The classifying space of an infinite totally ordered set is contractible I asked this question on math.stackexchange, but no one answered.
Let $(X,\le)$ be a totally ordered set. Regarding it as a category, it has a classifying space $B(X,\le)=|N_\bullet(X,\le)|$. This should be the (possible infinite) simplex with vertices $X$, hence I expect it to be contractible.
However, I was not able to explicitly prove contractibility starting from the definition $$B(X,\le)=(\coprod_{i\in\mathbb{N}_0}N_i(X,\le)\times\Delta^i)/\tilde{} $$ 
Can anyone help me?
 A: Let $P$ be any partially ordered set.  For any map $x\colon P\to [0,1]$, put $\sigma(x)=\{p\in P:x(p)>0\}$. Then $BP$ can be identified with the set of maps $x$ such that $\sigma(x)$ is finite and totally ordered, and $\sum_px(p)=1$.  Now suppose that $a$ is an element of $P$ which is comparable with every other element, and define $e\colon P\to [0,1]$ by $e(a)=1$ and $e(p)=0$ for $p\neq a$.  It is then easy to see that the map $h_t(x)=(1-t)x+te$ preserves $BP$ and gives a contraction.  This construction is most often used when $a$ is largest or smallest in $P$, but you really only need it to be comparable with every element of $P$.  In particular, if $P$ is nonempty and totally ordered then you can choose $a$ arbitrarily.
As another way to look at this, if $f,g\colon P\to Q$ are two poset maps, and $f(p)\leq g(p)$ for all $p$, then it is standard that $Bf$ is homotopic to $Bg$.  If $P$ is totally ordered, we can define $f,g,h\colon P\to P$ by $f(p)=p$ and $g(p)=\max(a,p)$ and $h(p)=a$.  Then $f\leq g\geq h$, so $Bf$, $Bg$ and $Bh$ are homotopic, but $Bf$ is the identity and $Bh$ is constant.
A: I am not sure if you consider this explicit, but here you go:
Choose a point $x_0\in X$, and let $F\colon X\to X$ be the functor that sends $x$ to itself if $x\geq x_0$, and to $x_0$ otherwise. There is a zig-zag of natural transformations
$$x_0\leq F(x)\geq x$$
between the constant functor $x_0$ and the identity functor, because $X$ is totally ordered. On classifying spaces this gives a homotopy between the constant map $x_0$ and the identity.
