Closure of topology sine $Y = \lbrace (x,\sin(\frac{1}{x})) : x > 0 \rbrace $

Question

Given the subspace $Y = \lbrace (x,\sin(\frac{1}{x})) : x > 0 \rbrace \subset \mathbb{R}^2$, I know that $Y$ is connected being the image of a connected of $\mathbb{R}$ (i.e. $(0,+\infty)$.), through a continuous function.

I know also that that the closure of a connected is connected, so $\overline{Y}$ is connected, but I'd like to prove that $\overline{Y} = Y \cup \lbrace (0,t) : \lvert t \rvert \leq 1 \rbrace$

Are there any simple method to prove it ?

Answer 1

Denote $Y_1= \lbrace (0,t) : \lvert t \rvert \leq 1 \rbrace$. We have $\overline{Y} = Y \cup Y_1$. Let's prove that $\mathbb R^2 \setminus \overline{Y}$ is open:

• $\mathbb R^2 \setminus \overline{Y} = \left(\mathbb R^2 \setminus Y\right) \cap \left(\mathbb R^2 \setminus Y_1\right)$. So proving that $\mathbb R^2 \setminus Y, \mathbb R^2 \setminus Y_1$ are open is sufficient.
• As $Y_1$ is closed, $\mathbb R^2 \setminus Y_1$ is open.
• Now let's consider $\mathbb R^2 \setminus Y$. For $0 <a<b$, $f(x) = \sin\left(\frac{1}{x}\right)$ is continuous on $[a,b]$. Hence $Y(a,b) = \{(x,\sin\left(\frac{1}{x}\right)) \mid x \in [a,b]\}$ is closed in $\mathbb R^2$ as the image of a compact by a continuous map is compact. Therefore for $P=(r,s) \in \mathbb R^2 \setminus Y$ with $r>0$, it exists an open ball centered on $P$ included in $\{(x,y) \in \mathbb R^2 \mid r/2 < x <2r \} \setminus Y$. This proves that $\mathbb R^2 \setminus Y$ is open.

and concludes the proof.

Answer 2

The open set $(\leftarrow,0)\times\Bbb R$ (i.e., the open left half-plane) shows that $Y$ has no limit points to the left of the $y$-axis. $\Bbb R$ is a Hausdorff space, and the sine function is continuous on $(0,\to)$, so $Y$ is a closed subset of $(0,\to)\times\Bbb R$; this shows that the only limit points of $Y$ to the right of the $y$-axis are the points of $Y$. Thus, all of the limit points of $Y$ not in $Y$ must be on the $y$-axis. The open sets $\Bbb R\times(1,\to)$ and $\Bbb R\times(\leftarrow,-1)$ are disjoint from $Y$, so any limit points of $Y$ on the $y$-axis must lie on the segment $\{0\}\times[-1,1]$. To complete the argument, for each $y\in[-1,1]$ find a sequence $\langle x_n:n\in\Bbb N\rangle$ in $\Bbb R$ such that $\big\langle\langle x_n,y\rangle:n\in\Bbb N\big\rangle$ converges to $\langle 0,y\rangle$.