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I would like to find an exercise of the type: Find some $x$ in $A\setminus B$. Solution: since $A$ is uncountable and $B$ is countable such $x$ exists...

Martin Sleziak
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Emanuele Paolini
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4 Answers4

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  • There is a function $f : \mathbb{N} \to \mathbb{N}$ not computable by a Turing machine. (Or there is a real number whose decimal expansion cannot be so computed.)
  • There is a non-Borel subset of $\mathbb{R}$. (Cardinality $\mathfrak{c} = 2^{\aleph_0}$ vs. cardinality $2^{\mathfrak{c}}$.)
user642796
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  • You are a responsible person, what with having kids and all that, so I appoint you to make sure that every KGRC member (student, or otherwise coughSycough) which is coming to YST2013 has at least two pickles for me! (I will reimburse everyone, of course) – Asaf Karagila Apr 04 '13 at 11:34
  • @Asaf: I've rounded up one person so far! (Quick question: would students of the Technische Universität Wien count as KGRC members if they attend most of the events here?) – user642796 Apr 04 '13 at 12:49
  • Yes. The more the merrier. – Asaf Karagila Apr 04 '13 at 12:56
  • Nice, thanks. In fact the first example can be stated also in a informal way as follows: "there are numbers which are not uniquely defined as the solution of a mathematical problem". – Emanuele Paolini Apr 04 '13 at 16:25
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    @Emanuele: That's a horrible way of stating this. – Asaf Karagila Apr 04 '13 at 16:59
  • @EmanuelePaolini: But any real number $a$ is the unique solution to the following mathematical problem: What number when subtracted from $a$ yields $0$? (You can, however, limit the sorts of operations allowed and what numbers can be explicitly referenced to get in something very close in spirit to the algebraic/transcendental dichotomy.) – user642796 Apr 04 '13 at 17:20
  • @Asaf: As long as it’s not this KGRC! – Brian M. Scott Apr 04 '13 at 21:08
  • @Brian: Obviously. The KGRC I am talking about is a wonderful place to get wonderful pickles! – Asaf Karagila Apr 04 '13 at 21:12
  • @ArthurFischer: The problem "What number when subtracted from a yields 0" is not well defined if you don't specify what number is a. – Emanuele Paolini Apr 05 '13 at 17:24
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The classical result presented by Cantor himself: Prove that there exists a real number that is not algebraic.

Remark: the fact that non-algebraic numbers exist was known before, but Cantor presented the proof of the uncountability of the reals and derived from it a very simple existence proof exactly using what you are asking about, using such a technique for the first time.

Ittay Weiss
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  • Yes, I already consider the countability of algebraics but I didn't thought to present the complementary result. Thanks. – Emanuele Paolini Apr 04 '13 at 10:22
  • However everybody knows that $\pi$ is trascendental (even if he maybe cannot prove it), so this example is not very surprising. Is there something else? I was thinking about continuous functions... or maybe something related to the Cantor set. – Emanuele Paolini Apr 04 '13 at 10:30
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    It is very hard to prove $\pi$ is transcendental. It's also quite hard to prove $e$ is. It's not too difficult giving an explicit transcendental number. But the cardinality argument is much simpler, it's an argument of a different flavour than any analytic proof, it's the first one ever presented, and it's exactly along the lines of what you are asking. I need to borrow Asaf's mind reading cap, but he left it in his other pants. – Ittay Weiss Apr 04 '13 at 10:35
  • @Emanuele: I gave an example related to the Cantor set. – Asaf Karagila Apr 04 '13 at 11:19
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Some ideas:

