# Baire Category Theorem and the Finite Intersection Property

A Baire space is a topological space in which the intersection of any countable family of open and dense sets is dense (equivalently every non-empty open subset is of second category). One version of the Baire category theorem states that every complete metric space is a Baire space. Another common version states that every compact Hausdorff space is a Baire space. Another version states that every locally compact Hausdorff space is a Baire space. The commonality among these versions is the finite intersection property (whenever a collection of a certain type of sets satisfies the property that any finite subcollection has non-empty intersection, the whole collection has non-empty intersection). For each of these classes of spaces, in addition to countably compact spaces and pseudocompact spaces, Baire category theorem is derived from having one specific form of the finite intersection property. In this post, we explore this relationship.

In each of the following theorem pairs, the B Theorem follows from the A theorem. The A theorem is a form of the finite intersection property and the B theorem is a version of Baire category theorem.

Another interesting observation is that the finite intersection properties discussed here can give a stronger property than being a Baire space. This stronger property is defined by the Banach-Mazur game.

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Theorem 1A
Let $(X, \rho)$ be a metric space. Then the following conditions are equivalent.

1. $(X, \rho)$ is a complete metric space.
2. For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$ such that the diameters of the sets $C_n$ converge to zero, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$.

Theorem 1B
Every complete metric space is a Baire space.

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Theorem 2A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

1. $X$ is a compact space.
2. For every family $\mathcal{F}$ consisting of non-empty closed subsets of $X$, if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.

Theorem 2B

• Every compact Hausdorff space is a Baire space.
• Every locally compact Hausdorff space is a Baire space.

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Theorem 3A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

1. $X$ is a countably compact space.
2. For every countable family $\mathcal{F}$ consisting of non-empty closed subsets of $X$, if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.
3. For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$.

Theorem 3B
Every countably compact Hausdorff space is a Baire space.

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Theorem 4A
Let $X$ be a regular space. The following conditions are equivalent:

1. The space $X$ is pseudocompact.
2. If $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $O_n \supset O_{n+1}$ for each $n$, then $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$.
3. If $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $\mathcal{V}$ has the finite intersection property, then $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$.

Theorem 4B
Every regular pseudocompact space is a Baire space.

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Remark
Theorem 1A (the Cantor Theorem) can be found in Engelking (page 269 in [1]). Theorem 2A and Theorem 3A can also be found in Engelking (they are also proved in this post). Theorem 4B is also found in Engelking (Theorem 3.10.23 in page 207 of [1]) and is proved this post.

We would like to explicitly point out that between Thoerem 1A and Theorem 2A, none of the two theorems implies the other. For example, even though both complete metric spaces and compact Hausdorff spaces are Baire spaces, complete metric spaces are not necessarily compact and there are compact spaces that are not even metrizable. However, the finite intersection property of Theorem 2A implies that of Theorem 3A, which in turn implies the finite intersection property of Theorem 4A.
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Baire Category Theorem

The proofs of all four B theorems are amazingly similar. It is a matter of exploiting the fact that whenever a decreasing sequence of open sets satisfying the condition that each closure is a subset of the previous open set (and satisfying some other condition), the sequence of open sets has non-empty intersection. For example, for complete metric space, make sure that the closures of the open sets have diameters going to zero. For any reader who is new to this material, it will be very instructive to walk through the arguments of these Baire category theorems. The proof of Theorem 1A can be found this post. We prove Theorem 4B.

Recall that $X$ is a Baire space if $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable family of open and dense sets in $X$, $\bigcap \limits_{i=1}^\infty U_i$ is dense in $X$, or equivalently every non-empty open subset of $X$ is of second category in $X$. For more background about the concepts of Baire space and category (see [1] or this post).

Proof of Theorem 4B
Let $X$ be a regular pseudocompact space. Let $\left\{U_1,U_2,U_3,\cdots \right\}$ be a countable family of open and dense sets in $X$. Let $O$ be a non-empty open subset of $X$. We show that $O$ has to contain points of $\bigcap \limits_{n=1}^\infty U_n$. We let $O_1=O \cap U_1$. We find open $O_2$ such that $O_2 \subset U_2$ and $\overline{O_2} \subset O_1$ (using regularity). Continue this inductive process, we have for each $n$, an open $O_n$ such that $O_n \subset U_n$ and $\overline{O_n} \subset O_{n-1}$. Then we have a decreasing sequence of open sets $O_n$ as in condition 2 of Theorem 4A. Then we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$. Since $\overline{O_{n+1}} \subset O_n$ for each $n$, we also have $\bigcap \limits_{n=1}^\infty O_n \ne \varnothing$. It is clear that $\bigcap \limits_{n=1}^\infty O_n \subset \bigcap \limits_{n=1}^\infty U_n$. $\blacksquare$

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Banach-Mazur Game

In proving the above versions of Baire category theorem, we can exploit the appropriate version of the finite intersection property – the situation that any nested decreasing sequence of open sets (under some specified conditions) has non-empty intersection. In fact, the finite intersection property offers more than just Baire category theorem; it can endow the space in question a type of completeness property stronger than Baire space. This completeness property is defined using the Banach-Mazur game.

