| Se $p\neq x$, seja $\delta:=|p-x|/2$; então existe $n_0\in \mathbb{N}$ tal que $|x_n-x|<\delta$ para todo $n>n_0$. Então $|x_n-p|\ge |x-p|-|x_n-x|>\delta$ para todo $n> n_0$; se $\delta':=\min \{|x_n-p|:n\le n_0,\ x_n\neq p\}$ e $\delta_0=\min\{\delta,\delta'\}$, temos $(p-\delta_0,p+\delta_0)\cap S=\{p\}$, donde $g\left((p-\delta_0,p+\delta_0)\cap S\right)=\{g(p)\}\subset \left(g(p)-\varepsilon,g(p)+\varepsilon\right)$ para todo $\varepsilon>0$.\\ | Se $p\neq x$, seja $\delta:=|p-x|/2$; então existe $n_0\in \mathbb{N}$ tal que $|x_n-x|<\delta$ para todo $n>n_0$. Então $|x_n-p|\ge |x-p|-|x_n-x|>\delta$ para todo $n> n_0$; se $\delta':=\min \{|x_n-p|:n\le n_0,\ x_n\neq p\}$ e $\delta_0=\min\{\delta,\delta'\}$, temos $(p-\delta_0,p+\delta_0)\cap S=\{p\}$, donde $g\left((p-\delta_0,p+\delta_0)\cap S\right)=\{g(p)\}\subset \left(g(p)-\varepsilon,g(p)+\varepsilon\right)$ para todo $\varepsilon>0$.\\ |
| Se $p=x$, dado $\varpsilon>0, existe $n_0\in \mathbb{N}$ com $|y_n-y|<\varepsilon$ para todo $n>n_0$. Se $\delta:=\min \{|x_1-p|,\dots,|x_{n_0}-p|\}$, então $(p-\delta,p+\delta)\cap S\subset \{p,x_{n_0+1},x_{n_0+2},\dots\}\Rightarrow g\left((p-\delta,p+\delta)\cap S\right)\subset g\left(\{p,x_{n_0+1},x_{n_0+2},\dots\}\right)=\{y,y_{n_0+1},y_{n_0+2},\dots\}\subset (y-\varepsilon,y+\varepsilon)=\left(g(p)-\varepsilon,g(p)+\varepsilon \right)$. | Se $p=x$, dado $\varepsilon>0$, existe $n_0\in \mathbb{N}$ com $|y_n-y|<\varepsilon$ para todo $n>n_0$. Se $\delta:=\min \{|x_1-p|,\dots,|x_{n_0}-p|\}$, então $(p-\delta,p+\delta)\cap S\subset \{p,x_{n_0+1},x_{n_0+2},\dots\}\Rightarrow g\left((p-\delta,p+\delta)\cap S\right)\subset g\left(\{p,x_{n_0+1},x_{n_0+2},\dots\}\right)=\{y,y_{n_0+1},y_{n_0+2},\dots\}\\ \subset (y-\varepsilon,y+\varepsilon)=\left(g(p)-\varepsilon,g(p)+\varepsilon \right)$.\\ |