$X = [0,1]$ and set $Y=\{1-\frac{1}{n}\ \vert\ n \in \mathbb{Z}_{>0}\}$; $\sim_X$ is the equivalence relation generated by $0 \sim 1$. I have to show that $Y/\sim_Y \rightarrow X/\sim_X$ is not a homeomorphism onto its image in $X/\sim_X$. $\sim_Y$ is the equivalence induced on $Y \subset X$ by $\sim_X$
I know the following facts:
$f:Y/\sim_Y \rightarrow X/\sim_X$ is continuous. So, I need to show $f$ is not an open map or a closed map onto it's image.
And clearly, $f(Y/\sim_Y)$ is closed, since $[1]=[0]\in f(Y/\sim_Y) \subset X/\sim_X$
How do I proceed forward?
Hint: $0$ is an isolated point of $Y/{\sim_Y}$ but is not an isolated point of $f(Y/{\sim_Y})$. Isolated points are mapped to isolated points under homeomorphisms (because a point $x$ is isolated iff $\{x\}$ is open and under a homeomorphism $h$, $\{h(x)\} = h(\{x\})$ is open if (and only if) $\{x\}$ is open).