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$\textbf{4.2}\,\,$ Theorem $\,\,$ Let $X,Y,E,f$, and $p$ be as in Definition $4.1$. Then $$\lim_{x\to p}f(x)=q\tag{4}$$ if and only if $$\lim_{n\to\infty}f(p_n)=q\tag{5}$$ for every sequence $\{p_n\}$ in $E$ such that $$p_n\ne p,\quad\lim_{n\to\infty}p_n=p.\tag{6}$$ *Proof*$\quad$ Suppose $\text{(4)}$ holds. Choose $\{p_n\}$ in $E$ is satisfying $\text{(6)}$. Let $\varepsilon>0$ be given. Then there exists $\delta>0$ such that $d_Y(f(x),q)<\varepsilon$ if $x\in E$ and $0<d_X(x,p)<\delta$. Also, there exists $N$ such that $n>N$ implies $0<d_X(p_n,p)<\delta$. Thus, for $n>N$, we have $d_Y(f(p_n),q)<\varepsilon$, which shows that $\text{(5)}$ holds.

$\quad\quad\quad\,\,$ Conversely, suppose $\text{(4)}$ is false. Then there exists some $\varepsilon>0$ such that for every $\delta>0$ there exists a point $x\in E$ (depending on $\delta$), for which $d_Y(f(x)),q)\ge\varepsilon$ but $0<d_X(x,p)<\delta$. $\color{yellow}{\boldsymbol{\underline{\color{black}{\text{Taking $\delta_n=1/n(n=1,\,2,\,3,\,\ldots)$, we thus find a sequence in $E$}}}}}$$\quad$ $\color{yellow}{\boldsymbol{\underline{\color{black}{\text{satisfying (6) for which (5) is false.}}}}}$


I'm having a hard time understanding the last line of proof (which I highlighted).

Which sequence satisfies $\text{(6)}$ but make $\text{(5)}$ false?

Thank you!

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From the previous sentence: for each $\delta$, there exists a point $x\in E$ s.t. $d_Y(f(x), q) \geq \varepsilon$ but $0 < d_X(x,p) < \delta$.

Take $(\delta_n)_{n\in\mathbb{N}}$ as defined, which is a sequence decreasing to $0$. For each $n$, consider the $x_n$ "corresponding to" $\delta_n$: $$x_n\in E,\qquad d_Y(f(x_n), q) \geq \varepsilon, \qquad 0 < d_X(x_n,p) < \delta_n$$ Then, since $\delta_n\xrightarrow[n\to\infty]{}0$, we have $\sf{(6)}$: $x_n\xrightarrow[n\to\infty]{}p$ (since $d_X(x_n,p)\to 0$). However, we can't have $\sf{(5)}$, since $$\forall n,\ d_Y(f(x_n), q) \geq \varepsilon > 0$$ i.e. $f(x_n)$ stays away from $q$ by a distance at least $\varepsilon$, and therefore cannot converge to $q$.

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