Closure and pathwise connected On $\mathbb R^2$ I consider the following set $G:=\{(x,\sin(\frac{1}{x}))| x\in (0,1]\}$ 
Well this set is pathwise connected (as a graph of a contionous function on an interval). The function is also oscillating on the left hand side so it makes sense to write the closure as $\bar{G}=G\cup(\{0\}\times[-1,1])$ but how can this be proved formally?
The next thing which is not clear to me is why $\bar{G}$ is connected but not pathwise connected, so there is no contin. way $c:[0,1]\rightarrow \bar{G}$ with starting point $c(0)=x$ and $c(1)=y$
 A: It seems to me that $G$ and the other set are not disjointed. If fact, every point of $(\{0\}\cup [-1,1])$ is an accumulation point of $G$. Now since both sets are the images of intervals which created continuously so the union of them is connected. I can't find a path connecting a point of $G$ to another point in another set which lie in the union, so it seems to me that it is not path-wise connected. If these points are not useful to make the problem clear, sign me to remove that. Thanks.
A: Take a point $(0,y)$ for $0\leq y\leq 1$, if $x=\arcsin(y)$ then we know that $\sin(x+2\pi n)=y$ for every $n\in\mathbb{N}$ so then $(\frac{1}{x+2\pi n},\sin(x+2\pi n))$ is a sequence (in $n$) of points in $G$ which converge to $(0,y)$ hence $(0,y)$ is in the closure of $G$. Then as the set $G\cup \{(0,y)\,:\, y\in [0.1]\}$ is a closed set containing all limit points, it is indeed the closure.
You can prove directly that if $S$ is any connected subset of $\mathbb{R}^n$ that for any $T$ satisfying $S\subseteq T\subseteq \overline{S}$ is connected. Indeed I am pretty sure this holds in general. (For instance let $f:T\rightarrow \{0,1\}$ be a continuous function, as $S\subseteq T$ is a subset of a connected component then $f$ is constant on $S$. As every point in $T\setminus S$ is a limit point of $S$ there is a sequence of points in $S$ converging to $T$. Apply $f$ to this sequence and continuity gives that $f$ maps the points in $T\setminus S$ to the same value as it does for $S$. Hence $f$ is constant)
For the failure of path connectedness notice that if you want to find a path $c$ from say $(2\pi,0)$ to $(0,1)$ then by continuity then every sequence of points $\{a_n\}_{n=1}^\infty$ in $[0,1]$ converging to $1$ has to satisfy $c(a_n)\rightarrow c(1)=(0,1)$ however if you take the troughs of the $(x,\sin(\frac{1}{x}))$ you can show this is not the case. To be more formal let $U$ be an open subset of $\overline{G}$ not containing any points of the $x$ axis. By continuity of $c$ we know that the image of $c$ is eventually contained in $U$ i.e. there exists $0<t<1$ so that $c([t,1])\subseteq U$ however it should be clear that $c$ has to go up the peaks and down the troughs of $G$ and so can't remain in $U$.
