difficulty in understanding the symbols?

Let $A$ denote the rational points of the interval $[0,1] \times 0$ of $\mathbb{R}^2$. Let $T$ denote the union of all line segments joining the point $p = 0 \times 1$ to points of $A$.

But i found difficulty in understanding the symbol

My confusion is that where $p$ has gone ?

• $B(p;\delta)$ is the open ball of radius $\delta$ centred at $p$ – A. Goodier Aug 28 '18 at 13:44
• @stupid You are right: that definition is weird, and they apparently assume $\;p=(0,1)\;$ , which doesn't appear anywhere in your question...What is written there up is something even weirder: $\;p=0\times1\;$ , which I guess could mean what I just wrote above. – DonAntonio Aug 28 '18 at 13:48
It's defined in that line, so understanding might not be the issue. But the $B(p;\delta)$ notation (or similar ones, like $B_\delta(p), B(p,\delta)$ etc.) is often used for the metric ball of radius $\delta$ around the point $p$. Here this seems to be the case for $p=(0,1)$ and the Euclidean distance in the plane. The distance of $(\xi,\eta)$ to $(0,1)$ is $\sqrt{(\xi-0)^2 + (\eta-1)^2}$ so that distance is smaller than $\delta$ iff its square $\xi^2 + (\eta-1)^2$ is smaller than $\delta^2$. Hence the equivalent reformulation.
$B(p,\delta)$ is the open ball centered at $p$ with radius $\delta$. In the notation after, $p$ is hiding by way of $$|(\xi,\eta)-p|^2=|(\xi,\eta)-(0,1)|^2=\xi^2+(\eta-1)^2$$ So $p$ isn't gone, entirely, but the author has inserted the definition of $p$ as $(0,1)$ and let that $0$ and $1$ go their separate ways.