A topological space is a Baire space if the intersection of any countable family of open and dense sets in is dense in (or equivalently, every nonempty open subset of is of second category in ). One version of the Baire category theorem implies that every complete metric space is a Baire space. The real line with the usual Euclidean metric is a complete metric space, and hence is a Baire space. The space of irrational numbers is also a complete metric space (not with the usual metric but with another suitable metric that generates the Euclidean topology on ) and hence is also a Baire space. In this post, we show that there are subsets of the real line that are Baire space but not complete metric spaces. These sets are called Bernstein sets.
A Bernstein set, as discussed here, is a subset of the real line such that both and intersect with every uncountable closed subset of the real line. We present an algorithm on how to generate such a set. Bernstein sets are not Lebesgue measurable. Our goal here is to show that Bernstein sets are Baire spaces but not weakly -favorable, and hence are spaces in which the Banach-Mazur game is undecidable.
Baire spaces are defined and discussed in this post. The Banach-Mazur game is discussed in this post. The algorithm of constructing Bernstein set is found in  (Theorem 5.3 in p. 23). Good references for basic terms are  and .
In constructing Bernstein sets, we need the following lemmas.
In the real line , any uncountable closed set has cardinality continuum.
In the real line, every uncountable subset of the real line has a limit point. In fact every uncountable subset of the real line contains at least one of its limit points (see The Lindelof property of the real line). Let be an uncountable closed set. The set has to contain at least one of its limit point. As a result, at most countably many points of are not limit points of . Take away these countably many points of that are not limit points of and call the remainder . The set is still an uncountable closed set but with an additional property that every point of is a limit point of . Such a set is called a perfect set. In Perfect sets and Cantor sets, II, we demonstrate a procedure for constructing a Cantor set out of any nonempty perfect set. Thus (and hence ) contains a Cantor set and has cardinality continuum.
In the real line , there are continuum many uncountable closed subsets.
Let be the set of all open intervals with rational endpoints, which is a countable set. The set is a base for the usual topology on . Thus every nonempty open subset of the real line is the union of some subcollection of . So there are at most continuum many open sets in . Thus there are at most continuum many closed sets in . On the other hand, there are at least continuum many uncountable closed sets (e.g. for ). Thus we can say that there are exactly continuum many uncountable closed subsets of the real line.
Constructing Bernstein Sets
Let denote the cardinality of the real line . By Lemma 2, there are only many uncountable closed subsets of the real line. So we can well order all uncountable closed subsets of in a collection indexed by the ordinals less than , say . By Lemma 1, each has cardinality . Well order the real line . Let be this well ordering.
Based on the well ordering , let and be the first two elements of . Let and be the first two elements of (based on ) that are different from and . Suppose that and that for each , points and have been selected. Then is nonempty since has cardinality and only less than many points have been selected. Then let and be the first two points of (according to ). Thus and can be chosen for each .
Let . Then is a Bernstein set. Note that meets every uncountable closed set with the point and the complement of meets every uncountable closed set with the point .
The algorithm described here produces a unique Bernstein set that depends on the ordering of the uncountable closed sets and the well ordering of .
Baire spaces are defined and discussed in this previous post. Baire spaces can also be characterized using the Banach-Mazur game. The following lemmas establish that any Bernstein is a Baire space that is not weakly -favorable. Lemma 3 is applicable to all topological spaces. Lemmas 4, 5, 6, and 7 are specific to the real line.
Let be a topological space. Let be a set of first category in . Then contains a dense subset.
Let be a set of first category in . Then where each is nowhere dense in . The set is a dense set in the space and it is contained in the complement of . We have:
We now set up some notaions in preparation of proving Lemma 4 and Lemma 7. For any set , let be the interior of the set . Denote each positive integer by . In particular, . Let denote the collection of all functions . Identify each by the sequence . This identification makes notations in the proofs of Lemma 4 and Lemma 7 easier to follow. For example, for , denotes a closed interval . When we choose two disjoint subintervals of this interval, they are denoted by and . For , refers to , refers to the sequence , and refers to the sequence and so on.
