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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 symbolenter image description here

My confusion is that where $p$ has gone ?

please help me

thanks in advance

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    $\begingroup$ $B(p;\delta)$ is the open ball of radius $\delta$ centred at $p$ $\endgroup$ – A. Goodier Aug 28 '18 at 13:44
  • $\begingroup$ @stupid Theee are meta posts which recommend avoiding use of phrases like "thanks in advance" on the grounds that the clutter up the question. Maybe you should consider editing them out? $\endgroup$ – The Long Night Aug 28 '18 at 13:44
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    $\begingroup$ @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. $\endgroup$ – DonAntonio Aug 28 '18 at 13:48
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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.

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$B(p,\delta)$ 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.

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  • $\begingroup$ thanks u arthur $\endgroup$ – jasmine Aug 28 '18 at 13:52

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