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Let $f\colon\mathbb{R}\to\mathbb{R}$. Prove that the set $$\{x \mid \mbox{if $y$ converges to $x$, then $f(y)$ converges to $\infty$}\}$$ is countable.

My book told me to consider $g(x)=\arctan(f(x))$, then it said "it is easy to see the set is countable." But I still can't understand what it mean.

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Please don't rely on the subject for content. The body of the message should be self-contained. –  Arturo Magidin Feb 16 '11 at 5:35
    
@Eric Naslund: Under the usual definition of the limit of a function, the argument $x$ itself is excluded from the sequences, so the limit of a function may well differ from the function value. See en.wikipedia.org/wiki/Limit_of_a_function, but in the German article, a "more recent limit definition" is discussed which allows the argument as a member of the sequence, as you do: de.wikipedia.org/wiki/…. The question was clearly written with the former definition in mind. –  joriki Feb 16 '11 at 6:03
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The nomenclature is all messed up anyway (functions don't converge to $\infty$, they diverge); is this being translated into English? –  Arturo Magidin Feb 16 '11 at 6:03
    
@Cheng Which book? Maybe some context will help. –  Glen Wheeler Feb 16 '11 at 18:04
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Do I not understand this question? If $f$ is not allowed asymptotes, (i.e. $\mathbb{R}$ is the regular real number line) then $f(x_i)$ can not become unbounded. If we instead consider the extended real number system then it may be false: put an asymptote at every point in the Cantor set. If we consider the standard reals but replace $\infty$ with `diverges' then maybe it is true. But we need context! –  Glen Wheeler Feb 16 '11 at 18:18

4 Answers 4

There was some discussion about the terminology in comments to the question -- I interpret the statement "if $y$ converges to $x$, then $f(y)$ converges to $\infty$" to mean "for all $M\in\mathbb{R}$ there is $\delta > 0$ such that $f(y)>M$ for all $y\neq x$ with $\lvert y-x\rvert < \delta$". Let $S$ be the set of all such points $x$, and call a set that contains uncountably many points of $S$ "uncountably divergent" for short.

I don't think that the $\arctan$ transformation is of any help, since it's just a homeomorphism between $\mathbb{R}$ and $(-\pi,\pi)$ and between $\mathbb{R}\cup\{\infty\}$ and $(-\pi,\pi]$ and doesn't change the substance of the required proof. [Edit, after seeing Nick Kirby's answer and following his links: The transformation does make a difference, in that the limit $\infty$ is not allowed as a function value here, whereas it is if it is finite. This difference is relevant with respect to the example of the function with removable discontinuities at all rationals given here, since this uses the limit at the rationals as the function value at the irrationals, which we can't do in the present case, so that this example doesn't yield an example of a function "converging to $\infty$" at the rationals -- in fact there is no such function; see the remark about nowhere dense sets below.]

Assume that $S$ is uncountable. First, observe that $[n,n+1]$ must then be uncountably divergent for some $n\in\mathbb{N}$. Without loss of generality, assume $n=0$, so $[0,1]$ is uncountably divergent.

Now we divide the interval in half. Either $[0,1/2]$ or $[1/2,1]$ or both are uncountably divergent. If only one of them is, we discard the other half and replace the interval by the uncountably divergent half and continue dividing it.

Assume that after each subdivision only one half is uncountably divergent. These halves form a chain of nested closed intervals whose intersection contains exactly one point. Thus, except possibly for one point, all points of $S$ in $[0,1]$ lie in one of the discarded halves that each contain at most countably many points of $S$. But there are only countably many of these halves, contradicting the fact that there are uncountably many points of $S$ in $[0,1]$. Hence the assumption is false and the subdivision must encounter an interval of which both halves are uncountably divergent.

Once we encounter such an interval, we denote it by $I$ and recursively apply the entire halving procedure to it (including the part where we discard intervals of which only one half is uncountably divergent). We call the two resulting intervals $I_0$ and $I_1$ and continue applying the halving procedure to them, resulting in intervals $I_{00}$, $I_{01}$, $I_{10}$, $I_{11}$, and so on, where each resulting interval has the property that both its halves are uncountably divergent.

