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Let $x_0$ be an accumulation point of the set $D \subset \mathbb{R}$. We say that $y$ is a limit point of a function $f:D \rightarrow \mathbb{R}$ in $x_0$ iff there exists a sequence $(x_n)$, where $x_n \in D\setminus \{x_0\}$ for $n\in \mathbb{N}$ and $x_n \rightarrow x_0$, such that $y=\lim_{n\rightarrow \infty} f(x_n)$.

I don't know how to prove the following lemma:

If $f: (0,c) \rightarrow \mathbb{R}$ be continuous then the set of all limits points of $f$ in $0$ is the interval $$[\liminf_{x\rightarrow 0}f(x), \limsup_{x\rightarrow 0}f(x)].$$

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Do you require that the sequence $x_n$ converge to $x_0$? –  Alex Becker Jul 18 '12 at 15:34
    
Sorry, it is needed –  L.T Jul 18 '12 at 15:39
    
Related question: $\limsup$ and cluster points. –  Martin Sleziak Jul 18 '12 at 15:52

2 Answers 2

up vote 3 down vote accepted

Note: I assume you require that $x_n\to x_0$ in your definition, otherwise this lemma is false.

Suppose $y$ is a limit point of $f$ in $0$, so we have a sequence $(x_n)$ in $(0,c)$ such that $x_n\to x$ and $f(x_n)\to y$. Then $$\liminf\limits_{n\to\infty} f(x_n)=y=\limsup\limits_{n\to\infty} f(x_n)$$ so we have that $$\liminf\limits_{x\to 0} f(x)\leq \liminf\limits_{n\to\infty} f(x_n)=y=\limsup\limits_{n\to\infty} f(x)\leq \limsup\limits_{x\to 0} f(x)$$ by definition. Thus $y\in [\liminf\limits_{x\to 0} f(x),\limsup\limits_{x\to 0} f(x)]$.

On the other hand, suppose $y\in [\liminf\limits_{x\to 0} f(x),\limsup\limits_{x\to 0} f(x)]$. Take sequence $(x_n)$ and $(x_n')$ in $(0,c)$ converging to $0$ such that $\lim\limits_{n\to\infty} f(x_n)=\liminf\limits_{x\to 0} f(x)$ and $\lim\limits_{n\to\infty} f(x_n')=\limsup\limits_{x\to 0} f(x)$. If $y$ is either $\liminf\limits_{x\to 0} f(x)$ or $\limsup\limits_{x\to 0} f(x)$ we are done. Otherwise, we have some $N$ such that $$n\geq N\implies f(x)<y<f(x_n')$$ so by the Intermediate value theorem we have some $y_n$ between $x_n$ and $x_n'$ such that $f(y_n)=y$. Since $x_n\to 0$ and $x_n'\to 0$ it follows that $y_n\to 0$, and clearly $f(y_n)\to y$, thus $y$ is a limit point of $f$ in $0$.

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The idea behind "why this lemma is morally true" is because it is trivial in the case where $f(x) \to C$ when $x \to 0$, and because if the limit $\lim_{x \to 0} f(x)$ doesn't exist, since the function is continuous, the function must be going back and forth between the sequence going towards the $\limsup$ and the other one going to the $\liminf$. So here's the idea :

I'll cover the (hard part of the) case where $\liminf < \limsup$. Choose $\liminf < \lambda < \limsup$. By assumption there exists $(y_n)$ such that $f(y_n) \to \liminf$ and $(z_n)$ such that $f(z_n) \to \limsup$. Now without loss of generality we can assume that $y_1 > z_1 > y_2 > z_2 > \dots > y_n > z_n > y_{n+1} > \dots$ since you can only take appropriate subsequences of your original sequences. Now if $N$ is large enough we know that for $n > N$ we have $f(y_n) < \liminf + \varepsilon$ and that $f(z_n) > \limsup - \varepsilon$.

Say we took $\varepsilon$ such that $\liminf + \varepsilon < \lambda < \limsup - \varepsilon$. By the mean value theorem this means there exists $z_n < c_n < y_n$ such that $f(c_n) = \lambda$. Letting $n \to \infty$ we obtain the appropriate sequence $c_n$.

Hope that helps,

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