Consider the sets
$X:= \left \{(0,0) \right\} ~~\bigcup ~\left \{(x,\sin \frac{1}{x}): x \in \mathbb{R} \setminus \{0\} \right \}$
$Y:= \left \{(0,0) \right\} ~~\bigcup ~\left \{(x,x\sin \frac{1}{x}): x \in \mathbb{R} \setminus \{0\} \right \}$
are metric spaces with metrics induced by the Euclidean metric of $\mathbb{R}^{2}$. Let $B_{X}$ and $B_{Y}$ be the open unit balls around (0,0) in $X$ and $Y,$ respectively. Then which of the following statements are true ?
- The closure of $B_{X}$ in $X$ is compact.
- The closure of $B_{Y}$ in $Y$ is compact.
$B_Y$ is closed and bounded in $\mathbb{R^2}$, hence it is compact. $B_X$ is bounded. Is it closed ?
Its limit points are $\{ (0,y)| y\in [-1,1]\}$
How to conclude is it a closed set or not ?
Thanks in advance.
Notice that $Y$ is a closed subset of $\mathbb R^2$. Indeed, let $(a_n,a_n\sin\frac{1}{a_n})_{n \in \mathbb N} $ be a convergent sequence of $Y$, with $$(a_n,a_n\sin\frac{1}{a_n}) \to (y_1,y_2).$$ It's not hard to see that $(y_1,y_2) \in Y$. (In the problematic case of $y_1 = 0$, we have that $y_2 = \lim_n a_n \sin \frac {1} {a_n}=0$ and so $(y_1,y_2) \in Y$). Therefore, $ \overline {B_Y}=Y \cap \overline {B_{\mathbb R} } $ is compact.
On the other hand, $ \overline {B_X}$ is not closed (and therefore not compact). Take for example the sequence $$ \Big (\frac {1/2}{2n\pi + \pi } , \sin (2n\pi +\pi/2) /2 \Big )_n \subset \overline {B_X}$$ Then,
$$\Big (\frac {1}{2n\pi + \pi/2 } , \sin (2n\pi +\pi/2) \Big )_n = \Big ( \frac {1}{2n\pi + \pi/2 }, 1 \Big )_n \to (0,1) \notin X .$$