Let $X = \{0\} \cup \{1, 1/2, 1/3, ... , 1/n, ... \}$ and let $Y$ be a countable discrete space. Show that $X$ & $Y$ don't have the same homotopy type 
Let $X = \{0\} \cup \{1, 1/2, 1/3, ... , 1/n, ... \}$ and let $Y$ be a countable discrete space. Show that $X$ and $Y$ do not have the same homotopy type.

I think that the author wants me to show a contradiction if I suppose that the spaces have the same homotopy type. So suppose that they have. Then for maps $f:X \to Y$ and $g: Y \to X$ there exists homotopies $H: g \circ f \simeq id_X$ and $H':f \circ g \simeq id_Y$. That is two maps $$H:X \times I \to X \text{ and } H':Y \times I \to Y$$ that satisfies $$H(x,0)=g(f(x)), H(x,1)=id_X \text { and } H'(x,0)=f(g(x)), H'(x,1)=id_Y$$
However it's not immediately obvious how to get a contradiction just from how $X$ and $Y$ are defined. I only have that $X$ is an infinite space and that $Y$ is a countable discrete space.
What is the "crux" here? Does it have something to do with the fact that we have chosen $X$ to be the set of points from the sequence $1/n$ which converges to $0$?
 A: Consider the homotopy $H' \colon Y \times I \to Y$. Fix an arbitrary $y_{0} \in Y,$ and consider the map
$$H'_{y_{0}} \colon I  \to Y,$$ defined by
$$H'_{y_{0}}(t) = H'(y_{0}, t)$$ for all $t \in I.$ The continuity of $H'_{y_{0}}$ follows from the continuity of $H'$ (do you see why?).
Note that $H'_{y_{0}}$ is a path connecting $H'_{y_{0}}(0)$ and $H'_{y_{0}}(1)$ in the space $Y$. However, since $Y$ is a discrete space, the path-components of $Y$ are singletons (sets containing only one point). This means that $H'_{y_{0}}(0)$ and $H'_{y_{0}}(1)$ must be the same point!
Unravelling the definition of $H'_{y_{0}}$ and using the fact that $H'$ is a homotopy between $f \circ g$ and $\text{id}_{Y}$, we have
$$f(g(y_{0})) = H'(y_{0}, 0) = H'_{y_{0}}(0) = H'_{y_{0}}(1) = H'(y_{0}, 1) = \text{id}_{Y}(y_{0}) = y_{0}.$$
Since $y_{0} \in Y$ was arbitrary, this proves that $(f \circ g)(y) = f(g(y)) = y$ for all $y \in Y.$ So, we have $f \circ g = \text{id}_{Y}.$
Now, we can use the same argument for $H \colon X \times I \to X.$ Even though $X$ is not discrete, you can still show that the path-components of $X$ are singletons. So, the above proof still goes through, and we can conclude that
$g \circ f = \text{id}_{Y}.$
So, we conclude that $f, g$ are homeomorphisms. Can you find a contradiction here?
