Does the Bolzano–Weierstrass Property Imply Countable Compactness?

The answer is yes under the usual hypotheses, but Bolzano–Weierstrass property is used in more than one way in topology. One common meaning is that every sequence has a convergent subsequence. Another is that every infinite subset has a limit point. These versions are related, but the precise implication depends on the definition and on separation assumptions.

Countable compactness means that every countable open cover has a finite subcover. The proofs below show directly why sequential compactness implies countable compactness, then explain the usual T1 version of the limit-point formulation.

If every sequence has a convergent subsequence

Suppose $X$ is sequentially compact: every sequence in $X$ has a convergent subsequence. Assume, for contradiction, that there is a countable open cover $U_1,U_2,\ldots$ with no finite subcover.

For each positive integer $n$, the finite union $U_1\cup\cdots\cup U_n$ does not cover $X$. Choose

$$x_n\in X\setminus(U_1\cup\cdots\cup U_n).$$

The sequence $(x_n)$ has a convergent subsequence $(x_{n_j})$, with limit $x$. Some member of the cover, say $U_k$, contains $x$. Since $U_k$ is open, convergence means that $x_{n_j}\in U_k$ for all sufficiently large $j$. But the indices $n_j$ eventually satisfy $n_j\ge k$, and the construction put $x_{n_j}$ outside $U_1\cup\cdots\cup U_{n_j}$, which includes $U_k$. This contradiction proves that every countable open cover has a finite subcover.

No T1 or Hausdorff assumption was used. Thus sequential compactness always implies countable compactness.

If every infinite subset has a limit point

A limit point of $A\subseteq X$ is a point $p$ such that every open neighborhood of $p$ meets $A\setminus\{p\}$. In a T1 space, the condition that every infinite subset has a limit point is equivalent to countable compactness.

Here is the direction relevant to the question. Suppose a countable open cover $U_1,U_2,\ldots$ has no finite subcover. Choose points outside its initial finite unions. They can be chosen distinct: if only finitely many points remained outside some initial union, each of those points could be covered by one additional cover member, producing a finite subcover. The chosen points therefore form an infinite set $A$.

By the limit-point property, $A$ has a limit point $p$. Choose $k$ such that $p\in U_k$. Because the space is T1, for each of the finitely many chosen points with index less than $k$ that is not equal to $p$, there is a neighborhood of $p$ that excludes that point. Intersect those neighborhoods with $U_k$ to get a neighborhood of $p$ that excludes all those early points other than $p$. The limit-point property says this neighborhood still meets $A\setminus\{p\}$. Since the chosen points are distinct, that intersection must contain a chosen point $x_n$ with $n\ge k$. Yet $x_n$ was chosen outside $U_k$, a contradiction.

The T1 condition matters in this argument because it lets you exclude finitely many points while keeping a neighborhood of $p$. In standard textbook usage, the equivalence between the limit-point formulation and countable compactness is stated for T1 spaces. If a text uses Bolzano–Weierstrass to mean sequential compactness instead, use the preceding proof, which has no such assumption.

Worked example: the real line

The interval $[0,1]$ is sequentially compact: every sequence in it has a convergent subsequence whose limit remains in $[0,1]$. Therefore it is countably compact. For instance, if countably many relatively open subsets cover $[0,1]$, some finite collection of them already covers the whole interval.

By contrast, the open cover of $\mathbb{R}$ given by $U_n=(-n,n)$ for $n=1,2,\ldots$ has no finite subcover. Any finite selection is contained in the largest selected interval and misses points farther from zero. Correspondingly, the sequence $x_n=n$ has no convergent subsequence in $\mathbb{R}$. This illustrates the implication: sequential compactness rules out countable covers of this kind.

Common mistakes

• Treating compactness and countable compactness as identical. Compactness applies to open covers of any cardinality; countable compactness only requires a finite subcover for countable covers.
• Using the limit-point definition without checking T1. The familiar equivalence is normally presented with a separation hypothesis. State which definition of Bolzano–Weierstrass you are using.
• Confusing a limit point with a limit of a sequence. In general topological spaces, a limit point need not be the limit of a sequence from the set. Sequential compactness is therefore a distinct condition, even though it does imply countable compactness.
• Assuming the converse. Countable compactness does not in general imply sequential compactness. Additional conditions, such as metrizability, can make the two properties equivalent.