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This is from GTM $202$, Introduction to Topological Manifolds, problem 4-6, 1st edition.

Long line is given by $\mathcal L= \big(Y\times[0,1)\big)-\{(a_0,0)\}$ with dictionary order.

$Y$ is an uncountable well-ordered set s.t. for any $p\in Y$, $\{q \in Y, q <p\}$ is countable. Its smallest element is $a_0$.

I'm going to prove $\mathcal L$ is path-connected.


For $x=(x_1,s), y=(y_1,t) \in \mathcal L\ $ with $x<y$.

(1) If $x_1=y_1$, then $s<t$ and $[x,y]=\{x_1\}\times[s,t]$ is homeomorphic to interval $[s,t] \in \Bbb R$, so $x,y$ can be connected by a path.

(2) If $x_1\not=y_1$, the long line interval $[x,y]$ is the union of $\{x_1\}\times[s,1)$, countably-many $\{z\}\times[0,1)$ with $x_1<z<y_1$ and $\{y_1\}\times[0,t]$.

Define $f:[x,y]\in\mathcal L\to[-1,2]\in \Bbb R$ in the following way:

$f$ maps $\big[(x_1,s),(x_1,1)\big)$ to $[-1,0)$, maps $\big[(y_1,0), (y_1,t)\big]$ to $[1,2]$ homeomorphically. For countably many $\{z\}\times[0,1)$ with $x_1<z<y_1$, they're 1-1 correspond to interval $[\frac{2^n-2}{2^n},\frac{2^n-1}{2^n}), n \in \Bbb N_+$ by order on $\mathcal L$.


My question:

How can I prove $f$ is continuous at $(y_1,0)\in \mathcal L$ and $f^{-1}$ is continuous at $1 \in \mathbb R$?

(Note:$f((y_1,0))=1$)

Thanks for your time and patience!


Update:

An alternative method: Problem 12, Sec. 24 of Munkres' "Topology," long line cannot be imbedded in reals

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    $\begingroup$ I suggest you look at this question and its answer for more clarity on how to show $L$ path-connected. $\endgroup$ – Henno Brandsma Sep 29 at 13:02
  • $\begingroup$ as a remark on your notation, it is accepted (i.e. it is a convention in set-theory) that an ordinal is the set of smaller ordinals $\alpha=\{\delta:\delta<\alpha\}$. This makes it particularly important to distinguish between $\alpha$ and the one-element set (singleton) $\{\alpha\}$. I believe that when you write $[x,y]=\alpha\times[s,t]$ you mean $[x,y]=\{\alpha\}\times[s,t]$. $\endgroup$ – Mirko Sep 29 at 13:44
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    $\begingroup$ Just a quick comment: you do not need the well-ordering theorem for $Y$ to exist. What you describe is just the ordinal $\omega_1$, which exists without the axiom of choice (which is equivalent to the well-ordering theorem). $\endgroup$ – Mark Kamsma Sep 29 at 15:16
  • $\begingroup$ Also see this note, of which you should only not trust the last statement: the long line is normal (despite being non-paracompact, but it is countably paracompact). The fact that the initial segments are homeomorphic to open intervals in $\Bbb R$, as shown, implies path-connectedness. See also this note which has many details. $\endgroup$ – Henno Brandsma Sep 29 at 18:03
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The following part of your construction need not work. You need a different construction. The way you define $f$, you may (and most of the time will) make $f$ discontinuous.

"And for countably many $\gamma\times[0,1)$ with $\alpha<\gamma<\beta$, they're 1-1 correspond to interval $[\frac{2^n-2}{2^n},\frac{2^n-1}{2^n}), n \in \Bbb N_+$ by order on $\mathcal L$."

For that matter, when you say "countably many" you mean countable or finite. You have not described what happens in the finite case. But, looking at the countable case, it is not necessary that (the set of) these countably many $\gamma$ is order-isomorphic to $\Bbb N_+$. (It is not clear to me what you mean when you say "by order on $\mathcal L$.")

Say you want to construct a path from $(0,\frac23)$ to $(\omega+1,\frac13)$, where $\omega=\{0,1,2,...\}$ is the first infinite ordinal (same order-type as $\Bbb N_+$, except $\omega$ starts at $0$, while $\Bbb N_+=\{1,2,3,...\}$ starts at $1$). In this example, the "countably many $\gamma\ $" in between would form the set $\{1,2,3,...,\omega\}$, a set which has a last element, unlike $\Bbb N_+$ which doesn't. So when you form a bijection with $\Bbb N_+$, this bijection cannot be order-preserving. It may look something like the following: $b:\{1,2,3,...,\omega\}\to\Bbb N_+$ defined by $b(\omega)=1$ and $b(n)=n+1$ for $1\le n<\omega$. In case I need it, let $c$ be the inverse bijection $c:\Bbb N_+\to\{1,2,3,...,\omega\}$ defined by $c(1)=\omega$ and $c(n)=n-1$ for $2\le n<\omega$.

