A core-compact space is one whose lattice of open subsets is continuous. A locally compact space is one where each point has a base of compact neighborhoods. Every locally compact space is core-compact (Theorem 5.2.9 in the book), and the converse holds if the space is sober (Theorem 8.3.10). In fact, the sobrification of a core-compact space is locally compact (Proposition 8.3.11).
But are there any core-compact spaces that would fail to be locally compact?
The answer is yes, as given in Exercise V-5.25 of the red book [1], following Hofmann and Lawson [2]. With Zhenchao Lyu, we did the exercise, and I realized it was harder than it looked. Hence—and sorry if I’m spoiling the fun of doing the exercise—I will make every step explicit—with a proviso: I will assume that you know what ordinals are, what a σ-algebra is, what the Borel σ-algebra of a topological space is (the smallest σ-algebra containing all the open sets), and that you know what Lebesgue measure is on R and some of its properties.
This month, we will give an explanation of the following sentence in the statement of the exercise:
By heavy use of the axiom of choice we pick a dense subset A of [0,1) such that A ∩ U is not a Borel set (or, if one prefers, not even Lebesgue measurable) for every nonempty open subset U of the unit interval.
As it turns out, a Bernstein set—a highly pathological subset of R—will fit the bill. My plan this month is to explain what a Bernstein set is, how we can construct one, and what its properties are. A good source of information is [3] (Examples 1.22 and 1.23). Another one is Dan Ma’s topology blog.
The continuum hypothesis, isolated points, and perfect sets
Let c be the cardinality of R. Georg Cantor had observed that the closed subsets of R can only be finite, countable, or of cardinality c. (The argument is entirely standard, and if you already know it, please proceed to the next section.)
That may not seem difficult to see: if you try to build an uncountable set, you will come up with something like R, or the set of all subsets of N, or the set of all infinite sequences of natural numbers, etc., and they all have cardinality c. But it might be that there are other closed uncountable sets of cardinalities different from c.
That does not happen, but requires a bit of effort to prove. The same property for arbitrary subsets of R (“every uncountable subset of R has cardinality c“) is… in a bizarre state. No, not wrong, but not right either. Let me explain.
Cantor had conjectured exactly that: every uncountable subset of R has cardinality c; equivalently, there is no set of cardinality strictly between that of N (=countable) and c. That is the continuum hypothesis, and is an extremely hard problem… to the point that it is independent of ZFC set theory: you cannot prove it, and you cannot prove its negation, as shown by Paul Cohen in 1963. (The first time I heard about it, I was a teen, and the text said something like [from memory] “Cantor spent a lot of effort on the question, then became insane, and remained so until his demise; one may wonder about the link between the two events”…)
Still, Cantor had proved that there is no closed set of any of those intermediate cardinalities, as I announced first. The usual way one proceeds is by showing that every closed subset C of R is the disjoint union of a countable set A and a so-called perfect set P.
A is obtained as the set of isolated points of C: those points x in C such that there is an open interval (x–ε,x+ε) with ε>0 whose intersection with C contains only x. Using the axiom of choice, we can pick a rational number qx in that interval, for each isolated point x in C. It is not hard to see that the map x∈A ↦ qx is injective, so A is countable.
We define P as C–A. By construction, P contains no isolated point. (Oops: no, mistake spotted on April 15, 2022. See below for the fix.) It is closed, too, because P is also equal to C minus the union of the open intervals (x–ε,x+ε) we just built, when x ranges over A. A closed set with no isolated point is called a perfect set.
(Oops, P may of course still contain some isolated point—mistake spotted on April 15, 2022. This is an entirely classical mistake, and I should not have made it. For example, if C consists of the points 1, 1/2, 1/4, …, 1/2n, …, plus 0, then its isolated points are all of them except 0, so P={0}, whose only point is isolated. The problem lies in the fact that, by removing some isolated points, some non-isolated points may become isolated. Anyway, the fix is as follows. Given a closed set C, define A as its set of isolated points as above, and C’, its derived set, as C–A. Then C’ is closed, and is included in C. This defines a map C ↦ C’ which maps closed sets to smaller closed sets. It has a largest fixed point included in C, which happens to be P. Showing that C–P is countable requires a bit more work: typically one shows that the fixed point is obtained by ordinal induction up to ε0, removing only countably many points at each step, and noticing that ε0 is countable.)