The Banach-Mazur game is a two-person game played on a topological space. Let $X$ be a space. There are two players, $\alpha$ and $\beta$. They take turn choosing nested decreasing nonempty open subsets of $X$ as follows. The player $\beta$ goes first by choosing a nonempty open subset $U_0$ of $X$. The player $\alpha$ then chooses a nonempty open subset $V_0 \subset U_0$. At the nth play where $n \ge 1$, $\beta$ chooses an open set $U_n \subset V_{n-1}$ and $\alpha$ chooses an open set $V_n \subset U_n$. The player $\alpha$ wins if $\bigcap \limits_{n=0}^\infty V_n \ne \varnothing$. Otherwise the player $\beta$ wins. For more detailed discussion of the game, see this post.

One interesting point that we like to make about the finite intersection property ranging from Theorem 1A to Theorem 4A is that the player $\alpha$ can always win the Banach-Mazur game as long as he/she plays the game according to each specific version of the finite intersection. For example, playing the game in a complete metric space, player $\alpha$ always wins as long as he/she makes the diameters of the closures of the open sets going to zero. In a regular pseudocompact space, player $\alpha$ can always win by making the closure of each of his/her open sets a subset of the previous move of other player.

A topological space in which the player $\alpha$ has a winning strategy is said to be a weakly $\alpha$-favorable space. Thus complete metric spaces, compact Hausdorff spaces, locally compact Hausdorff spaces, countably compact Hausdorff spaces, regular pseudocompact spaces are all weakly $\alpha$-favorable.

There is characterization of Baire spaces in terms of the Banach-Mazur game. A space $X$ is a Baire space if and only if the player $\beta$ has no winning strategy in the Banach-Mazur game played on the space $X$ (see theorem 1 in this post). If the player $\alpha$ can always win, then player $\beta$ can never win. In terms of game terminology, if player $\alpha$ has a winning strategy, then the other player (player $\beta$) has no winning strategy. Thus a space is weakly $\alpha$-favorable implies that it is a Baire space. But the implication is not reversible (see example in this post).

So all the spaces discussed from Theorem 1A to Theorem 4A are all weakly $\alpha$-favorable, a property stronger than Baire spaces. These observations are summarized in the following theorems.

Theorem 1C
Every complete metric space is a weakly $\alpha$-favorable space.

Theorem 2C

• Every compact Hausdorff space is a weakly $\alpha$-favorable space.
• Every locally compact Hausdorff space is a weakly $\alpha$-favorable space.

Theorem 3C
Every countably compact Hausdorff space is a weakly $\alpha$-favorable space.

Theorem 4C
Every regular pseudocompact space is a weakly $\alpha$-favorable space.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# A Question About The Rational Numbers

Let $\mathbb{R}$ be the real line and $\mathbb{Q}$ be the set of all rational numbers. Consider the following question:

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Question

• For each nonnegative integer $n$, let $U_n$ be an open subset of $\mathbb{R}$ such that that $\mathbb{Q} \subset U_n$. The intersection $\bigcap \limits_{n=0}^\infty U_n$ is certainly nonempty since it contains $\mathbb{Q}$. Does this intersection necessarily contain some irrational numbers?

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While taking a real analysis course, the above question was posted to the author of this blog by the professor. Indeed, the question is an excellent opening of the subject of category. We first discuss the Baire category theorem and then discuss the above question. A discussion of Baire spaces follow. For any notions not defined here and for detailed discussion of any terms discussed here, see [1] and [2].

In the above question, the set $\bigcap \limits_{n=0}^\infty U_n$ is a $G_\delta$ set since it is the intersection of countably many open sets. It is also dense in the real line $\mathbb{R}$ since it contains the rational numbers. So the question can be rephrased as: is the set of rational numbers $\mathbb{Q}$ a $G_\delta$ set? Can a dense $G_\delta$ set in the real line $\mathbb{R}$ be a “small” set such as $\mathbb{Q}$? The discussion below shows that $\mathbb{Q}$ is too “thin” to be a dense $G_\delta$ set. Put it another way, a dense $G_\delta$ subset of the real line is a “thick” set. First we present the Baire category theorem.
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Baire Category Theorem

Let $X$ be a complete metric space. For each nonnegative integer $n$, let $O_n$ be an open subset of $X$ that is also dense in $X$. Then $\bigcap \limits_{n=0}^\infty O_n$ is dense in $X$.

