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In my research I came across the following question :

Is it true that for every real function $f:\mathbb{R}\to\mathbb{R}$, there exists a real sequence $(x_n)_n$, taking infinitely many values, converging to some real number $c$, such that the sequence $(f(x_n))_n$ converges to $f(c)$ ?

Micah
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Dattier
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    The question looks interesting, because $f$ is not assumed to have any property (such as continuity). Perhaps prove first this. There is $a$ so that for every $\epsilon > 0$, the set ${x ; : ; |f(x) - f(a)|<\epsilon }$ is uncountable. – Gerald Edgar Aug 24 '18 at 10:31
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    The difficulty is to have, $f(x_n)$ converged to $f(a)$ and $x_n$ converged to $a$. – Dattier Aug 24 '18 at 10:49
  • @GeraldEdgar : for your question, you can show that with compacity argument – Dattier Aug 24 '18 at 10:52
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    I had to read it like four times before I convinced myself that this is not an immediately false triviality. – Qfwfq Aug 24 '18 at 19:36

3 Answers3

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Suppose $f$ were a counterexample. Then, for any $c$, we could find a little interval $(a,b)$ containing $c$ and we could find some $\varepsilon>0$ such that all points $x\in(a,b)$ except $c$ have $|f(x)-f(c)|>\varepsilon$. (Otherwise, by taking smaller and smaller intervals and $\varepsilon$'s, we could produce a sequence converging to $c$ with $f$-image converging to $f(c)$.) By shrinking things a little, we can arrange for $a$, $b$, and $\varepsilon$ to always be rational. So uncountably many $c$'s must have the same $a$, $b$, and $\varepsilon$. Fix such $a$, $b$, and $\varepsilon$. So you've got uncountably many $c$'s, all lying in $(a,b)$ and (therefore) all having their $f(c)$'s separated by a distance of at least $\varepsilon$ --- a contradiction because there isn't that much room in $\mathbb R$.

Andreas Blass
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    A very quick way to say essentially the same idea: Since $\mathbb R^2$ is second countable, not every point in the graph of $f$ can be isolated. – Monroe Eskew Aug 24 '18 at 18:21
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    what about the case : $(x_n)_n$ to $c$ and $\frac{f(x_n)-f(c)}{x_n-c}$ converged ? @MonroeEskew – Dattier Aug 26 '18 at 11:29
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    https://mathoverflow.net/questions/309158/does-every-real-function-have-this-weak-derivation-property – Dattier Aug 26 '18 at 12:42
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This has already been answered satisfactorily, but let me mention that it also follows from the much stronger (and noteworthy) fact that for every $f\colon\mathbb{R}\to\mathbb{R}$ there exists a dense $D\subseteq\mathbb{R}$ such that $f|_D$ is continuous (Blumberg's theorem): see here for references. Given that this is true, take $c\in D$ and $x_n$ injective converging toward $c$ (which exists since $D$ is dense), and we have $f(x_n)$ converging to $f(c)$ by continuity of $f|_D$.

Gro-Tsen
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  • what about the case when $f(\mathbb R)=\mathbb N$, with $\forall n \in \mathbb N,f^{-1}({n})$ is not dense in the real. – Dattier Aug 24 '18 at 13:03
  • I can't think about a function continuous in D (dense in the real) with $f(D)=\mathbb N$, can you give an example. – Dattier Aug 24 '18 at 13:06
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    @Dattier Blumberg's theorem is not that $f$ is continuous on $D$, it is that $f|_D$ is continuous. The two are quite different. (For some reason, every person who hears about Blumberg's theorem for the first time seems to fall for the same confusion.) – Gro-Tsen Aug 24 '18 at 13:07
  • what the topologie on D ? (métrique no ?) – Dattier Aug 24 '18 at 13:09
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    The topology on $D$ is the induced topology. (And as for an example of $f\colon D\to\mathbb{Z}$ surjective continuous, consider $f(x)=\lfloor x/\sqrt{2}\rfloor$ on $D=\mathbb{Q}$.) – Gro-Tsen Aug 24 '18 at 13:11
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    The floor function is continuous on $\mathbb R \setminus \mathbb N$. – Monroe Eskew Aug 24 '18 at 17:05
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    @Gro-Tsen: The function $x\mapsto \lfloor \sqrt 2x\rfloor$ defined on $\mathbb \Q$ is discontinuous at 0 for what it’s worth. – Anthony Quas Aug 24 '18 at 18:04
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    @AnthonyQuas Oh, yeah, right. So remove $0$ from the domain or replace $\lfloor x/\sqrt{2}\rfloor$ by $\lfloor x-\sqrt{2}\rfloor$. – Gro-Tsen Aug 24 '18 at 19:15
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It has "been known" since 1908 that for any such function, this holds for all but countably many real numbers $c$, even when we additionally require all the sequences to approach $c$ from the same side.

In May 1908 William Henry Young presented several results for general functions from $\mathbb R$ to ${\mathbb R},$ including a result implying that, given any such function, all but countably many real numbers $c$ have the property you're asking about. These results (for more about them, see my answer here) may have been joint work with his wife, Grace Chisholm Young, and the results were published in the 1908 paper cited below.

Young showed that for co-countably many real numbers $c$ we have

$$f(c) \in C^{-}(f,c) \;\; \text{and} \;\; f(c) \in C^{+}(f,c)$$

Definition: Given a function $f: {\mathbb R} \rightarrow {\mathbb R}$ and $c \in {\mathbb R}$, we let $C^{-}(f,c)$ be the set of all extended real numbers $y$ (i.e. $y$ can be $-\infty$ or $+\infty$) for which there exists a sequence $\left\{x_{k}\right\}$ such that for each $k$ we have $x_k < c,$ and we have $x_{k} \rightarrow c$ and $f(x_k) \rightarrow y.$ In other words, $C^{-}(f,c)$ is the set of all numbers (including $-\infty$ and $+\infty$) that can be obtained as a limit of $f$-values when using some sequence converging to $c$ from the left. The right version, $C^{+}(f,c),$ is defined analogously.

Incidentally, the requirement in this definition that each $x_k < c$ (and also each $x_k > c)$ allows you to find such sequences converging to $c$ that have infinitely many values.

William Henry Young, Sulle due funzioni a più valori costituite dai limiti d'una variabile reale a destra e a sinistra di ciascun punto [On the two functions of multiple values that are determined by the left and right limits of a real variable at each point], Atti della Accademia Reale dei Lincei. Rendiconti. Classe di Scienze fisiche, Matematiche e Naturali (5) 17 #9 (1st semestre) (1908), 582-587. [Paper given at session dated 3 May 1908.]

Qfwfq
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Dave L Renfro
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