Set $S$ which is path-connected, but $\overline{S}$ is not path-connected

Question

Suppose the set

$$S := \left\lbrace x+i \sin \left( \frac{1}{x} \right) \Bigg\vert x \in (0,1]\right\rbrace \subseteq \mathbb{C}. $$

I want to show that $S$ is path-connected but $\overline{S}$ is not path connected.

To show that $S$ is path-connected I have to show that for all points $p,q \in S$ there exists a continuous function $\gamma \colon [0,1] \rightarrow S$ with $\gamma(0)=p$ and $\gamma(1)=q$.

I know that the limit $\lim_{x \to 0} \sin(1/x)$ does not exist. So I am not sure how to calculate $\overline{S}$. My guess is that $\overline{S}=S \cup \left\lbrace 0 \right\rbrace$.

Answer 1

Hint:$f(x)=\sin(\frac {1}{x})$ is continuous on $(0,1]$. So intermediate value theorem.

Write is as $S=\{(x,f(x)):x\in(0,1]\}$. $f$ is continuous , and $S$ is the graph of $f$.

Answer 2

$\overline{S} = S \bigcup \{(0, y): y \in [-1, 1]\}$. Suppose $\gamma: [0, 1] \rightarrow \overline{S}$ is a one-one path from say $\gamma(0) = (0, 1)$ to $\gamma(1) = (\pi/2, 1)$. Argue that the diameter of $\gamma([0, \epsilon])$ must be at least $1$ for all $\epsilon > 0$. This contradicts the continuity of $\gamma$.

Answer 3

For path connected, each point p, q is characterized by $x_1,x_2\in (0,1]$ such that $p=x_1+i\sin (\frac 1 {x_1})$, $q=x_2+i\sin (\frac 1 {x_2})$. Go for the natural path, rescaling $x_1$ to 0 and $x_2$ to 1, i.e. a parameterization. So $\gamma(t)=((1-t)x_1+tx_2)+i\sin(\frac 1 {(1-t)x_1 +tx_2)})$. Continuity comes automatically from composition of continuous functions.

Now, for the closure of S, the only point where you have limit behavior is as $x\to 0$ Here, we approach the points with a real coordinate of 0 and an imaginary component anywhere between $[-1,1]$. Indeed, you can show any point of the form $0+\alpha i$ is a limit point as there are an infinite amount of the points inside any epsilon neighborhood of this point from S. What's more, if you aren't on the set itself and your real component ISN"T 0, then you won't have an infinite number of points in any epsilon neighborhood (you can bound the set around the real component part and show only a finite number of the points are nearby). Similarly, if the imaginary component is outside [-1,1] you aren't in the closure.

So, the closure of S is equal to $S\cup {ti:t\in [-1,1]}$. Now, to show this isn't path connected, take one point in the limit set and one point not in the limit set. So, say $x_1=0$ and $x_2=1+i\sin 1$. There is no continuous path on the closure of S to connect these points in finite time, as the you have infinite oscillations as your real part approaches 0