I'm doing this exercise in Munkres' "Topology" (~paraphrasing). I have two motivations for reading this book: 1, topology is really pretty, but the main one, 2, is that I am not yet an undergraduate; thus have never taken any sort of 'proofs-based' course and would like to learn how to write readable proofs. I would appreciate:
- A verification of my proof
- Any advice about how to make my proof more readable.
Recall that $S_{\Omega}$ denotes the minimal uncountable well-ordered set. Let L denote the ordered set $S_{\Omega}\times[0,1)$ in the dictionary order, with its smallest element deleted. The set L is a classical example in topology called the long line.
Theorem: The long line is path connected and locally homeomorphic to $\mathbb{R}$, but it cannot be imbedded in $\mathbb{R}$.
The structure of the proof is given by Munkres.
(a) Let X be an ordered set; let $a<b<c$ be points of X. Show that $[a,c)$ has the order type of $[0,1)$ iff both $[a,b)$ and $[b,c)$ have the order type of $[0,1)$.
Assume $[a,b)$ and $[b,c)$ have the order type of $[0,1)$. Then there exist order-isomorphisms $f:[a,b)\rightarrow [0,\frac{1}{2})$ and $g:[b,c)\rightarrow [\frac{1}{2},1)$, because $[0,1)$ has the same order type as $[0,1/2)$. Define $h: [a,c) \rightarrow[0,1)$ as follows:
$$h(x) = \begin{cases}
f(x) \text{ if } x\in [a,b) \\
g(x) \text{ if } x\in [b,c)
\end{cases}$$
$h$ is an order-isomorphism, as $p<q$ is true iff $h(p)<h(q)$.
Conversely, assume $[a,c)$ has the order type of $[0,1)$. Then there exists an order-isomorphism $f:[a,c)\rightarrow [0,1)$. Restricting the domain to $[a,b)$ gives a new order-isomorphism $g:[a,b)\rightarrow [0,d)$, where $d\in (0,1)$. $[0,d)$ is of the order type of $[0,1)$; multiplying by $1/d$ is an order isomorphism. Similarly, restricting the domain of $f$ to $[b,c)$ gives order-isomorphism $h:[b,c) \rightarrow [d,1)$, where $d\in (0,1)$.
(b) Let X be an ordered set. Let $x_0 < x_1 < …$ be an increasing sequence of points of X; suppose $b=\sup \{x_i\}$. Show that $[x_0,b)$ has the order type of $[0,1)$ iff each interval $[x_i,x_i+1)$ has the order type of $[0,1)$.
Proving that $[x_0,b)$ having order type $[0,1)$ implies the same for the given intervals is simple, already having proved $(a)$. Begin with $i=1$: $[x_0,x_1), [x_1,b)$ is a partition of a set of the same form as we saw in $(a)$; thus each must have the same order type as $[0,1)$. We proceed by induction, considering now the set $[x_1,b)$ and the sequence $x_1<x_2<…$ The proof is the same.
edit: Conversely, assume each interval $[x_{i},x_{i+1})$ has order type $[0,1)$. We seek to construct an order-isomorphism $f:[x_0,b)\rightarrow [0,1)$ by assigning each $[x_{i},x_{i+1})$ to a unique section of $[0,1)$, as $[x_0,b)=\sqcup_{i\in \mathbb{Z}^{+}}[x_{i},x_{i+1})$. Define $U_0=[0,\frac{1}{2})$. Then inductively define $U_{i+1}=[\sup \{ U_{i}\},\sup \{ U_{i}\}+\frac{1}{2^{i+1}})$, each of which clearly have order type $[0,1)$ and are pairwise-disjoint. Note that the upper bounds of $U_i$ converge to $1$. Thus we can define order-isomorphisms $f_i:[x_i,x_{i+1})\rightarrow U_i$, and thereby order-isomorphism $f:[x_0,b)\rightarrow [0,1)$ by the union of all $f_i$.