We discard any points of $S$ lying in the discarded halves, as well as all points lying on a boundary of one of the intervals. Since there are only countably many such points, each interval that was uncountably divergent remains so. Note that each remaining point of $S$ in $[0,1]$ now lies in a chain of nested intervals $I$, $I_{i_1}$, $I_{i_1i_2}$, ... with lengths converging to $0$, and the intersection of that chain contains exactly that point.

We now begin with the interval $I$, choose some point $s$ of $S$ in it and find $\delta$ such that $f(y) > 1$ for all $y\neq s$ with $y \in (s-\delta,s+\delta)$. Now $s$ lies in a chain of nested closed intervals with lengths converging to $0$, so one of these intervals is entirely contained in $(s-\delta,s+\delta)$. Denote this interval by $I_i$, where $i$ is a string of binary digits, and choose the one of $I_{i0}$ and $I_{i1}$ which does not contain $s$, so that $f>1$ on this entire subinterval. We now apply the same procedure to this interval, choosing a point of $S$ in it but now choosing $\delta$ such that $f(y) > 2$.

Iterating this procedure, we obtain a chain of nested closed intervals whose lengths converge to $0$, and the intersection of these intervals contains exactly one point $t$. But now $f(t)>n$ for all $n\in\mathbb{N}$, which is impossible. Hence $S$ must be countable.

[Edit: Note that uncountability was only used to make sure that we have points in both halves of an interval at every level of subdivision -- thus, we can use the same proof to show that $S$ is nowhere dense (except for the "removable technicality" that we discarded the set of boundary points, i.e. the set of points with finite binary representation, which is itself dense -- this was just to ensure that each point of an interval is contained in exactly one of the closed subintervals, but we can alternatively ensure that by choosing one of the four sub-subintervals, at most two of which contain the given point.). The same is not true for the case of finite limits (i.e. limits which are allowed as function values elsewhere), as the example of the function with removable discontinuities at all rationals shows. The crucial difference is in the last step of the proof, since $f(t)>n$ for all $n\in\mathbb{N}$ is impossible, whereas the corresponding statement for the limit $0$, $\lvert f(t)\rvert<1/n$ for all $n\in\mathbb{N}$, is consistent.]

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There is a question:When we encounter the situation that both halves have uncountable points in S,can we say any subinterval in I also has uncountable points in S ?But S can be nowhere dense.(see the 6th paragraph.) –  Cheng Feb 17 '11 at 4:03
    
@Cheng: No, not any subinterval in $I$ -- I don't think I'm using that (I hope not :-). All I'm using is that each half has uncountably many points in $S$ and therefore we can apply the same reasoning to each of them as we did originally to $[0,1]$, namely that we can subdivide them and this subdivision must at some point, not necessarily immediately, yield two halves with uncountably many points in $S$. This is what "recursively apply the entire halving procedure to it (including the part where we discard intervals of which only one half is uncountably divergent)" was intended to express. –  joriki Feb 17 '11 at 5:44
    
@Cheng: Perhaps I should emphasize for further clarification that in the rest of the proof I never refer to any of the intermediate intervals of which only one half contained uncountably many points in $S$ (except to say that the points in the other half get discarded) -- all the intervals $I$, $I_0$, $I_1$, $I_{00}$, ... that are used in the rest of the proof are by definition intervals at which the subdivision encountered two halves with uncountably many points in $S$. Thus, not all intervals of the form $[m/2^n,(m+1)/2^n]$ necessarily occur here, but all of these intervals are of this form. –  joriki Feb 17 '11 at 5:51
    
But when we encounter $I$ which both halves have uncountable points in $S$,we cannot make sure all the subintervals we get by applying the halving procedure have points in $S$,let alone uncountable points in $S$.so in 8th and 9th paragraph,we cannot make sure there is a point of $S$ in the interval you said.I don't know if I understood what you mean correctly.But the other proof is really effective. –  Cheng Feb 17 '11 at 13:51
    
