Q:

Filtered colimit of a topological space

Suppose that $X$ is a space filtered by closed subspaces $X_{1}\subset X_{2}\subset \dots$. 
As topological space $X=\operatorname{colim}_{n}X_{n}$.
We define $Y_{n}=X_{n+1}/X_{n}$, and consider the induced maps $Y_{n}\rightarrow Y_{n+1}$. Let define $Y=\operatorname{colim}_{n}Y_{n}$. 
Question: $X$ is homeomorphic to $Y$?

A:

As far as I can tell, this is not true in general.
Consider simply $X_n=X$. In this case $Y_n$ (and consequently $Y$) will be a singleton.

