# Oscillation of $f$ (at $x_0$) equals zero iff $f$ is continuous.

I'm reading a proof of that claim mentioned in the title and I have some difficulties understanding it.

Statement:

$$O_f(x_0)=0 \iff f \ \ \text{is continuous on} \ \ x_0$$

(Where oscillation, $O_f(x):= \displaystyle\lim_{n\to\infty}\text{diam}f(D(x,\frac{1}{n})$ and where $D(x,\frac{1}{n})$ is an open ball with center $x$ and radius $\frac{1}{n}$)

Proof:

Suppose that $O_f(x_o)=0$.

Let $V$ be some neighborhood of $f(x_0)$. Exist $\epsilon>0$ which for him $D(f(x_0,\epsilon))\subseteq V$. From $O_f(x_o)=0$, we know that there is $k\in\mathbb{N}$, that from that $k$ and on, $\text{diam}f(D(x,\frac{1}{n}))<\epsilon$ we choose $n>k$. For all $x\in D(x_0,\frac{1}{n})$ we have: $$d(f(x),f(x_0))\leq \text{diam}f(D(x,\frac{1}{n})<\epsilon$$ and therefor(which I can't understand): $$f(D(x_0,\frac{1}{n}))\subseteq D(f(x_0),\epsilon)$$

That last claim completes the proof(of first direction)

So, again -- how from $d(f(x),f(x_0))\leq \text{diam}f(D(x,\frac{1}{n})<\epsilon$ implies that $f(D(x_0,\frac{1}{n}))\subseteq D(f(x_0),\epsilon)$?

Thank you!

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The diameter of a set in a metric space is the supremum of the distances between all its points, so in your definition of oscillation we have this: $$\mathcal O_f(x_0)=\lim_{n\to\infty}\sup_{x,y\in D\left(x_0,1/n\right)}d\left(f(x),f(y)\right)$$ and since the following inequality is true $$d\left(f(x),f(x_0)\right)\leq \textrm{diam} f\left(D\left(x_0,\frac{1}{n}\right)\right)$$ it also holds for the supremum, from which the inclusion you ask about follows.
Because we took any $\,x\in D(x_0,1/n)\,$ and we got $$d(f(x),f(x_0))\leq \text{diam} f(D(x_0,1/n))<\epsilon\Longrightarrow$$ and this means $\,f\,$ maps any element of $D(x_0,1/n)\,$ , namely $\,x\,$ , to the element $\,f(x)\,$ which is within a distance of $\,\epsilon\,$ from $\,f(x_0)\,$ , i.e. to the open ball $\,D(f(x_0),\epsilon)\,$...!
In other words, all the elements in $\,D(x_0,1/n)\,$ are mapped by $\,f\,$ to elements within a distance of $\,\epsilon\,$ from $\,f(x_0)\,$ , i.e. are mapped into the ball $\,D(f(x_0),\epsilon)\,$
Yes, that is clear to me. And maybe I was unclear in may problem. I'm asking, why $f(D(x_0,\frac{1}{n}))\subseteq D(f(x_0),\epsilon)$ follows from that inequality..? –  Salech Alhasov Aug 9 '12 at 4:16