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 :(

  • $\begingroup$ Hint: 1. Show that if $Y$ is discrete and $f:Y\rightarrow Y$ is any function, then any homotopy of $F$ is constant in time. 2. As you noticed, if $f:X\rightarrow Y$ is continuous, then $f(X)$ is a finite subset of $Y$. Now, if $g:Y\rightarrow X$ then $f\circ g$ has finite image. Now apply 1... ( Feel free to write up your own answer. And note that this proof works for any infinite $Y$, but does not work if $Y$ is finite.) $\endgroup$ – Jason DeVito Feb 2 '16 at 2:34
  • $\begingroup$ @JasonDeVito Than you, nice solution, I was intended to fill you sketch, but Najib Idrissi came first. $\endgroup$ – O Empalador de Cabras Feb 2 '16 at 19:27

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.


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