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!