Now we observe that every non-empty perfect set P has cardinality c. If P contains a closed interval [a,b] with a<b, then this is clear since [a,b] already has cardinality c. Hence it suffices to consider those non-empty perfect sets P with empty interior.
There are indeed some non-empty perfect sets with empty interior. The prototypical example is Cantor’s middle-third-out set. That is obtained from a closed interval, say C0=[0,1], by removing the middle interval (1/3,2/3). From each of the two remaining intervals [0,1/3] and [2/3,1] (whose union we call C1), we carve out the middle interval again, yielding four intervals [0,1/9], [2/9,1/3], [2/3,7/9] and [8/9,1] (whose union we call C2). And we continue in the same way. The intersection of the sets Cn is Cantor’s set. It is closed, since it is an intersection of closed sets. It contains exactly the numbers between 0 and 1 that can be written in base 3 using only the digits 0 and 2 but not 1, from which we obtain that Cantor’s set has cardinality 2 to the cardinality of N (that is equal to c), has no isolated point, and has empty interior.
Let us return to our investigation of non-empty perfect sets with empty interior. Note that:
- A non-empty perfect set must be infinite: because every finite set consists of isolated points only.
- Between any two points x<y of a set P with empty interior, there must be a point z that is not in P: otherwise the interval [x,y] would be included in P.
- The intersection of a perfect set P with an open interval (a,b) has no isolated point: if there were an isolated point x in P ∩ (a,b), some open interval (x–ε,x+ε) would meet P ∩ (a,b) at x only; replacing ε by min(ε,x–a,b–y), we can assume that (x–ε,x+ε) is entirely included in (a,b), and then it can meet P at x only, showing that x is isolated in P.
- The intersection of a perfect set P with empty interior with a closed interval [a,b] with a<b, and such that neither a nor b is in P, is again perfect and has empty interior. It is clearly closed, and has empty interior. It also perfect, because it is also equal to the intersection of (a,b) with P (owing to the fact that a and b are outside P), and using the previous point.
Now we pick a non-empty perfect set P with empty interior, and we show that it contains a set of cardinality c as follows. The construction looks a lot like the construction of Cantor’s set, and for good reasons.
- Since non-empty perfect sets are infinite, pick three points x<y<z<t from P. Because of the second item above, we can find three points a, b, c outside P such that x<a<y<b<z<c<t. We define P0 as P ∩ [a,b] and P1 as P ∩ [b,c]. We have just seen that those are again perfect sets with non-empty interior. They are disjoint because b is not in P, and they are non-empty because they contain y, resp. z.
- We repeat the process, and find two disjoint, non-empty perfect sets with non-empty interior P00 and P01 included in P0, and similarly, two disjoint, non-empty perfect sets with non-empty interior P10 and P11 included in P1. We repeat again, building P000, P001, P010, P011, P100, P101, P110, and P111, then again, building P0000, …, P1111, and so on.
- Given any infinite sequence of bits (indices in {0, 1}) b1b2…bn…, the sets Ps, where s ranges over the finite prefixes b1b2…bn of that sequence, form a descending sequence of non-empty closed sets. Since they are all included in the compact interval [x,t] (where x and t were chosen when we built P0 and P1), the intersection of those sets is non-empty. Pick one element from the intersection. This yields a point for each infinite sequence of bits, and any two distinct sequences of bits must yield distinct point (those two sequences must differ at some finite rank, namely one must start with a prefix s of the form b1b2…bn0 while the other one starts with the prefix t=b1b2…bn1; then realize that Ps and Pt are disjoint).
- The points we have found therefore form a family whose cardinality is 2 to the powerset of N, and that is c. Those points are all in P, so P has cardinality c, and we are done.
