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I am trying to prove that only contractible discrete space is the one-point space. The proof of it can be seen here show that discrete space of $X$ is contractible if have one point and here Proving that a three-point discrete space is not contractible.

I believe I have proof, but I do not use connectedness of the unit interval, so I'm asking for verification of my proof.

My proof

Suppose discrete space $X$ was contractible. Any contractible space is path-connected. Indeed, take a homotopy $H \colon X \times [0,1] \to X$ from identity on $X$ to $cost$ on $x_1$. Then for any $x_2 \in X$, $H(x_2, -) \colon [0,1] \to X$ gives a path from $x_1$ to $x_2$ as $H$ being continuous implies $H$ is continuous in each variable.

Now I claim that any discrete space with more than 2 points is not connected. To show this, note that if there is a continuous and non-constant map $X \to \{0,1\}$, then $X$ is not connected. As $X$ has more than 2 points, we can find a non-constant set map $X \to \{0,1\}$. And as $X$ is discrete, this non-constant set map must be continuous.

So if $X$ has more than 1 point and contractible, we get that $X$ is path connected but not connected. This is impossible, so $X$ must have one point if it were to be contractible.

Thanks!

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  • $\begingroup$ You ''secretly" used the connectivity of $[0, 1]$ for the fact "path-connected $\implies$ connected" which was necessary to reach a contradiction. $\endgroup$
    – 0XLR
    Jan 16 at 4:45
  • $\begingroup$ @OXLR Thanks so much! $\endgroup$
    – Phil
    Jan 16 at 4:47

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