{"id":1640,"date":"2019-01-21T20:24:40","date_gmt":"2019-01-21T19:24:40","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1640"},"modified":"2022-11-19T15:09:15","modified_gmt":"2022-11-19T14:09:15","slug":"bernstein-subsets-of-r","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1640","title":{"rendered":"Bernstein subsets of R"},"content":{"rendered":"

A core-compact space is one whose lattice of open subsets is continuous. \u00a0A locally compact space is one where each point has a base of compact neighborhoods. \u00a0Every locally compact space is core-compact (Theorem 5.2.9 in the book<\/a>), and the converse holds if the space is sober (Theorem 8.3.10). \u00a0In fact, the sobrification of a core-compact space is locally compact (Proposition 8.3.11).<\/p>\n

But are there any core-compact spaces that would fail to be locally compact?<\/p>\n

The answer is yes, as given in Exercise V-5.25 of the red book [1], following Hofmann and Lawson [2]. \u00a0With Zhenchao Lyu<\/a>, we did the exercise, and I realized it was harder than it looked. \u00a0Hence\u2014and sorry if I’m spoiling the fun of doing the exercise\u2014I will make every step explicit\u2014with a proviso: I will assume that you know what ordinals<\/a> are, what a\u00a0\u03c3-algebra is, what the Borel\u00a0\u03c3-algebra of a topological space is (the smallest\u00a0\u03c3-algebra containing all the open sets), and that you know what Lebesgue measure<\/a> is on\u00a0R<\/strong> and some of its properties.<\/p>\n

This month, we will give an explanation of the following sentence in the statement of the exercise:<\/p>\n

By heavy use of the axiom of choice we pick a dense subset\u00a0A<\/em> of [0,1) such that\u00a0A<\/em>\u00a0\u2229<\/em>\u00a0U<\/em> is not a Borel set (or, if one prefers, not even Lebesgue measurable) for every nonempty open subset\u00a0U<\/em> of the unit interval.<\/p><\/blockquote>\n

As it turns out, a Bernstein set<\/a>\u2014a highly pathological subset of\u00a0R<\/strong>\u2014will fit the bill. \u00a0My plan this month is to explain what a Bernstein set is, how we can construct one, and what its properties are. \u00a0A good source of information is [3] (Examples 1.22 and 1.23). \u00a0Another one is Dan Ma’s topology blog<\/a>.<\/p>\n

The continuum hypothesis, isolated points, and perfect sets<\/h2>\n

Let\u00a0c<\/strong> be the cardinality of\u00a0R<\/strong>. \u00a0Georg Cantor had observed that the closed subsets of\u00a0R<\/strong> can only be finite, countable, or of cardinality\u00a0c<\/strong>. \u00a0(The argument is entirely standard, and if you already know it, please proceed to the next section.)<\/p>\n

That may not seem difficult to see: if you try to build an uncountable set, you will come up with something like\u00a0R<\/strong>, or the set of all subsets of\u00a0N<\/strong>, or the set of all infinite sequences of natural numbers, etc., and they all have cardinality\u00a0c<\/strong>. \u00a0But it might be that there are other closed uncountable sets of cardinalities different from c<\/strong>.<\/p>\n

That does not happen, but requires a bit of effort to prove. \u00a0The same property for arbitrary subsets of\u00a0R<\/strong>\u00a0(“every uncountable subset of R<\/strong> has cardinality\u00a0c<\/strong>“) is… in a bizarre state. \u00a0No, not wrong, but not right either. \u00a0Let me explain.<\/p>\n

Cantor had conjectured exactly that: every uncountable subset of R<\/strong> has cardinality\u00a0c<\/strong>; equivalently, there is no set of cardinality strictly between that of\u00a0N<\/strong> (=countable) and\u00a0c<\/strong>. \u00a0That is the\u00a0continuum hypothesis<\/a>, 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. \u00a0(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”…)<\/p>\n

Still, Cantor had proved that there is no\u00a0closed<\/em> set of any of those intermediate cardinalities, as I announced first. \u00a0The usual way one proceeds is by showing that every closed subset C<\/em>\u00a0of\u00a0R<\/strong> is the disjoint union of a countable set A<\/em>\u00a0and a so-called perfect set<\/em> P<\/em>.<\/p>\n

