Homotopy equivalence between $X=\{0\}\cup \{\frac{1}{n},n\in \mathbb{N}\}$ and a discrete space Consider the space  $X=\{0\}\cup \{\frac{1}{n},n\in \mathbb{N}\}$ with the topology induced by the real line. 
Is $X$ homotopy equivalent to some enumerable discrete space $Y$?
My try was the following: Let $Y$ a enumerable discrete space. If $X$ and $Y$  are homotopy equivalent then there exists maps $f:X\to Y$ and $g:Y\to X$ s.t  $g\circ f:Y\to Y$ is homotopic to  $Id_Y$.  But $X$ is compact, so the image of $g\circ f$ must be finite.
However I can't explore this ideia :(
 A: Suppose that you have a homotopy equivalence $f : X \to Y$ with homotopy inverse $g : Y \to X$, where $Y$ is some discrete space (of any cardinality, finite or infinite). The space $X$ is compact, thus $f(X) \subset Y$ is compact; but the only compact subspaces of a discrete space are finite, so $f(X)$ is finite.
Since $f$ and $g$ are inverse homotopy equivalences, $f \circ g$ is homotopic to $\operatorname{id}_Y$, i.e. there is a homotopy $H : Y \times [0,1] \to Y$ with $H(y,0) = f(g(y))$ and $H(y,1) = y$. But $Y$ is discrete, so $H$ has to be constant on connected subspaces, and $\{y\} \times [0,1]$ is connected, thus $f(g(y)) = y$. But the image of $f$ is finite, whereas the image of the identity is all of $Y$, so $Y$ is finite.
But $X$ has an infinite number of path-connected components (every singleton is a path-connected component), whereas a finite discrete space only has a finite number of path-connected components. Since a homotopy equivalence preserves the number of path-connected components, this is a contradiction.
