Proof about one theorem about $\lim f(x)$ and $\lim f(p_n)$ Could you help me understand the parts in yellow box?
What does $\delta_n$ signify?
Why is it specifically $1/n$? 
Which is a subsequence in $E$ satisfying (6) but not (5)?

 A: Given that $\lim_{x\rightarrow p} f(x) \neq q$, we are trying to construct a sequence $p_n$ in $E$ such that $p_n \neq p$ for all $n$ and $p_n \rightarrow p$ but with the property that $f(p_n)$ does not converge to $f(p)$ in $Y$. Negating the $\epsilon-\delta$ definition of limit (in the context of metric spaces), we see that $\lim_{x\rightarrow p} f(x) \neq q$ means that there exists an $\epsilon>0$ such that for all $\delta>0$ we do not have the implication $0<d_X(x,p)<\delta \Rightarrow d_Y(f(x),q)<\epsilon$, i.e. there exists an $\epsilon$ such that for each $\delta>0$ we have a 'bad' point $x_\delta$ in $E$ such that $0<d_X(x_\delta,p)<\delta$ and $d_Y(f(x_\delta),q) \ge \epsilon$. Lets make a sequence out of these 'bad' points. Consider a decreasing sequence of $\delta$'s, say $\delta_n = \frac{1}{n}$ (or $\frac{1}{2^n}$, or $\frac{1}{n!}$, or $\frac{1}{n+1}$, it doesn't matter so long as $\delta_n \rightarrow 0$). For each $n$, there exists a point $p_n$ in $E$ not equal to $p$ such that $0<d_X(p_n,p)<\frac{1}{n}$ and $d_Y(f(p_n),q)\ge \epsilon$. Since $d_X(p_n,p)<\frac{1}{n}$ for each $n$, the distance between $p_n$ and $p$ becomes arbitrarily small as $n\rightarrow \infty$ (this is why we chose $\delta_n$ the way we did). It follows that $p_n \rightarrow p$. However, $d_Y(f(p_n),q) \ge \epsilon$ for all $n$, which means that $f(p_n)$ cannot converge to $q$. Thus we have a found a sequence $p_n\neq p$ in $E$ such that $p_n$ converges to $p$ and $f(p_n)$ does not converge to $f(p)$.
A: There is no $\epsilon_{n}$ but there is a $\delta_{n}$ and this simply says that choice of delta (which is the size of the neighbourhood around $p$) is dependent on $n$ where $n\in\mathbb{N}$. There is nothing special about choossing $\delta_{n}=\frac{1}{n}$. All we want is that $\lim_{n\to\infty}\delta_{n}=0$ so that the neighbourhoods get smaller around $p$. The author is constructing a sequence that converges to $p$ (but is never equal to $p$) by first assuming that $\lim_{x\to p}f(x)\neq q$. This assumption implies that there is an $\epsilon$ such that for every $\delta>0$ we can find $x\in E$ such that $d(f(x),q)\ge\epsilon$ but $0<d(x,p)<\delta$. Since this is true of any $\delta>0$ we may choose neighbourhoods as we please. First take $\delta_{1}=1$. By the above there exists a point $x_{1}\in E$ such that $d(f(x_{1}),q)\ge\epsilon$ but $0<d(x_{1},p)<1$. We choose the next $\delta$ neighbourhood as $\delta_{2}=\frac{1}{2}$. By above again there exists a point $x_{2}\in E$ such that $d(f(x_{2}),q)\ge\epsilon$ but $0<d(x_{2},p)<\frac{1}{2}$. It could be that $x_{1}=x_{2}$ but as we continue to make $\delta$ smaller and smaller this will eventually not be possible. Proceeding in this way we construct a sequence that converges to $p$ but $d(f(x_{n}),q)\ge\epsilon$. Note that $x_{n}\neq p$ for all $n$ since $0<d(x_{n},p)$ for all $n$. We conclude that $\lim_{n\to\infty}f(x_{n})\neq q$.
A: 1) What does $\delta_n$ signify?
$\delta_n = 1/n$ is just a specific positive number for the general $\delta$.
2) Why is it specifically $1/n$?
It is chosen to make the sequence $\{p_n\}$ such that $0 \lt d_X(p_n, p) \lt 1/n$ converge to $p$.
We can choose any sequence $\{a_n\}$ of positive numbers converging $0$ instead of the sequence $\{1/n\}$.
3) Which is a suquence in E satisfying (6) but not (5)?
It is the sequence $\{p_n\}$ such that $d_Y(f(p_n), q) \ge \epsilon$ and $0 \lt d_X(p_n, p) \lt 1/n$.