(c) Let $a_0$ denote the smallest element of $S_{\Omega}$. For each element $a$ of $S_{\Omega}$ different from $a_0$, show that the interval $[a_0 \times 0, a\times 0)$ of $L$ has the order type of $[0,1)$. [Hint: Proceed by transfinite induction. Either a has an immediate predecessor in $S_{\Omega}$, or there is an increasing sequence $a_i$ in $S_{\Omega}$ with $a = \sup \{ a_i \}$]
$[a_0,a)$ is countable by the definition of $S_{\Omega}$. Following the hint: if $a$ has an immediate predecessor, $b$, then $[b \times 0, a \times 0)$ has the order type of $[0,1)$ (all points in this set are of the form $b\times x$, where $x\in [0,1)$).
By backwards induction to $a_0$ any finite section of $S_{\Omega}$, of which $a$ is the largest element, is order isomorphic to $[0,1)$.
If $a$ is not the largest element of a finite section of $S_{\Omega}$, then there is an increasing sequence $a_i$ in $S_{\Omega}$ with $a = \sup \{ a_i \}$, this sequence being exactly every element of $S_{\Omega}$ strictly less than $a$, in order (which must be countable). Each $a_{i+1}$ in this sequence has an immediate predecessor, namely $a_1$, so each $[a_{i}\times 0,a_{i+1}\times 0)$ must be order isomorphic to $[0,1)$ by the above argument. So, by $(b)$, so must $[a_0\times 0,a\times 0)$.
edit: See JunderscoreH's answer for a neater version of these arguments.
(d) Show that $L$ is path connected.
Extending $(c)$ to include sets of the form $[a,b), a,b\in L$ is trivial – simply add/subtract sets from the ends – I will assume this. This implies that $[a,b]$ has the same order type as $[0,1]$. So between any two points $a,b\in L$, there exists an order-isomorphism $f:[0,1]\rightarrow [a,b]$; which is necessarily a homeomorphism and thereby a continuous map.
edit: See JunderscoreH's answer for a more general version of this argument.
(e) Show that every point of $L$ has a neighborhood homeomorphic with an open interval in $\mathbb{R}$.
Let $x$ be our point, assume $x\neq a_0$. Then there exist some points, $a,b\in L$, such that $a<x<b$. By $(d)$, $L$ is path connected, so there exists some homeomorphism $f:[c,d]\rightarrow [a,b]$ such that $f(c)=a$, $f(d)=(b)$. Then by restricting the range of $f$ to $(c,d)$ we obtain a homeomorphism $f':(c,d)\rightarrow (a,b)$.
edit: Note that this is the purpose of 'deleting' the smallest element of $S_{\Omega} \times [0,1)$ to get $L$.
And finally…
(f) Show that L cannot be imbedded in $\mathbb{R}$, or indeed in $\mathbb{R}^n$ for any $n$. [Hint: any subspace of $\mathbb{R}^n$ has a countable basis for its topology]
edits sprinkled throughout: Assume there existed an imbedding $f:L\rightarrow \mathbb{R}^n$, thereby a homeomorphism $f':L\rightarrow Y$, $Y\subset \mathbb{R}^n$ obtained by restricting the range of $f$. $Y$ must have a countable basis because $\mathbb{R}^n$ is metrizable, so $L$ must also have a countable basis for its topology, being homeomorphic with $Y$.
Assume $L$ had a countable basis, call this basis the collection $X = \{U_{i}:i\in \mathbb{Z}^{+}\}$. Every open set of $L$ must contain, entirely, at least one element of $X$. Let $Y = \{ x_{\alpha}: x_{\alpha}=\alpha \times \frac{1}{2}, \alpha \in S_{\Omega} \}$. $Y$ is uncountable, being indexed by $S_{\Omega}$.
Then by (e), there is some uncountable collection $Z = \{ V_{\alpha} \}$ such that each $V_{\alpha}$ is a neighborhood of $x_{\alpha}$. Note that we can pick, more strongly, $Z$ such that it is pairwise-disjoint: for example, $V_{\alpha} = \{ \alpha \}\times (0,1)$ works.
Then each contains some $U_{i}$; thus there exists some injective function $f:Z\rightarrow X$ assigning $V_{\alpha}$ to $U_{i}$ if the former contains the latter. $f$ is injective, because $Z$ is pairwise-disjoint. But $Z$ is indexed by $S_{\Omega}$, $X$ by $\mathbb{Z}^{+}$, implying the existence of an injection $g:S_{\Omega}\rightarrow \mathbb{Z}^{+}$, contradicting the uncountability of $S_{\Omega}$.