So using this bijection, and your definition, it looks to me that you would match the interval $\{\omega\}\times[0,1)=\big[(\omega,0),(\omega,1)\big)$ with $[0,\frac12)$. Also, you would match $\{1\}\times[0,1)=\big[(1,0),(1,1)\big)$ with $[\frac12,\frac14)$. Could you see that the result would not be a continuous functions?

There are (at least) two ways around this problem. You could use transfinite recursion to define suitable paths, when larger-and-larger sets of $\gamma$'s get involved. Alternatively, you could use that every (finite or) countable ordinal is order isomorphic to a subset of $[0,1]$, and use that order-isomorphisim to define your continuous path.

By way of example, say again I want to construct a path from $(0,\frac23)$ to $(\omega+1,\frac13)$. As you do, you could start with a (hmm, you edited and changed your $\alpha$'s to $x_1$'s, but I will stick with $\alpha$'s, what was wrong with them), so start with an $f$ that maps $\big[(0,\frac23),(0,1)\big)$ to $[-1,0)$, then $f$ maps $\big[(\omega+1,0), (\omega+1,\frac13)\big]$ to $[1,2]$ homeomorphically. Next, map each $\big[(n,0),(n,1)\big)$ to $[\frac{2^n-2}{2^{n+1}},\frac{2^n-1}{2^{n+1}}),n\in\Bbb N_+$. Also, map $\big[(\omega,0),(\omega,1)\big)$ to $[\frac12,1)$. Then verify that this works.

I believe I have posted an answer on MSE, in the past (I couldn't find it, but I found this related question, and I added an extra answer there), showing that every countable ordinal is order-isomorphic to a subset of the real line (or a subset or $[0,1]$). Here is a sketch. For finite ordinals this is clear (just take a finite subset of $[0,1]$ with the same number of points). Now let $\nu=\{\gamma:\gamma<\nu\}$ be an infinite countable ordinal. Map $\gamma=0$ to $0\in[0,1]$. In general, fix a bijection $b:\{\gamma:\gamma<\nu\}\to\Bbb N_+$, and map each $\gamma<\nu$ to $\sum\limits_{\delta<\gamma}\frac{1}{2^{b(\delta)}}$.

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I think you are on the right track. You want to construct an order-preserving bijection between $ x=(x_1,s),y=(y_1,t)$, which will necessarily be continuous. You can proceed as follows, working in generalities before applying them to this specific case:

$1).\ $ You can chop up intervals so that the subintervals have the same order type as the original: for any ordered topological space $X,\ [x,y)$ has the same order type as $[0, 1)$ if and only if $[x, z)$ and $[z, y)$ do. Then, show that if $(x_n)$ is a sequence in $X$, and $s=\sup (x_n),\ [x_0, s)$ has the same order type as $[0, 1)$ if and only if every interval $[x_i, x_{i+1})$ does.

$2). $ Show that for every $x\in S_{\Omega}\setminus \{a_0\},$ the interval $[a_0\times 0, x\times 0)$ has the same order type as $[0,1).$

Hints:

For $1).$ try the following functions: if $f$ is an order-preserving bijection (from suitably chosen intervals),

$ f_i(x) =\frac{f(x)−f(x_i)}{f(x_{i+1})−f(x_i)}$

$g(x)=\begin{cases} 1−2^{−i}+2^{−(i+1)}f_i(x) & x_i\le x<x_{i+1}\\ 1 & x=s \end{cases}$

For $2).\ $ define $S=\{x\in S_{\Omega}\setminus\{a_0\}: [a_0×0, x×0)\ \text{has the same order type of [0, 1)}\}$ and use transfinite induction to prove that $S= S_{\Omega}\setminus\{a_0\}$.

Now note that if $x<y \in \mathcal L$, then there is an $s$ such that $x,y\in [a_0×0, s×0)$ and use the above to conclude.

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  • $\begingroup$ I mean $X$ is a space with an order topology. $\endgroup$ – Matematleta Sep 29 at 16:41
  • $\begingroup$ For instance $[x,y):=\{z\in X: x\le z<y\}$ I'm sorry I do not understand your comment. $\endgroup$ – Matematleta Sep 29 at 16:51
  • $\begingroup$ I guess you don't mean $[a,z)$ then, but $[x,z)$. And $[x,y)$ has OT $[0,1)$ iff $[x,z)$ and $[z,y)$ do whenever $x<z<y$. But in your (1) you wrote $[a,z)$ and this makes your sentence confusing. $\endgroup$ – Mirko Sep 29 at 16:55
  • $\begingroup$ I see. Yes. OK. Thanks $\endgroup$ – Matematleta Sep 29 at 18:08

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