Say we have the following subset of $R^2$:
$E=(x,sin\frac1x ):x\in (0,1]$
E is just the graph of $g(x)=sin\frac1x$ along the interval $(0,1]$.
How can we determine its closure $E^-$? And how then would one show connectedness for the closure $E^-$?
MY SILLY THOUGHTS INCOMING:
I can see the graph $g(x)=sin\frac1x$ oscillates infinitely, and quite wildly near $x=0$. So I know $sin$ is continuous, and I'm trying to think what limits points are not contained in the set...
would the closure $E^-$ just be $(x,sin\frac1x ):x\in (0,1] \cup [0,1]$, so that its closed on both sides?
First let us try to find the closure of $E$ denoted by $\bar{E}$. Consider the following $\{ a_k \}_{k>0}$ defined by $$a_k = \frac{1}{(4k+1)\pi/2} $$ then this sequence is in $(0,1]$, thus the sequence $\{ (a_k,b_k)\}_{k>0}$ defined by $$b_k= \sin(\frac{1}{a_k}) $$ is in $E$ . Moreover, $b_k=\sin((4k+1)\pi/2 )=1$ for all $k>0$, and we have $$\lim_{k \rightarrow \infty } a_k=0 $$ Hence the point $(0,1)$ is in $\bar{E}$.
Similarly if you take $a_k=\frac{1}{(4k+3)\pi/2}$ and $b_k=\sin(\frac{1}{a_k})$ then $\{ (a_k,b_k)\}$ is in $E$ and $b_k=-1$ for all $K>0$, and $a_k$ has limit zero at infinity. Thus the point $(0,-1)$ is also in $\bar{E}$.
On the other hand, for any $x\in ]-1,1[$, let $\theta= \arcsin(x)$ (choose $\theta $ positive), and let $a_k=\frac{1}{\theta+2k\pi}$ then $a_k \in (0,1]$ for all $K>0$ and considering $b_k=\sin(\frac{1}{a_k})$, then the sequence $\{ (a_k,b_k)\}$ is in $E$, and its limit as $k$ approaches infinity is $ (0,x)$ . Thuhs $(0,x) \in \bar{E}$ for all $x \in ]-1,1[$.
Therefore $\bar{E}= \{ (x, \sin(\frac{1}{x})) , x \in (0,1]\} \cup \{ 0 \} \times [-1,1]$.
Now back to connectedness, As you set $E$ is connected thenthe closure of $E$ is connected , for this result you may see this. $E$ is clearly connected since it is the graph of a continuous function defined on a connected set $(0,1]$.
By the way, the set you are asking about has a lot of properties it is Topologist sine curve.