$\therefore$
Best Answer
Some initial remarks: your writing style is remarkably good for someone who's not even an undergraduate. Kudos!
Also, the notation here is a little odd at times. For example, you'd usually see $\omega_1$ instead of $S_\Omega$. Maybe he wants to avoid the confusion of $0$ the ordinal with $0$ the real number, but some of the other notation is weird too: $\mathbb{Z}^+$ for $\mathbb{N}$; $x\times y$ used both for the cartesian product of $x$ and $y$, and for the ordered pair $\langle x,y\rangle$; and so on.
Regardless, I have more substantive comments too.
To show that $h$ is an order-isomorphism, we need both that $p<q$ implies $h(p)<h(q)$, but also that $h(p)<h(q)$ implies $p<q$. As defined, $h$ is not an order-isomorphism (it's not even injective). I believe you want $g$ to be a function from $[b,c)$ to $[\frac{1}{2},1)$ instead of $[0,\frac{1}{2})$. This would fix the problem.
You've only proven one direction: that if $[x_0,b)$ has order type $[0,1)$, then each $[x_i,x_{i+1})$ has order type $[0,1)$. It's the other direction that requires work.
What you're doing here is arguing by induction on $a$. For $a=b+1$ a successor, $[a_0,a)$ can be decomposed into $[a_0,b)$ and $[b,a)$. $[b,a)$ has order type $[0,1)$, and $[a_0,b)$ has the same order type by the inductive hypothesis. Hence by (c) $[a_0,a)$ has order type $[0,1)$.
Here is the same sort of argument as before. If $a\not = a_0$ is not a successor, it's a (countable) limit $a=\sup\{a_i\}$. By the inductive hypothesis, $[a_0,a_i)$ has order type $[0,1)$ for each $i$, and thus so too does $[a_i,a_{i+1})$ for each $i$. Hence by (c) $[a_0,a)$ has order type $[0,1)$.
Yes it looks good, and I think it's pretty clear, but it may help to be explicit: saying how you're extending (c), or just stating that it easily follows that for each $a,b\in L$, $[a,b)$ also has order type $[0,1)$.
If you're looking for a more expanded argument, $a\not = a_0$ and note that $[a,b)$ can be written as $[a_0,b)\setminus [a_0,a)$. $[a_0,b)$ has order type $[0,1)$ by (c), so take $f:[a_0,b)\rightarrow[0,1)$ an order isomorphism. Since $[a_0,a)$ will be mapped to an initial segment $[0,d)$, where $0<d<1$, it follows that $[a,b)$ is order isomorphic to $[d,1)$ which clearly has order type $[0,1)$.
Hence for all $a,b\in L$, $[a,b)$ has order type $[0,1)$ and so $[a,b]$ has order type $[0,1]$. The isomorphism witnessing this is necessarily continuous. As $a,b\in L$ were arbitrary, $L$ is path connected.
$U_i$ isn't the basis, it's a member of it. More precisely, $\{U_i:i\in\mathbb{Z}^+\}$ is a countable basis.
I think it would be better to write "let $X=\{a\times 0:a\in S_\Omega\}$ which is uncountable since $S_\Omega$ is". Again, $x_\alpha$ isn't this set; you're thinking of it as an element. As a minor note, $X$ isn't really indexed by anything at this point: you just have an association between $\alpha\in S_\Omega$ and $\alpha\times 0$ that you're indicating with $x_\alpha$.
Again, you're confusing the set with the elements of the set. You get for each $x_\alpha\in X$ a neighborhood $V_\alpha$ of $x_\alpha$ ($V_\alpha$ itself is not an uncountable collection of neighborhoods around $x_\alpha$). Similarly, you don't want to say that $V_\alpha$ is pairwise-disjoint, but that the collection of $V_\alpha$s is, or (more succinctly) that "they are pairwise-disjoint".
I think you should probably say something about why you can assume these $V_\alpha$ are pairwise-disjoint. It's not really difficult, but it's not super immediate just from what has been shown so far, and it's really the crux for the final step.
I think you should say why $f$ as defined is injective. It doesn't need to be much, and otherwise I think it's fine.