@Cheng: Could you please point out more specifically where/how you think a problem might occur? Again, I'm not claiming that all subintervals obtained simply by halving are uncountably divergent, only the subintervals obtained when you apply the entire procedure, including discarding halves which are not uncountably divergent. The reasoning that ensures that in this way we will obtain suitable $I_0$ and $I_1$ is precisely the same reasoning that allowed us to obtain $I$, since the starting point in both cases is the same: an uncountably divergent set. –  joriki Feb 17 '11 at 14:06

If we allow $f$ to be a partial function, that is, where the domain is only a subset of $\mathbb{R}$ rather than all of $\mathbb{R}$, then there is an interesting counterexample.

Namely, let $C\subset\mathbb{R}$ be the usual Cantor set, and let $U=\mathbb{R}-C$ be the complement, an open set. For any real $x\in U$, let $$f(x)=\frac{1}{d(x)},$$ where $d(x)$ is the distance from $x$ to $C$, which is nonzero for all $x\in U$. Note that for any sequence $y_n\in U$ with $y_n\to x\in C$, we have $d(y_n)\to 0$ and hence $f(y_n)\to\infty$. (Note also that every $x\in C$ is a limit of points in $U$.) Since the Cantor set has uncountable size continuum, we have therefore an uncountable set of points where $f$ diverges to infinty. Furthermore, $f$ is continuous on its domain $U$.

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I see now that Glen Wheeler had the same idea up in the comments, which I hadn't noticed earlier. –  JDH Feb 17 '11 at 2:49
    
I guess I should have posted it as an answer, although honestly I was just looking for clarification. (Since without clarification it seems false.) –  Glen Wheeler Feb 17 '11 at 6:59

For completeness, here is a direct proof without the $\arctan$ transformation using the ideas in the book cited in Nick Kirby's answer (where the proof of the part of the theorem relating to removable discontinuities is left as an exercise).

Let $S_{mn}$ with $m,n\in \mathbb{N}$ be the set of all points $x$ such that $f(x) < m$ and $f(y) > m$ for all $y\neq x$ with $x-1/n<y<x+1/n$. For each point $x$ at which $f$ "converges to $\infty$" (in the sense explicated in my other answer), we can choose $m > f (x)$, and then choose $n$ such that $f(y) > m$ for all $y\neq x$ with $x-1/n<y<x+1/n$. Thus each point at which $f$ "converges to $\infty$" is contained in one of the sets $S_{mn}$. But none of these sets has an accumulation point, since that would imply that there are two points which are less than $1/n$ apart, both of whose function values are below $m$ and both of which have a neighbourhood of radius $1/n$ where the function values are above $m$, which is impossible. Hence each $S_{mn}$ is countable, and the set of points at which $f$ "converges to $\infty$", being the countable union of these countable sets, is also countable.

[Edit: Note that this proof substantially uses countability and can't be adapted to nowhere-denseness as immediately as the proof in my other answer. This is because although each $S_{mn}$ is not only countable but also nowhere dense, this property, unlike countability, is not necessarily inherited in a countable union. Thus, we cannot exclude on the basis of this proof alone that the union of all $S_{mn}$ might be dense in $\mathbb{R}$, as demonstrated by the example of the function with removable discontinuities at all rationals. However, the union of all $S_{mn}$ with bounded $m$ cannot be dense anywhere, and it is this necessary unbounded growth of $m$ that in the present case precludes denseness, since we cannot have $\infty > m$ for all $m\in \mathbb{N}$ as a function value, whereas we can have $0 < 1/m$ for all $m\in \mathbb{N}$ as a function value in the finite case.]

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A very brilliant proof !!! Thank you! –  Cheng Feb 17 '11 at 4:10

Notice that $g(x)=\arctan(f(x))$ has a removable discontinuity at points $x$ such that $\lim_{t\to x} f(t)=\infty$. Therefore, the countability of the set of interest can be seen as a consequence of the fact discussed here: Is there a function with a removable discontinuity at every point? and particularly, Chandru1's link: Chandru1's link to Theorem 5.63, which proves that the set of removable discontinuities is at most countable.

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