Proof
Let $A=\bigcap \limits_{n=0}^\infty O_n$. Let $V_0$ be any nonempty open subset of $X$. We show that $V_0$ contains some point of $A$.

Since $O_0$ is dense in $X$, $V_0$ contains some point of $O_0$. Let $x_0$ be one such point and choose open set $V_1$ such that $x_0 \in V_1$ and $\overline{V_1} \subset V_0 \cap O_0 \subset V_0$ with the additional condition that the diameter of $\overline{V_1}$ is less than $\displaystyle \frac{1}{2^1}$.

Since $O_1$ is dense in $X$, $V_1$ contains some point of $O_1$. Let $x_1$ be one such point and choose open set $V_2$ such that $x_1 \in V_2$ and $\overline{V_2} \subset V_1 \cap O_1 \subset V_1$ with the additional condition that the diameter of $\overline{V_2}$ is less than $\displaystyle \frac{1}{2^2}$.

By continuing this inductive process, we obtain a nested sequence of open sets $V_n$ and a sequence of points $x_n$ such that $x_n \in V_n \subset \overline{V_n} \subset V_{n-1} \cap O_{n-1} \subset V_0$ for each $n$ and that the diameters of $\overline{V_n}$ converge to zero (according to some complete metric on $X$). Then the sequence of points $x_n$ is a Cauchy sequence. Since $X$ is a complete metric space, the sequence $x_n$ converges to a point $x \in X$.

We claim that $x \in V_0 \cap A$. To see this, note that for each $n$, $x_j \in \overline{V_n}$ for each $j \ge n$. Since $x$ is the sequential limit of $x_j$, $x \in \overline{V_n}$ for each $n$. It follows that $x \in O_n$ for each $n$ ($x \in A$) and $x \in V_0$. This completes the proof of Baire category theorem. $\blacksquare$

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Discussion of the Above Question

For each nonnegative integer $n$, let $U_n$ be an open subset of $\mathbb{R}$ such that that $\mathbb{Q} \subset U_n$. We claim that the intersection $\bigcap \limits_{n=0}^\infty U_n$ contain some irrational numbers.

Suppose the intersection contains no irrational numbers, that is, $\mathbb{Q}=\bigcap \limits_{n=0}^\infty U_n$.

Let $\mathbb{Q}$ be enumerated by $\left\{r_0,r_1,r_2,\cdots \right\}$. For each $n$, let $G_n=\mathbb{R}-\left\{ r_n \right\}$. Then each $G_n$ is an open and dense set in $\mathbb{R}$. Note that the set of irrational numbers $\mathbb{P}=\bigcap \limits_{n=0}^\infty G_n$.

We then have countably many open and dense sets $U_0,U_1,U_2,\cdots,G_0,G_1,G_2,\cdots$ whose intersection is empty. Note that any point that belongs to all $U_n$ has to be a rational number and any point that belongs to all $G_n$ has to be an irrational number. On the other hand, the real line $\mathbb{R}$ with the usual metric is a complete metric space. By the Baire category theorem, the intersection of all $U_n$ and $G_n$ must be nonempty. Thus the intersection $\bigcap \limits_{n=0}^\infty U_n$ must contain more than rational numbers.

It follows that the set of rational numbers $\mathbb{Q}$ cannot be a $G_\delta$ set in $\mathbb{R}$. In fact, the discussion below will show that the in a complete metric space such as the real line, any dense $G_\delta$ set must be a “thick” set (see Theorem 3 below).
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Baire Spaces

The version of the Baire category theorem discussed above involves complete metric spaces. However, the ideas behind the Baire category theorem are topological in nature. The following is the conclusion of the Baire category theorem:

$(*) \ \ \ \ X$ is a topological space such that for each countable family $\left\{U_0,U_1,U_2,\cdots \right\}$ of open and dense sets in $X$, the intersection $\bigcap \limits_{n=0}^\infty U_n$ is dense in $X$.

A Baire space is a topological space in which the condition $(*)$ holds. The Baire category theorem as stated above gives a sufficient condition for a topological space to be a Baire space. There are plenty of Baire spaces that are not complete metric spaces, in fact, not even metric spaces. The condition $(*)$ is a topological property. In order to delve deeper into this property, let’s look at some related notions.

Let $X$ be a topological space. A set $A \subset X$ is dense in $X$ if every open subset of $X$ contains a point of $A$ (i.e. $\overline{A}=X$). A set $A \subset X$ is nowhere dense in $X$ if for every open subset $U$ of $X$, there is some open set $V \subset U$ such that $V$ contains no point of $A$ (another way to describe this: $\overline{A}$ contains no interior point of $X$).