A<\/em> is obtained as the set of isolated points<\/em> of\u00a0C<\/em>: those points\u00a0x<\/em> in\u00a0C<\/em> such that there is an open interval (x<\/em>\u2013\u03b5,x<\/em>+\u03b5) with\u00a0\u03b5>0 whose intersection with\u00a0C<\/em> contains only\u00a0x<\/em>. \u00a0Using the axiom of choice, we can pick a rational number q<\/em>x<\/em><\/sub>\u00a0in that interval, for each isolated point x<\/em> in\u00a0C<\/em>. \u00a0It is not hard to see that the map\u00a0x<\/em>\u2208A<\/em>\u00a0\u21a6 q<\/em>x<\/em><\/sub>\u00a0is injective, so\u00a0A<\/em> is countable.<\/p>\n

We define\u00a0P<\/em> as\u00a0C<\/em>\u2013A<\/em>. \u00a0By construction,\u00a0P<\/em> contains no isolated point. \u00a0(Oops: no, mistake spotted on April 15, 2022. \u00a0See below for the fix.) \u00a0It is closed, too, because\u00a0P<\/em> is also equal to\u00a0C<\/em> minus the union of the open intervals\u00a0(x<\/em>\u2013\u03b5,x<\/em>+\u03b5) we just built, when\u00a0x<\/em> ranges over\u00a0A<\/em>. \u00a0A closed set with no isolated point is called a\u00a0perfect set<\/em>.<\/p>\n

(Oops,\u00a0P<\/em> may of course still contain some isolated point\u2014mistake spotted on April 15, 2022. \u00a0This is an entirely classical mistake, and I should not have made it. \u00a0For example, if\u00a0C<\/em> consists of the points 1, 1\/2, 1\/4, …, 1\/2n<\/sup><\/em>, …, plus 0, then its isolated points are all of them except 0, so\u00a0P<\/em>={0}, whose only point is isolated. \u00a0The problem lies in the fact that, by removing some isolated points, some non-isolated points may become isolated. \u00a0Anyway, the fix is as follows. \u00a0Given a closed set\u00a0C<\/em>, define\u00a0A<\/em> as its set of isolated points as above, and\u00a0C’<\/em>, its\u00a0derived set<\/em>, as\u00a0C<\/em>\u2013A<\/em>. \u00a0Then\u00a0C’<\/em> is closed, and is included in\u00a0C<\/em>. \u00a0This defines a map\u00a0C<\/em> \u21a6\u00a0C’<\/em> which maps closed sets to smaller closed sets. \u00a0It has a largest fixed point included in\u00a0C<\/em>, which happens to be\u00a0P<\/em>. \u00a0Showing that\u00a0C<\/em>\u2013P<\/em> is countable requires a bit more work: typically one shows that the fixed point is obtained by ordinal induction up to \u03b50<\/sub>, removing only countably many points at each step, and noticing that \u03b50\u00a0<\/sub>is countable.)<\/p>\n

Now we observe that every non-empty perfect set P<\/em>\u00a0has cardinality c<\/strong>. \u00a0If\u00a0P<\/em> contains a closed interval\u00a0[a<\/em>,b<\/em>] with\u00a0a<\/em><b<\/em>, then this is clear since\u00a0[a<\/em>,b<\/em>] already has cardinality\u00a0c<\/strong>. \u00a0Hence it suffices to consider those non-empty perfect sets\u00a0P<\/em> with\u00a0empty interior<\/em>.<\/p>\n

There are indeed some non-empty perfect sets with empty interior. \u00a0The prototypical example is\u00a0Cantor’s middle-third-out set<\/a>. That is obtained from a closed interval, say\u00a0C<\/em>0<\/sub>=[0,1], by removing the middle interval (1\/3,2\/3). \u00a0From each of the two remaining intervals [0,1\/3] and [2\/3,1] (whose union we call C<\/em>1<\/sub>), 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 C<\/em>2<\/sub>). \u00a0And we continue in the same way. \u00a0The intersection of the sets Cn<\/sub><\/em>\u00a0is Cantor’s set. \u00a0It is closed, since it is an intersection of closed sets. \u00a0It 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\u00a0N<\/strong> (that is equal to\u00a0c<\/strong>), has no isolated point, and has empty interior.<\/p>\n

Let us return to our investigation of non-empty perfect sets with empty interior. \u00a0Note that:<\/p>\n