Let $A$ be a nonempty subset of $\mathbb R$ and $f: A\rightarrow \mathbb R$. Suppose $c$ is a cluster point of $A$.

  1. Suppose the limit of $f(x)$ at $c$ does not exist. Show that there exists $\varepsilon>0$ and two sequences $(x_n)$ and $(y_n)$ in $A\setminus \{c\}$, both converging to $c$, such that $|f(x_n)-f(y_n)|\geq\varepsilon$ for all $n\in\mathbb N$.

  2. limit of $f(x)$ exists if and only if for all $\varepsilon>0$, there exists $\delta>0$ such that if $x,y\in A$ with $0<|x-c|$ ,$|y-c|<\delta$, then $|f(x)-f(y)|<\varepsilon$.

I have made some efforts on this but failed.

My plan for question 1: the limit of $f(x)$ at $c$ doesn't exist means for every $L$ in $\mathbb R$ there is some $\varepsilon>0$ such that for any $\delta>0$ there is a $x_\delta≠c $ in the $\delta$-neighborhood of $c$ such that $|f(x_\delta)-f(c)|≥\varepsilon$.

Then, since we can find an arbitrary sequence $(y_n)$ in $A\setminus \{c\}$ converging to $c$, by letting $L_n=f(y_n)$ and $\delta_n=1/n$, we can find $\varepsilon_n>0$ so that there is a $x_n≠c$ in the $1/n$-neighborhood of c such that $|f(x_n)-f(y_n)|≥\varepsilon_n$. And it also follows that $(x_n)$ converges to c.

So we just need to let $\varepsilon=inf${$\varepsilon_1, \varepsilon_2,...,\varepsilon_n,...$}. Here the question comes that I cannot assure $\varepsilon>0$.

Can anybody give a hand?

Thanks very much!!

  • 2
    $\begingroup$ Thank you for your question! It is good practice on this site to add a bit of information on the context your question came up in, and to share your own work on it. It's also fine if you state that you're completely lost -- the information is helpful for answerers to gauge their answer on. For more information on asking a good question on this site, see here. $\endgroup$ – Lord_Farin Nov 16 '13 at 9:37

ad 1: As you saw yourself, the assumption that the $\lim$ doesn't exist is difficult to handle, since there are an unfinite number of values $\eta$ to which $f(x)$ doesn't converge. In fact we need some heavier tools to prove the claim.

If $f$ is unbounded in a neighborhood of $c$ then construct a sequence $(x_n)_{n\geq0}$ recursively as follows: Choose $x_0$ arbitrarily, and for each $n\geq1$ choose an $x_n$ with $|x_n-c|<{1\over n}$ and $|f(x_n)|\geq |f(x_{n-1})|+1$. Then the $x_n$ converge to $c$, as do the $y_n:=x_{n-1}$ $\>(n\geq1)$. Furthermore we have $|f(x_n)-f(y_n)|\geq1$ for all $n\geq1$.

When $f$ is bounded choose an arbitrary sequence $(x_n)_{n\geq0}$ converging to $c$. The sequence $a_n:=f(x_n)$ is a bounded sequence of real numbers and so has an accumulation point $\alpha\in{\mathbb R}$. Passing to a subsequence we may assume $\lim_{n\to\infty} f(x_n)=\alpha$. As, by assumption, the $\lim_{x\to c} f(x)$ does not exist there is an $\epsilon>0$ and a sequence $(y_n)_{n\geq0}$ converging to $c$ with $|f(y_n)-\alpha|\geq\epsilon$ for all $n$. It follows that we have $|f(x_n)-f(y_n)|\geq{\epsilon\over2}$ for all large enough $n$.

A second way to prove 1. would be to prove 2. first.

ad 2: This is the "Cauchy criterion for functions". As with the Cauchy criterion for sequences one direction is trivial. So assume that for all $\epsilon>0$ there is a $\delta>0$ such that $\ldots$

Choose an arbitrary sequence $(x_n)_{n\geq0}$ converging to $c$. Then it is easy to see that $\bigl(f(x_n)\bigr)_{n\geq0}$ is a Cauchy sequence in ${\mathbb R}$ and therefore converges to a number $\alpha\in{\mathbb R}$. Given an $\epsilon>0$ there is a $\delta>0$ with $$|f(y)-f(x_n)|<\epsilon\tag{1}$$ for all $y\in \dot U_\delta(c)$ and all large enough $n$. For fixed $y$ letting $n\to\infty$ in $(1)$ we conclude that $$|f(y)-\alpha|\leq\epsilon\quad\forall \ y\in \dot U_\delta(c)\ .$$


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