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The two-point Sierpinski space is usually defined as follows:

Let $X =\{x,y\}$ be the two-point space where the only open sets are $X, \varnothing, \{x\}$. I think from this it can be inferred that $X, \varnothing, \{y\}$ are closed, correct? Then this information can be used to show that $X$ is in fact connected. I believe you can also show that it is path connected; a path from $0$ to $1$ in $X$ is given by the function: $f(0) = 0$ and $f(t) = 1$ for $t > 0$. The function $f : I → X$ is continuous since $f^{−1}(1) = (0,1]$, which is open in $I$. Now I know that a space $Y$ is contractible provided that we have id$_Y \simeq e_{y_0}$, i.e. the identity map of $Y$ is homotopic to the constant map $e_{y_0}$ (definition of being nulhomotopic). My question is how can you show that the two-point Sierpinski space is contractible? Doesn't that just require "tweaking" the function $f$ that was used to show that $X$ is path connected? Also, recall that a space is contractible provided that it is both path connected and simply connected. How would this approach go?

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It may be worth pointing out that there are a ton of spaces which are path connected and simply connected, but not contractible. The simplest example is probably $S^2$, the two dimensional sphere sitting in $\mathbb{R}^3$. – Jason DeVito Dec 25 '12 at 3:32
up vote 4 down vote accepted

The easiest way is probably to find a working homotopy between the identity map and a constant function.

Define $H:X\times [0,1]\to X$ so that $H(z,0)=z$ for all $z\in X$, and $H(z,t)=x$ for all $z\in X$ and $t\in]0,1]$. Now $H^{-1}(\{x\})=\{(x,0)\}\cup X\times (0,1]$, which is an open subset of $X\times [0,1]$, as a complement of a closed set if you wish. This shows that $H$ is continuous. Also, since $H(z,0)=\mathrm{id}_{X}(z)$ and $H(z,1)=e_{x}(z)$ for all $z\in X$, then this shows that $H:\mathrm{id}_{X}\cong e_{x}$.

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This was exactly what I had in mind! – Libertron Dec 25 '12 at 0:15

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