If we had worked a bit harder, we could have made sure that Ps, is included in an interval of width at most 1/2n, for every finite sequence s of length n. In this way, we would not have to pick a point in the intersection of Ps, where s ranges over the finite prefixes of any given infinite sequence of bits: that intersection would contain exactly one point.
The set of points we have built is not just of cardinality c, it is homeomorphic to Cantor’s middle-third-out set. That applies to any non-empty perfect subset P of R, even if P has non-empty interior: in that case, P contains an interval [a,b] with a<b, and the latter already contains a copy of Cantor’s set. We will however not need those observations in the sequel.
Bernstein sets
A Bernstein set is defined as a set of real numbers B such that, whichever uncountable closed subset C of R you take, B intersects C but does not contain it (equivalently, B meets both C and its complement).
This is relatively easy to build, by ordinal induction. In brief, we enumerate all uncountable closed subsets of R, pick two points in each that we have not picked in any of the previously enumerated closed sets, and add the first one but not the second one to B, starting from the empty set.
A bit more formally, we need to be clear about the cardinalities of various collections.
We have seen that every uncountable closed subset of R has cardinality exactly c. Good.
Also, there are exactly c uncountable closed subsets of R. Indeed, the topology of R has a base of open intervals with rational endpoints. Hence every open subset of R is a (countable) union of such open intervals. There are c countable lists of pairs of rational numbers, so there are at most c open subsets of R. Hence there are at most c closed subsets of R, hence also at most c uncountable closed subsets of R. There are at least c uncountable closed subsets of R, too, for example the closed intervals [x–1, x+1], x∈R, already form a subcollection of cardinality c.
It follows that we can enumerate all the uncountable closed subsets of R as Cα, where α ranges over the ordinals <c. (A cardinal number such as c is also an ordinal. In fact the usual definition is that a cardinal number is an ordinal β such that no ordinal α<β is in bijection with β—remember that ordinals are sets, so bijections make sense.)
We can also enumerate all the real numbers as xα, α<c. This defines a total ordering ≺ on R by xα≺xβ if and only if α<β. (In the theory of ordinals, we actually first show that every set can be ordered by a total, well-founded ordering—this is Zermelo’s theorem—and then one shows that every total well-founded poset is isomorphic to a unique cardinal.)
By induction on α, we build two sets Bα and Nα. Bα will be the set of points taken step α, and Nα will be the set of points already picked but not taken at step α. The union of the sets Bα will be B. Importantly, those two sets will all have cardinality strictly less than c. Otherwise, the induction would fail miserably at some point. We actually have to show a more precise invariant: that there are bijections between α (as a set), Bα, and Nα. I will say later how, but this is a technical formality that would obscure the argument somehow.
We start by letting B0 and N0 be empty. At step α+1, we realize that Cα is uncountable, and being closed, that it has cardinality c. Since Bα and Nα both have cardinality strictly less than c, the cardinality of Bα ∪ Nα is also strictly less than c. Therefore, there is a point in Cα that is not in Bα ∪ Nα. In fact, adding countably many points to Bα ∪ Nα would not allow us to obtain a set of cardinality c, so there are at least countably infinitely many points in Cα that are not in Bα ∪ Nα. (In fact there are c such points.) Hence there are at least two. Take the first two in our enumeration xα of real numbers that are in Cα but not in Bα ∪ Nα. Add the first one to Bα, yielding Bα+1, and the second one to Nα, yielding Nα+1.
At step β, where β is a limit ordinal, define Nβ as the union of the sets Nα with α<β, and Bβ as the union of the sets Bα with α<β.
I have said earlier that we have to establish a stronger invariant that just saying that Bα and Nα are of cardinality <c. It is even stronger than what I said: we must say (at least) that, for each α, we have collections of bijections fγ:γ→Bγ and gγ:γ→Nγ for all γ≤α, which behave well, in the sense that for all ordinals γ<δ≤α, the restriction of fδ to γ is exactly fγ, and similarly for gδ and gγ. I’ll let you do that if you are into checking every detail.
Finally, let B be the union of the sets Bα with α<c. This will be our Bernstein set, as we now argue.
