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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.2011-02-16
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    @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 http://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: http://de.wikipedia.org/wiki/Grenzwert_(Funktion)#Neuerer_Grenzwertbegriff. The question was clearly written with the former definition in mind.2011-02-16
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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?2011-02-16
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    @Cheng Which book? Maybe some context will help.2011-02-16
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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!2011-02-16
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    @Glen: A function can be defined on all of $\mathbb{R}$ and take values only on $\mathbb{R}$, and yet have vertical asymptotes; e.g., $f(x) = 1/x$ if $x\neq 0$, $f(0)=0$. Here, $f$ would be defined at $0$, but $\lim\limits_{y\to 0} f(x) =\infty$. So there is no problem with allowing $f$ to have asymptotes and yet require that all values be real, and be defined on all reals.2011-02-16
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    @Glen: But if you "put an asymptote at every point in the Cantor set", then since every point in the Cantor set can be approximated by a sequence of points all of which are in the Cantor set, then unless those values go to $\infty$ themselves, the points of the Cantor set will not be contained in the set under consideration. For instance, if you try taking $f(x)=\frac{1}{d(x,C)}$ for $x\notin C$ ($C$ the cantor set, $d(x,C)$ the distance from $x$ to the Cantor set), and $f(x)=0$ for $x\in C$, then the set given is empty, because taking $y_n\in C$ with $y\to x_0$ gives the sequence $f(y_n)=0$.2011-02-16
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    @Arturo: Thanks, yes, of course! I find it very difficult to drop the habit of continuity. The function $f(x) = 1/d(x,C)$ for $x\notin C$ and $f(x) = \infty$ otherwise was actually the candidate I was thinking of, because I was considering the extended reals. But as you say, if we define $f(x) = 0$ for $x\in C$ then $C$ is not contained in the set. However, I still do not see why another sequence of points $x_n\rightarrow y$ where $x_n \notin C$ and $y\in C$ cannot exist.... is my intuition completely dead tonight??2011-02-16
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    @Glen: No, your intuition is correct, such sequences exist; but the set consists of all points for which, given **any** sequence $y_n$ with $y_n\to x$, you have $f(y_n)\to\infty$. That is, in order for a point $x$ to belong to the set, it's not enough for there to exist at least one sequence that converges to $x$ and where $f$ takes values that go to infinity, you need $f$ to take values that go to infinity at *all* sequences that converge to $x$. But you're not allowed to have the function itself take the value $\infty$ (otherwise, your example would give a counterexample to the statement).2011-02-16
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    @Arturo: I understand. So then about your earlier example of $f(x) = 1/x$, $f(0) = 0$, zero would not be in the set? The sequence $0 = x_n \rightarrow 0$ and $f(x_n) \rightarrow f(0) = 0$. That's a bit of a shame. I guess one should disallow trivial sequences...2011-02-16
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    @Glen: I'm pretty sure that here "sequence converging to $x$" requires that $y_n\neq x$ for all sufficiently large $x$. Basically, you want to take a real valued function, and consider the set of all $x$ for which $\lim\limits_{y\to x}f(y) = \infty$, with the usual notion of limits (so that we don't consider $y=x$ allowable).2011-02-16

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