The Greek letter denotes the first infinite ordinal. We equate it as the set of all nonnegative integers . Let denote the set of all functions from to .
Let be a dense set. Let be a nonempty open subset of . Then contains a Cantor set (hence an uncountable closed subset of the real line).
Let where each is an open and dense subset of . We describe how a Cantor set can be obtained from the open sets . Take a closed interval . Let . Then pick two disjoint closed intervals and such that they are subsets of the interior of and such that the lengths of both intervals are less than . Let .
At the step, suppose that all closed intervals (for all ) are chosen. For each such interval, we pick two disjoint closed intervals and such that each one is subset of and each one is subset of the interior of the previous closed interval and such that the lenght of each one is less than . Let be the union of over all .
Then is a Cantor set that is contained in .
Let . If is not of second category in , then contains an uncountable closed subset of .
Suppose is of first category in . By Lemma 3, the complement of contains a dense subset. By Lemma 4, the complement contains a Cantor set (hence an uncountable closed set).
Let . If is not a Baire space, then contains an uncountable closed subset of .
Suppose is not a Baire space. Then there exists some open set such that is of first category in . Let be an open subset of such that . We have where each is nowhere dense in . It follows that each is nowhere dense in too.
By Lemma 3, contains , a dense subset of . By Lemma 4, there is a Cantor set contained in . This uncountable closed set is contained in .
Let . Suppose that is a weakly -favorable space. If is dense in the open interval , then there is an uncountable closed subset of such that .
Suppose is a weakly -favorable space. Let be a winning strategy for player in the Banach-Mazur game . Let be an open interval in which is dense. We show that a Cantor set can be found inside by using the winning strategy .
Let . Let . Let and . We take as the first move by the player . Then the response made by is . Let .
Choose two disjoint closed intervals and that are subsets of the interior of such that the lengths of these two intervals are less than and such that and satisfy further properties, which are that and are open in . Let and be two possible moves by player at the next stage. Then the two possible responses by are and . Let .
At the step, suppose that for each , disjoint closed interval have been chosen. Then for each , we choose two disjoint closed intervals and , both subsets of the interior of , such that the lengths are less than , and:
- and ,
- and are open in ,
We take and as two possible new moves by player from the path . Then let the following be the responses by player :
The remaining task in the induction step is to set .
Let , which is a Cantor set, hence an uncountable subset of the real line. We claim that .
Let . There there is some such that . The closed intervals are associated with a play of the Banach-Mazur game on . Let the following sequence denote this play:
Since the strategy is a winning strategy for player , the intersection of the open sets in must be nonempty. Thus .
Since the sets , and since the lengths of go to zero, the intersection must have only one point, i.e., for some . It also follows that . Thus . We just completes the proof that contains an uncountable closed subset of the real line.
Conclusions about Bernstein Sets
Lemma 6 above establishes that any Bernstein set is a Baire space (if it isn’t, the complement would contain an uncountable closed set). Lemma 7 establishes that any Bernstein set is a topological space in which the player has no winning strategy in the Banach-Mazur game (if player always wins in a Bernstein set, it would contain an uncountable closed set). Thus any Bernstein set cannot be a weakly favorable space. According to this previous post about the Banach-Mazur game, Baire spaces are characterized as the spaces in which the player has no winning strategy in the Banach-Mazur game. Thus any Bernstein set in a topological space in which the Banach-Mazur game is undecidable (i.e. both players in the Banach-Mazur game have no winning strategy).
One interesting observation about Lemma 6 and Lemma 7. Lemma 6 (as well as Lemma 5) indicates that the complement of a “thin” set contains a Cantor set. On the other hand, Lemma 7 indicates that a “thick” set contains a Cantor set (if it is dense in some open interval).
- Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
- Oxtoby, J. C., Measure and Category, Graduate Texts in Mathematics, Springer-Verlag, New York, 1971.
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