# Prove the long line is not contractible.

Given the following definition of the long line:

Let $\omega_1$ be the first uncountable ordinal and consider $[0,1)$ as an ordinary set. Define the long ray to be the ordered set $\omega_1 \times [0,1)$ taken in lexicographical order. As a space, it is given the order topology. Define the long line to be the space obtained by gluing together two long rays together at their initial points.

Prove the long line is not contractible.

An outline of a proof can be given as follows:

Assume it is contractible. Then, denoting $L$ to be the long line, there exists a homotopy $H: L \times [0,1] \to L$ such that, for $x \in L$, $H(x,0) = c$, a constant, and$H(x,1) = id_L$, the identity map. For each $t \in [0,1]$, $\forall x \in L$, $H(x,t)$ would therefore be an interval, since L is connected. Define $A = \left\{ t \in [0,1] : H(x,t) \text{ is bounded} \right\}$. If I can somehow show that $A$ is clopen, then since $0 \in A$, it would be the case that $A = [0,1]$. But this is impossible, since $1 \notin A$.

Any thoughts on showing how $A$ is both closed and open?

HINT: What you’ve written isn’t quite right. $H(x,t)$ is simply a point of $L$; I suspect that you mean that for each $t\in[0,1]$, the set

$$H[L\times\{t\}]=\{H(x,t):x\in L\}$$

is an interval in $L$. Call it $I_t$. Then presumably $A=\{t\in[0,1]:I_t\text{ is bounded in }L\}$.

Fix $t\in A$, and let $b\in L$ be such that $I_t\subseteq[-b,b]$. (Here I’m using $-b$ for the copy of $b$ in the copy of the long ray that is to the ‘left’ of the common point of the two long rays.)

• Show that in fact $I_t\subseteq(-b,b)$.

Let $U=H^{-1}[(-b,b)]$.

• Show that $U$ is an open nbhd of $L\times\{t\}$ in $L\times[0,1]$, and conclude that for each $x\in L$ there are an $n_x\in\Bbb N$ and points $u_x,v_x\in L$ such that $u_x<x<v_x$ and $(u_x,v_x)\times B(t,2^{-n_x})\subseteq U$, where $(u_x,v_x)$ is an open interval in $L$, and $$B(t,2^{-n_x})=(t-2^{-n_x},t+2^{-n_x})\cap[0,1]\;.$$

For $n\in\Bbb N$ let $L_n=\{x\in L:n_x=n\}$, and let $V_n=\bigcup\limits_{x\in L_n}(u_x,v_x)$.

• Verify that $V_n\times B(t,2^{-n})\subseteq U$ for each $n\in\Bbb N$.

Let $\mathscr{V}=\{V_n:n\in\Bbb N\}$; $\mathscr{V}$ is a countable open cover of $L$.

• Show that $L$ is countably compact. (If you’ve not already done so, show first that $\omega_1$ with its order topology is countably compact.) Use a finite subcover of $\mathscr{V}$ to show that there is an $n\in\Bbb N$ such that $L\times B(t,2^{-n})\subseteq U$. Conclude that $A$ is open.

Showing that $A$ is closed is easier. Let $\langle t_n:n\in\Bbb N\rangle$ be a convergent sequence in $A$ with limit $t\in[0,1]$. For each $n\in\Bbb N$ there is a $b_n\in L$ such that $I_{t_n}\subseteq[-b_n,b_n]$. Now use the fact that $\{b_n:n\in\Bbb N\}$ must be bounded in $L$.

• That's very helpful, but following that proof, why wouldn't that work if you just substituted the real line for the long line. Obviously the same proof doesn't work, but it's not clear to me where it fails. It doesn't seem to fail at the point when you proof that the set is closed, since closrd intervals of the reals do contain their limit points. But I don't see where the proof would fail in the open argument either. The reals are countable compact too, so it wouldn't fail there. Can you provide any insight? Thanks. – nycguy92 May 21 '15 at 13:21
• @nycguy92: The first part would fail because $\Bbb R$, unlike $L$, is not countably compact. The second would fail because the union of countably many bounded subsets of $\Bbb R$ need not be bounded. – Brian M. Scott May 21 '15 at 22:47