Openness/Closeness of components of the topologist's sine curve Let $A = \{(0, y)\ | \ y ∈ [−1, 1]\}$, $B = \{(x,\text{sin}(\frac{1}x)) \ | \ x \in (0, 1]\}$ and $X = A \cup B$ as subspaces of $\mathbb{R}^2$.
How do I determine whether $A$ and $B$ are open in $X$ or not?
 A: I suppose that $X$ inherits its topology from the canonical topology of $\Bbb R^2$.
Then $A$ is not open (but closed), and $B$ is open in $X$.
Proposition. $A$ is not open in $X$.
Proof. Suppose that $A$ were open in $X$. Then 
$$\exists \text{open } O\subset \Bbb R^2: A = O\cap X.$$ 
So $0\in O$. Since $O$ is open, 
$$\exists \varepsilon>0: B_\varepsilon(0)\subset O,$$ where  $B_\varepsilon(0)$ denotes the circle of radius $\varepsilon$ around $0$.
Note that the topologist's sine curve gets arbitrarily close to $0$, so $O$ can't be open (since we have for every $\varepsilon>0: B_\varepsilon(0)\cap B = (B_\varepsilon(0)\cap X)\setminus A\neq\emptyset$.)
Contradiction. $\square$
$B$ is open in $X$, since $B=X\cap\Omega$ where $\Omega := ]0,2[ \times ]-2,2[$ is open in $\Bbb R^2$. It follows aswell that $A$ is closed in $X$.
A: $B$ is open in $X$ since $B = X \cap \big((0, +\infty)\times \mathbb{R}\big)$.
$A$ is not open in $X$ because $X = \overline{B}$ in $\mathbb{R}^2$ (see e.g. here) so for every $a \in A$ there exists a sequence $(b_n)_n$ in $B$ such that $b_n \to a$. Hence $A$ cannot contain any open balls around $a$.
