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 $A$.
i found the answer here Prob. 5, Sec. 25 in Munkres' TOPOLOGY, 2nd ed: Is there a connected set that is locally connected at none of its points?
But i found difficulty in understanding the symbol
My confusion is that where $p$ has gone ?
please help me
thanks in advance
On
$B(p,\delta)$ 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.
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.