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

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 $\overline{S}=S \cup \left\lbrace 0 \right\rbrace$.

• Hint: On any interval, $(M, \infty)$, $\sin(x)$ achieves all values in $[-1, 1]$ infinitely many times. Oct 12 '14 at 13:49
• The problem is that we cannot define $\gamma$ in the way $\gamma(t) := p+t(q-p)$ for $t \in [0,1]$ because we need $\gamma([0,1]) \subseteq S$. Oct 12 '14 at 14:01
• My hint was supposed to help find $\bar{S}$, but doesn't (directly) help with the path-connectedness issue. Oct 12 '14 at 14:05

$\overline{S} = S \bigcup \{(0, y): y \in [-1, 1]\}$. 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 $\gamma$.

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 $f$.

• And how can I find $\gamma$ to show that $S$ is path-connected? Oct 12 '14 at 14:32
• It's a continuous image of the path connected $[0,1)$ under $x \rightarrow (x, \sin(\frac{1}{x}))$, which is continuous. Oct 12 '14 at 16:33

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

• I am not sure if the definition of $\gamma$ is well-defined because we need that $\gamma([0,1]) \subseteq S$. And $S$ is a kind of graph of a function and definitely not convex. So I think we cannot take the natural path. Oct 12 '14 at 14:37
• @user144697 the parameterization then gets plugged into the graph of the function, so it's not a straight line path, it's a path along the function.
– Alan
Oct 12 '14 at 14:40
• I am sorry, you are right of course. Oct 12 '14 at 14:41