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I've asked the question below before with no answer, but I would like to stress that this time it is not a homework question (and also that I've spent hours trying to come up with a solution).

This is the question:

Let f be a function defined around $x_o$. For every $\epsilon>0$ there's some $\delta>0$ such that if $0<|x-x_0|<\delta$ and $0<|y-x_0|<\delta$ then $|f(x)-f(y)|<\epsilon$.

And what's needed to be proven is that $\lim_{x\to x_0}f(x)$ exists.

I've been told that there are two ways to do so: One is quite easy and requires Cauchy sequences (I haven't learned sequences yet, but I think I'll look it up sometime soon and try to solve it this way).

The second way is a direct way, which I've been told is cumbersome and unrecommended, but since this is the way I tried solving it so far, I am really curious as to how the proof goes and this is the way I'm asking about. I tried applying all kinds of inequalities but with no success.

Even a little hint/direction would be swell. Thank you in advance.

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S/He is asking for a proof without the use of Cauchy sequences. – Amr Nov 11 '12 at 16:27
I've said that just a little hint would be fine, no need for a complete answer. Furthermore, it has nothing to do with homework right now. Also, the old question would probably never be seen again, except maybe through search queries. It did have an answer mentioning Cauchy sequences (which is a good answer to the question, but I'm curious about other answers). – Py42 Nov 11 '12 at 16:31
I guess you can show that $\liminf_{x \to x_0}f(x)=\limsup_{x \to x_0}f(x)$. – Siminore Nov 11 '12 at 16:40
I believe any proof will use the completeness of the reals. – Amr Nov 11 '12 at 16:46
@Py42: What do you mean by "a direct way"? In other words: What are the rules of the game? Are we allowed to use inf and sup, the convergence of bounded monotone sequences? Or do you expect a proof in terms of Dedekind cuts? – Christian Blatter Nov 11 '12 at 21:02

The following proof is essentially taken from Zorich, Mathematical analysis I.

Let $\mathcal{B}$ denote the family of all deleted neighborhoods of $x_0$. For each $B \in \mathcal{B}$, define $$ m_B = \inf_{x \in B} f(x), \quad M_B=\sup_{x \in B}f(x). $$ Since $$ m_{B_1} \leq m_{B_1 \cap B_2} \leq M_{B_1 \cap B_2} \leq M_{B_2} $$ for any elements $B_1$ and $B_2$ of $\mathcal{B}$, the axiom of completeness of $\mathbb{R}$ implies the existence of some $A \in \mathbb{R}$ that separates the numerical sets $\{m_B\}_B$ and $\{M_B\}_B$ as $B$ ranges over $\mathcal{B}$. By assumption, given $\varepsilon>0$ there exists $B \in \mathcal{B}$ such that $M_B-m_b < \varepsilon$. Then $|f(x)-A| < \varepsilon$ whenever $x \in B$.

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Where did the proof use what was given? (or it didn't?) – Py42 Nov 11 '12 at 19:00
Since $|f(y)-f(x)|<\varepsilon$ for all $x$ and $y$ close to $x_0$, you deduce that $M_B-m_B<\varepsilon$ for some $B \in \mathcal{B}$. – Siminore Nov 12 '12 at 9:02

O.k.; here is a proof not using sequences:

For each $\epsilon>0$ choose a point $x_\epsilon$ with $0<|x_\epsilon-x_0|<\delta$, where $\delta=\delta(\epsilon)$ is described in the statement. Put $$a_\epsilon:=f(x_\epsilon)-\epsilon\ ,\quad b_\epsilon:=f(x_\epsilon)+\epsilon\ .$$ Given $\epsilon$, $\epsilon'$ we have $a_\epsilon\leq f(x)\leq b_{\epsilon'}$ for any $x$ with $0<|x-x_0|<\min\{\delta,\delta'\}$. Keeping $\epsilon'$ fixed we see that $$a:=\sup_{\epsilon>0}a_\epsilon\leq b_{\epsilon'}\ ,$$ and since this is true for all $\epsilon'>0$ we conclude that $$b:=\inf_{\epsilon'>0} b_{\epsilon'}\geq a\ .$$

Now for each $\epsilon>0$ we have $$f(x_\epsilon)-\epsilon=a_\epsilon\leq a\leq b\leq b_\epsilon=f(x_\epsilon)+\epsilon\ .\qquad(*)$$ Therefore $0\leq b-a\leq 2\epsilon$, and since this is true for all $\epsilon>0$ we conclude that $a=b$.

Consider now any $x$ with $0<|x-x_0|<\delta$. Then $f(x)$ lies also within the two bounds $f(x_\epsilon)\pm\epsilon$ given in $(*)$. Therefore $|f(x)-a|\leq 2\epsilon$. Since $\epsilon>0$ was arbitrary we have proven that $\lim_{x\to x_0} f(x)=a$.

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