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At the end of the mathematical high-school education I usually introduce the easiest facts of set theory: counbtability and Cantor's proof as the basis of modern mathematics. Now my brightest student has opposed by mentioning that set theory contradicts the limit notion of analysis. His argument: In every interval (0, n] the ratio of not enumerated fractions and natural numbers is 0. We can consider this as a sequence: 0, 0, 0, ... which in analysis can only have the limit 0. According to set theory the limit is $\infty$ since no fraction in the interval (0, $\omega$) is not enumerated. This makes set theory useless as a basis of analysis. I could not disagree at the moment. What is a valid counter argument?

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    $\begingroup$ I think that the first thing you might want to do is very precisely define "ratio of not enumerated fractions and natural numbers". I am a little unsure about precisely what "not enumerated fractions" is supposed to mean. As it stands, the limit is obviously zero of this sequence, but so what? The limit of the $n$ in analysis is not $\omega$, but $\infty$, which are not the same thing - one might say that in analysis $\infty$ isn't a thing at all, in fact. $\endgroup$ – kcrisman Jul 10 '17 at 14:44
  • $\begingroup$ I agree with @kcrisman. For example (but I would agree that what I'm about to say is getting a bit too extreme with details), regarding your comment "which in analysis can only have the limit $0,$" if we're talking about the reals as a topological space in which the only open sets are $\emptyset$ and ${\mathbb R},$ then $0,\,0,\,0,\,\ldots$ has EVERY real number as a limit. $\endgroup$ – Dave L Renfro Jul 10 '17 at 15:31
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    $\begingroup$ Clarification of your intent in the language? "... the ratio of not-enumerated factions and natural numbers is $0$" means that the ratio of the cardinaliry of such numbers to the totality of numbers in the interval is $0$? The next sentence might make sense except for the "in analysis": what is intended? And "according to set theory the limit is $\infty$ since no rational number is not enumerated" means that the count of such things is obviously infinite? (If this interpretation is correct, then the seeming problem is resolved by noting that there are different sizes of infinity...?) $\endgroup$ – paul garrett Jul 10 '17 at 22:28
  • $\begingroup$ Richard Lipton: "I find this intriguing: what makes this proof so disputed? What makes it the most discussed proof on the web? In short why is it so hard to believe?" rjlipton.wordpress.com/2011/10/21/… $\endgroup$ – Daniel R. Collins Jul 11 '17 at 2:42
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    $\begingroup$ I have a sneaking suspicion that new user Ibrahim Abd el Faruk-Shaik from Germany, new user Hilbert7, self-published author Wolfgang Mückenheim (cited in Hilbert7's deleted answer), and the "brightest student" may possibly all be the same person. This question may need to get protected at some point. $\endgroup$ – Daniel R. Collins Jul 11 '17 at 2:46
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Not every property is preserved by limits.

Here is a more basic situation in which the same reasoning is used:

For each natural number $n$, there are only finitely many natural numbers in the interval $[0,n]$. However, letting $n \to \infty$, there are infinitely many natural numbers in the interval $[0,\infty)$. There is clearly no contradiction here, but the argument is of the same form as offered by the student. The function which assigns the word "finite" or "infinite" to sets is not continuous under limits. So why should his function be continuous?

I would also point out that the statement that "set theory is useless as a basis for analysis" is just plain false. It is the basis for analysis. He just isn't using it correctly. It is similar to saying that ``hammers are useless for driving nails'' after attempting once to drive a nail into a piece of wood by hitting it with the handle.

I wouldn't recommend being this blunt with a student though. For constructive advice would be to focus on the point about ``continuity''.

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  • $\begingroup$ This is really a great, incisive answer. $\endgroup$ – Daniel R. Collins Jul 9 '17 at 16:47
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The problem is not one of reconciling analysis and set theory at all; it's one of understanding analysis properly (and how things go sideways when dealing with the infinite). One of the first and primary questions of analysis is, "when can limits and operators be interchanged?"

To see why this is not trivial, consider the following exampe, which is a standard demonstration of the need for the dominating function in the Dominated Convergence Theorem.