A set is dense if its points can be found in every nonempty open set. A set is nowhere dense if every nonempty open set has an open subset that misses it. For example, the set of integers $\mathbb{N}$ is nowhere dense in $\mathbb{R}$.

A set $A \subset X$ is of first category in $X$ if $A$ is the union of countably many nowhere dense sets in $X$. A set $A \subset X$ is of second category in $X$ if it is not of first category in $X$.

To make sense of these notions, the following observation is key:

$(**) \ \ \ \ F \subset X$ is nowhere dense in $X$ if and only if $X-\overline{F}$ is an open and dense set in $X$.

So in a Baire space, if you take away any countably many closed and nowhere dense sets (in other words, taking away a set of first category in $X$), there is a remainder (there are still points remaining) and the remainder is still dense in $X$. In thinking of sets of first category as “thin”, a Baire space is one that is considered “thick” or “fat” in that taking away a “thin” set still leaves a dense set.

A space $X$ is of second category in $X$ means that if you take away any countably many closed and nowhere dense sets in $X$, there are always points remaining. For a set $Y \subset X$, $Y$ is of second category in $X$ means that if you take away from $Y$ any countably many closed and nowhere dense sets in $X$, there are still points remaining in $Y$. A set of second category is “thick” in the sense that after taking away a “thin” set there are still points remaining.

For example, $\mathbb{N}$ is nowhere dense in $\mathbb{R}$ and thus of first category in $\mathbb{R}$. However, $\mathbb{N}$ is of second category in itself. In fact, $\mathbb{N}$ is a Baire space since it is a complete metric space (with the usual metric).

For example, $\mathbb{Q}$ is of first category in $\mathbb{R}$ since it is the union of countably many singleton sets ($\mathbb{Q}$ is also of first category in itself).

For example, let $T=[0,1] \cup (\mathbb{Q} \cap [2,3])$. The space $T$ is not a Baire space since after taking away the rational numbers in $[2,3]$, the remainder is no longer dense in $T$. However, $T$ is of second category in itself.

For example, any Cantor set defined in the real line is nowhere dense in $\mathbb{R}$. However, any Cantor set is of second category in itself (in fact a Baire space).

The following theorems summarize these concepts.

Theorem 1a
Let $X$ be a topological space. The following conditions are equivalent:

1. $X$ is of second category in itself.
2. The intersection of countably many dense open sets is nonempty.

Theorem 1b
Let $X$ be a topological space. Let $A \subset X$. The following conditions are equivalent:

1. The set $A$ is of second category in $X$.
2. The intersection of countably many dense open sets in $X$ must intersect $A$.

Theorem 2
Let $X$ be a topological space. The following conditions are equivalent:

1. $X$ is a Baire space, i.e., the intersection of countably many dense open sets is dense in $X$.
2. Every nonempty open subset of $X$ is of second category in $X$.

The above theorems can be verified by appealing to the relevant definitions, especially the observation $(**)$. Theorems 2 and 1a indicate that any Baire space is of second category in itself. The converse is not true (see the space $T=[0,1] \cup (\mathbb{Q} \cap [2,3])$ discussed above).

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Dense G delta Subsets of a Baire Space

In answering the question stated at the beginning, we have shown that $\mathbb{Q}$ cannot be a $G_\delta$ set. Being a set of first category, $\mathbb{Q}$ cannot be a dense $G_\delta$ set. In fact, it can be shown that in a Baire space, any dense $G_\delta$ subset is also a Baire space.

Theorem 3
Let $X$ be a Baire space. Then any dense $G_\delta$ subset of $X$ is also a Baire space.

Proof
Let $Y=\bigcap \limits_{n=0}^\infty U_n$ where each $U_n$ is open and dense in $X$. We show that $Y$ is a Baire space. In light of Theorem 2, we show that every nonempty open set of $Y$ is of second category in $Y$.

Suppose that there is a nonempty open subset $U \subset Y$ such that $U$ is of first category in $Y$. Then $U=\bigcup \limits_{n=0}^\infty W_n$ where each $W_n$ is nowhere dense in $Y$. It can be shown that each $W_n$ is also nowhere dense in $X$.

Since $U$ is open in $Y$, there is an open set $U^* \subset X$ such that $U^* \cap Y=U$. Note that for each $n$, $F_n=X-U_n$ is closed and nowhere dense in $X$. Then we have:

$\displaystyle (1) \ \ \ \ \ U^*=\bigcup \limits_{n=0}^\infty (F_n \cap U^*) \cup \bigcup \limits_{n=0}^\infty W_n$

$(1)$ shows that $U^*$ is the union of countably many nowhere dense sets in $X$, contracting that every nonempty open subset of $X$ is of second category in $X$. Thus we can conclude that every nonempty open subset of $Y$ is of second category in $Y$. $\blacksquare$

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Revised July 3, 2019