The book “Probability with martingales” by David Williams provides a proof of Baire’s category theorem in p. 203-204. This blog post reproduces the proof in greater detail to clarify some of the statements that are left to the book reader to confirm.

Baire’s category theorem is stated as follows:

*If a complete metric space can be written as a union of a countable sequence of closed sets , then at least one of these closed sets contains an open ball.*

The theorem was proved for by René-Louis Baire in his PhD thesis “Sur les fonctions de variable réelles“.

Assume a complete metric space with some metric , which need not be necessarily the Euclidean metric space .

Start by assuming that none of , contains an open ball. Moreover, assume that none of the complements , is empty. If there exists an empty complement , then , which is a trivial case of finite union that will be treated at the end of the proof separately.

Since is closed, is open, i.e. . Due to being non-empty, there exists . Hence, there exists so that the open ball with center and radius satisfies .

By assumption, contains no open ball, so , which means that . Furthermore, is an open set as the intersection of the open sets and . Thus, there exist and with such that . Note that .

Inductively, choose , to introduce the n-th ball . This way, a countable sequence of nested balls is defined with . It will be shown that the centers of these balls make up a Cauchy sequence . According to the Archimedean property, for any , there exists such that . Let be natural numbers with and . Using the triangle inequality, . Obviously, . Since , it follows that , so . Apparently, implies . As and , it follows that . Similarly, , which leads to , so the ball centers constitute the Cauchy sequence . The metric space has been assumed to be complete, so the Cauchy sequence is convergent, therefore its limit exists in .

The next step is to show that the limit of the ball centers belongs to the intersection of these balls, i.e. . Let and with . By applying the triangle inequality, . Notice that yields , hence , which leads to . Observe that is a subsequence of , which means that . So, taking the limit as produces , whence for any , thus .

It is obvious from that . Consequently, . The contradiction has been reached, which completes the proof.

The case of finite union is simpler, as it is not needed to construct an infinite sequence of open balls and then take the limit of their centers. Instead, it suffices to observe that the center of the smallest ball satisfies , leading to the analogous contradiction .

On a final note, it is emphasized that the existence of the countable sequence of ball centers is based on the axiom of choice.