Prove that the following map is not a quotient map

Question

Let $X= \bigcup_{n \in \mathbb{N}} (\mathbb{R} \times \{n \})$ and let $Y=\bigcup_{n \in \mathbb{N}} \{(x,y):y=nx \}$. Suposse both $X$ and $Y$ have the subspace topology induced by usual topology on $\mathbb{R}^2$. Define $f: X \to Y$ by $p((x,n))=(x,nx)$ for each $(x,n) \in X$. Show that $p$ is not a quotient map. This is my idea let $A= \{ (x,y): y=x \} \subset Y$, and $p^{-1} (A) =\{(x,n): x=0 \} \cup \{ (x,1): x \in \mathbb{R} \}$, but I don't know if $p^{-1} (A)$ is open in $X$

Answer 1

$A$ is not open in $Y$, since every open nbhd of $\langle 0,0\rangle$ contains points of $Y\setminus A$, and $p^{-1}[A]$ is not open in $X$, since every open nbhd of $\langle 0,2\rangle$ contains points of $X\setminus p^{-1}[A]$.

Try this instead. Let

$$S=\left\{\langle x,nx\rangle\in Y:|x|<\frac1{2^n}\right\}\,,$$

and show that $S$ is not open in $Y$, but $p^{-1}[S]$ is open in $X$. If you have too much trouble showing that $S$ is not open in $Y$, check the spoiler below.

Show that the sequence $\left\langle\left\langle \frac1{2^n},\frac{n}{2^n}\right\rangle:n\in\Bbb N\right\rangle$ that bounds $S$ in the first quadrant converges to the origin.