How to prove that this map is not a quotient map This question was aksed in my topology quiz and I was unable to solve it.

Let $X=\bigcup_{n\in \mathbb{N}} (\mathbb{R} \times \{n\})$ and let $Y = \bigcup_{n\in \mathbb{N}} \{(x,y)\in \mathbb{R}^2 : y=nx\}$. Suppose both $X$ and $Y$ have the subspace topology induced by usual topology on $\mathbb{R}^2$. Define $p:X \to Y$ by $p((x,n))=(x,nx)$ for each $(x,n)\in X$.
Show that $p$ is not a quotient map.

I have proved the map to be surjective but I am not able to get an idea on how to obtain acontradiction if the map is assumed to be quotient map.
Kindly help me.
 A: Notice that $p$ is almost a bijection: it sends all of the points $\langle 0,n\rangle$ to $\langle 0,0\rangle$, but everywhere else it’s bijective. Thus, the place to look for problems is around $\langle 0,0\rangle$. What if we can find a sequence $\langle y_n:n\in\Bbb Z^+\rangle$ in $Y$ that converges to the origin and has just one point on each line $y=nx$? Then $\{y_n:n\in\Bbb Z^+\}$ won’t be closed in $Y$, because it’s missing the limit point $\langle 0,0\rangle$, but its preimage under $p$ will have just one point on each line $\Bbb R\times\{n\}$ and will be closed. If $p$ were a quotient map, this would be impossible: $p^{-1}[C]$ would be closed in $X$ if and only if $C$ were closed in $Y$.
Since we want the sequence to have just one point on each line $\Bbb R\times\{n\}$, we must have $y_n=p(\langle x_n,n\rangle)$ for some $x_n\in\Bbb R$, so that $y_n=\langle x_n,nx_n\rangle$. If these points are to converge to $\langle 0,0\rangle$ in $Y$, clearly the points $x_n$ must converge to $0$ in $\Bbb R$, and they must do so fast enough that the points $nx_n$ also converge to $0$ in $\Bbb R$. An easy way to ensure this is to let $x_n=\frac1{n^2}$, so that $nx_n=\frac1n$. Then

*

*the set $A=\left\{\left\langle\frac1{n^2},\frac1n\right\rangle:n\in\Bbb Z^+\right\}$ is not closed in $Y$, since it has $\langle 0,0\rangle$ as a limit point, but

*the set $p^{-1}[A]=\left\{\left\langle\frac1{n^2},n\right\rangle:n\in\Bbb Z^+\right\}$ is closed in $X$.