Let $f_n(x)=n\chi_{_{(0,1/n]}}$. Then $$\int{f_n}=1$$ for every $n$ whence $$\lim\int f_n =1$$ $\textrm{}$

But now let's consider the pointwise limit of our sequence of functions. Clearly we have that $$\lim f_n=0$$ which means that $$\int\lim f_n=0$$ and thus we can conclude that $$\lim\int f_n=1\ne 0= \int\lim f_n$$

Consider that the argument in play from your student is analogous. The claim is that the limit and the taking of the ratios commute, i.e. the limit of the ratio is the same as the ratio of the limit (the "limit" of the sets is itself a measure theory notion, not simply a set theoretic one!). But there is no justification for that.

On a side note (and perhaps I've misinterpreted the tone of things), this could be a teachable moment for your student. While bright, the knowledge and understanding that they possess is but a sliver of the knowledge that any one of thousands of mathematicians across the world hold at this moment in time, to say nothing of those throughout history. Topics like these are not new and have been fleshed out through over a hundred years of painstaking work. We want our students to ask questions and to think critically, but when it comes to fairly settled topics in math, they will likely learn much more with a, "my argument leads me to a seemingly unreasonable result; where did I go wrong?" than with the thought that they have shown those who have come before to be wrong.

Edit: The original claim itself wasn't formally written out, so I'm adding this as an addendum so that we can see the claim made explicit.

Let the set $A_n=\mathbb{Q}\cap (0,n]$ and $B_n=\mathbb{N}\cap (0,n]$. Then define $f(n)=\frac{\lvert B_n\rvert}{\lvert A_n\rvert}$ (note that in order for this function to be defined, we are working with the extended real numbers). Then $$\lim_{n\to\infty} f(n)=\lim_{n\to\infty} 0=0$$ But $$f\left(\lim_{n\to\infty}n\right)=f(\omega)=1$$ The claim is then that $$\lim_{n\to\infty} f(n)=f\left(\lim_{n\to\infty}n\right)$$ which would indicate a contradiction.

Except for the limit commuting with the operator (the operator is $f$) is not a priori allowed. In this case, the property allowing for the commutation would be continuity (as noted in Steven Gubkin's answer), for which there has been no justification.

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  • $\begingroup$ I assume you mean the measure for $A_n$? What is $|A_n|$ otherwise? $\endgroup$ – kcrisman Jul 10 '17 at 18:24
  • $\begingroup$ This all sort of reeks of constructive methods to me too, but I'm not knowledgeable enough about them to know whether that could be made precise to resolve this to everyone's satisfaction ... $\endgroup$ – kcrisman Jul 10 '17 at 18:25
  • $\begingroup$ Yes. I'm using the common notation for cardinality of a set which is the counting measure. I used that notation because simply saying "measure" is ambiguous. $\endgroup$ – rnrstopstraffic Jul 10 '17 at 18:29
  • $\begingroup$ In which sense is $f(\omega)=1$? $\endgroup$ – Dag Oskar Madsen Jul 11 '17 at 16:23
  • $\begingroup$ @DagOskarMadsen I was simply writing out what appears to be the argument (based on the unclear claims) of the OP. That would be the apparent ratio of natural numbers to enumerated fractions in that interval. $\endgroup$ – rnrstopstraffic Jul 11 '17 at 16:28
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Here is another attempt.

Consider the function $f:\mathbb{N} \to \mathbb{R}$ defined by $f(n) = \textrm{ The number of elements in the set $\{n,n+1,n+2,...2n\}$}$.

In other words $f(n) = n+1$.

Clearly $$\lim_{n \to \infty} f(n) = \infty$$

However, the limit of the sets $\{n,n+1,n+2,...2n\}$ is the empty set, since no element would be a member of such a limit. So the "set theoretic'' limit would be $0$. Is mathematics in jeopardy because of this "contradiction''? Or perhaps there is no contradiction at all.

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    $\begingroup$ This may not be persuasive to someone like the questioner. I hate to say that I looked at it, but the self-published work cited in the deleted answer uses an equivalent sequence ("enumerated dollars") to, indeed, argue that mathematics is in jeopardy due to a contradiction ("set theory requires the belief in useless, discontinuous, and unmathematical properties of 'the infinite'"). $\endgroup$ – Daniel R. Collins Jul 11 '17 at 5:51