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I asked this questionthis 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?

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?

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?

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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?