ in any open set that contains $x$.

• Tom Collinge over 6 years
Many thanks, so having the B-W property is just another name for limit point compact ?
• Tom Collinge over 6 years
• 