  1. Prove that there exists an irrational number.
  2. Prove that there exists a sequence of irrational numbers converging to any real number.
  3. Prove that there exists a subset of the integers which is neither finite nor co-finite.
  4. Prove that there exists a function $f\colon\Bbb{R\to R}$ which is discontinuous everywhere (replace "countable" with "size continuum" and uncountable with "larger than the continuum").
  5. There exists a number in the Cantor set which is not the endpoint of an interval disjoint from the Cantor set (the complement of the Cantor set can be written as a countable union of disjoint intervals, so there are only a countable number of endpoints which are elements of the Cantor set).
  6. There exists a normal number.
  7. There exists a linear functional on $(\Bbb R[x])^\ast$ which is not an evaluation functional (the dimension of $\Bbb R[x]$ is countable, and therefore the dimension of the evaluation functionals is countable; but $(\Bbb R[x])^{**}$ has an uncountable dimension).
Martin Sleziak
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Asaf Karagila
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  • 1 is usually achieved by showing the irrationality of $\sqrt 2$... I wouldn't use cardinals there. The same for 3 where a simple example is given by even numbers. In 2 also I would make an explicit example... I don't see how to do it with cardinals. – Emanuele Paolini Apr 04 '13 at 10:25
  • @Emanuele: Let me quote your question "I would like to find an exercise of the type: Find some $x$ in $A\setminus B$. Solution: since $A$ is more than countable and $B$ is countable such $x$ exists...", you can prove the existence of irrational numbers by cardinality arguments. That is a fact. You can also prove $2$ and $3$ directly. But you can also prove them by a cardinality argument. – Asaf Karagila Apr 04 '13 at 10:26
  • If you want ideas that are more easily or commonly proved by cardinality games, write that in the question. I left my mind reading cap in my other pants. – Asaf Karagila Apr 04 '13 at 10:27
  • you actually have a mind reading cap?!?!? I want one! – Ittay Weiss Apr 04 '13 at 10:28
  • @Ittay: There exists only one, and it (including the schematics for building one - and an anti-mind reading cap) are in my other pants. – Asaf Karagila Apr 04 '13 at 10:29
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    Re 4: I don't think a cardinality argument works here. There are $2^\mathfrak{c}$ functions $\mathbb{R} \to \mathbb{R}$ that are continuous at $0$ (outside, say, $(-1,1)$ the function can be arbitrarily behaved). – user642796 Apr 04 '13 at 10:36
  • @Arthur: Yeah, you're correct. I must have been thinking something else. – Asaf Karagila Apr 04 '13 at 10:37
  • @AsafKaragila: May I ask to have your solution in http://math.stackexchange.com/a/350933/8581 visible. Please let us to have it again. Sorry for asking here. – Mikasa Apr 04 '13 at 10:45
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    Re 6: The usual proof of the existence of normal numbers a Lebesgue measure argument. (Since being a non-normal number just requires non-normality with respect to a single base, the reals in $[0,1]$ which have decimal expansions using only the digits $4,5,6,7$ would all be non-normal, and there are $4^{\aleph_0} \geq 2^{\aleph_0} = | \mathbb{R} |$ many of these.) – user642796 Apr 04 '13 at 12:56
  • @Arthur: Hm yeah. I remember there was something countable there, but I suppose that was the countable union of null sets. On a side note, do you claim that the set of numbers in $[0,1]$ whose decimal expansion has only $4,5,6,7$ has a strictly larger cardinality than the continuum? :-) – Asaf Karagila Apr 04 '13 at 13:27
  • @Asaf: my need was to find some examples which are well suited to explain why arguments using cardinality might be useful. This maybe was not clear in the question. I like 5. While 7 is maybe outside my knowledge. – Emanuele Paolini Apr 04 '13 at 16:29
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    Asaf, here are a couple of more which I'll let you write up (because I'm at work and busy): (1) See the paragraph beginning with Incidentally, one way to answer Rudin's question in my answer at this post. comment continues – Dave L. Renfro Apr 04 '13 at 16:30
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    continuation (2) In elementary calculus we can easily find examples to show that every subset of ${0,,1}$ can be a convergence set for a power series in $\mathbb R$ centered at $x=0$ and with a radius of convergence equal to $1$. What about power series in $\mathbb C$? Note there are $c$ many power series in $\mathbb C$ and $2^c$ many subsets of the boundary of the unit disk. For results more precise than those obtainable by cardinality, see this math overflow post. – Dave L. Renfro Apr 04 '13 at 16:31
  • Of course, in my last comment ${0,,1}$ should be ${-1,,1} \ldots$ – Dave L. Renfro Apr 04 '13 at 16:37
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One example which I quite like is this:

Show that there are points in the plane which cannot be constructed from unit segment using just compass and straightedge.

The cardinality based proof is simple: We are starting with two points, at each step of construction we can only add finitely many new points and we employ constructions which only have finitely many steps. So the set of all constructible points is countable. The set of all points in the plane has cardinality of continuum, it is uncountable.

The students will probably see in some algebra course1 proofs showing that some particular lengths (such as $\sqrt[3]2$ or $\cos\pi/9$ cannot be constructed). But this proof can be considered simple and it also works for other constructions of similar type. (I.e. constructions with finitely many steps where only finitely many new objects are added in each step.)

1 I would guess that typically cardinalities appear in a curriculum before the course of algebra which includes result needed for proving impossibility of trisecting an angle or doubling a cube. But perhaps somewhere these topics are taught in a different order.

Martin Sleziak
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