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I was wondering if there is a function $f: \mathbb{R} \to \mathbb{R}$ such that its limit at every point is infinite.

I guess not, because what would its graph look like, but then again, I don't know how to prove it.

Did
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user4201961
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  • While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $\mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context. – PrincessEev Jan 08 '19 at 08:31
  • My intuition tells me that such function doesn't exist. I like to be proven wrong. –  Jan 08 '19 at 08:40
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    @EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $\lim_{x \to p} f(x) = \infty$ while $f(p)$ takes a (finite) value. – Anders Kaseorg Jan 08 '19 at 08:44
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    No, there does not exist such $f$. In fact, the set $$ {x\in\mathbb{R};|;\lim_{y\to x}f(y)=\infty} $$ is a countable subset of $\mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post https://math.stackexchange.com/questions/3060529/how-can-we-see-that-the-proof-of-this-following-real-analysis-problems-is-reason/3060591#3060591. – Myunghyun Song Jan 08 '19 at 08:44
  • Hmmm, I see. Thanks for the insight @AndersKaseorg – PrincessEev Jan 08 '19 at 08:45
  • No: notice that the set of x for which f(x) is smaller than a fixed real number $n$, consists solely of isolated points. Hence it is countable. In particular by taking union over all $n$, one sees that the set of points at which f is finite, is countable. This would contradict the fact that f is real valued. Note that this argument subtly uses AC. Try to find argument which doesn’t use Choice (or prove none exists ;). – shalop Jan 08 '19 at 09:13
  • @Song: (and others) Sets such that there exists a function $f$ in which $f$ has an infinite limit at each point of that set are a lot smaller than just being countable in general. For more details see this 30 November 2007 sci.math post and footnote 12 on p. 9 of these notes. – Dave L. Renfro Jan 08 '19 at 13:29
  • @DaveL.Renfro Though I couldn't catch up all the arguments, I appreciated your knowledge. Thank you, sir! – Myunghyun Song Jan 08 '19 at 14:49

2 Answers2

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If $f$ is such a function and $M>0$ then $$ \forall x_0\in \! [0,1] \qquad \exists \delta_{x_0}\ \qquad \ 0<|x-x_0|<\delta_{x_0} \implies f(x)>M.$$ But $$\bigcup_{x_0\in [0,1]} B(x_0,\delta_{x_0})\setminus \{x_0\}\supseteq [0,1]$$ and $[0,1]$ is compact So there exists $x_1,\ldots , x_n$ such that $$B(x_1,\delta_{x_1})\cup \ldots \cup B(x_n,\delta_{x_n}) \setminus \{x_1,\dotsc,x_n\}\supseteq [0,1]$$ So $f(x)>M$ for all $x\in [0,1]\setminus\{x_1,\ldots,x_n \} $. So $ \{ x: f(x)\le M \} $ is at most countable and $$\bigcup\limits_{M\in \mathbb{N}} \{ x:f(x)\le M \}$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.

SvanN
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Darman
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Let us show that such a function $f$ does not exist.

Choose any function $f:\mathbb R\to\mathbb R$ and, for every positive integer $n$, consider the set $$A_n=\{x\in[0,1]\,;\,|f(x)|\leqslant n\}$$ then, since $f(x)$ is finite for every $x$, $$\bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.

Pick a sequence $(x_k)$ such that $x_k\ne x_j$ for every $k\ne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X=\{x_k\,;\,k\geqslant0\}$ is infinite and such that $X\subseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $\lim\limits_{k\to\infty}x_k=p$ and that $x_k\ne p$ for every $k$.

Now, $|f(x_k)|\leqslant n$ for every $k$ hence $\lim\limits_{x\to p}|f(x)|=+\infty$ is impossible because $\lim\limits_{k\to\infty}x_k=p$ with $x_k\ne p$ for every $k$ and $\limsup\limits_{k\to\infty}|f(x_k)|\leqslant n$.

Did
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