Every closed uncountable subset of R is a Cα for some α<c. Of the two points we picked in Cα to construct Bα+1 and Nα+1, one is in B (being in Bα+1) and the other is not (being in Nα+1 instead). Hence B is a Bernstein set, as promised.
A few properties of Bernstein sets
Fact. A Bernstein set B has no isolated point.
Indeed, imagine it had one, x. For some ε>0, (x–ε,x+ε) intersects B at x only. This is impossible, since for example [x+ε/3,x+2ε/3], which is an uncountable closed set, must intersect B.
Fact. A Bernstein set B is dense in R, and its complement is dense too.
Indeed, consider any non-empty open subset of R. It contains an open interval (x–ε,x+ε), hence the uncountable closed subset [x–ε/2,x+ε/2], which must intersect B, and also the complement of B.
Fact. The complement of a Bernstein set is Bernstein.
(That is obvious.)
One can show that a Bernstein set is not Borel, and in fact more:
Proposition. The intersection of a Bernstein set B with any Borel subset A of R of positive Lebesgue measure λ(A)>0 is not Borel.
Wise and Hall [3, Example 1.23] even show the more general statement obtained by replacing “Borel” by “Lebesgue measurable”, and cite Oxtoby as their source.
Proof. Henceforth let B be a Bernstein set, and A be a Borel subset of R.
Given any closed subset C of R included in B, C must be countable. Indeed, if it were uncountable, it would intersect the complement of B, by definition of Bernstein sets.
Since Lebesgue measure λ is σ-additive, λ(C) is the countable sum of λ({x}), x in C, and all those summands are zero, so every closed subset C included in B has Lebesgue measure 0.
Lebesgue measure has the marvelous property that it is inner regular (or tight): for every Borel subset S of R, λ(S) is the supremum of the values λ(K), where K ranges over the compact subsets of S. (In fact, every σ-finite measure on a Polish space is inner regular.) Since the compact subsets of R are its closed bounded subsets, all the compact subsets of B have Lebesgue measure 0. Hence, if S is any Borel subset of R included in B, then λ(S)=0.
The same holds for the Borel subsets of the complement of B, since that complement is also Bernstein.
Now imagine that B ∩ A were Borel. Then λ(B ∩ A)=0 (because B ∩ A is Borel, and included in B) and λ(A – B)=0 (because A – B is Borel, and included in the complement of B), so λ(A)=λ(B ∩ A)+λ(A – B)=0, contradicting our assumption that λ(A)>0. ☐
From this proposition, we retrieve that Bernstein sets are dense: the intersection of a Bernstein set B with any non-empty open set U (which is Borel and contains an open interval of non-zero Lebesgue measure) cannot be Borel… and therefore cannot be empty.
We can now prove that there exists a dense subset A of [0,1) such that A ∩ U is not Borel for any non-empty open subset U of the unit interval, as promised at the beginning of this post: we just pick the intersection of a Bernstein set B with the (Borel) interval [0,1) for A. For every non-empty open set U of the unit interval, U contains a point x in (0,1), and then a small interval (x–ε,x+ε) around it. We may even require ε>0 to be so small that (x–ε,x+ε) is included in (0,1). Then A ∩ (x–ε,x+ε) = B ∩ (x–ε,x+ε) is not Borel, by the Proposition above. That A is dense in [0,1) is immediate: for every non-empty open subset U of [0,1), A ∩ U is not Borel, and can therefore not be empty.
Next month, we will see how the rest of Exercise V-5.25 of [1] can be done. The mathematics will be very different, and much more domain-theoretic.
- Gerhard Gierz, Karl Heinrich Hofmann, Klaus Keimel, Jimmie D. Lawson, Michael W. Mislove, and Dana S. Scott. Continuous Lattices and Domains. Number 93 in Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 2003.
- Karl H. Hofmann and Jimmie D. Lawson. The Spectral Theory of Distributive Continuous Lattices. Transactions of the American Mathematical Society 246 (Dec. 1978), pages 285- 310.
- Gary L. Wise and Eric B. Hall. Counterexamples in probability and real analysis. Oxford University Press, 1993.
— Jean Goubault-Larrecq (January 21st, 2019)