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.

$h$ is an order-isomorphism, as $p<q$ implies $h(p)<h(q)$.

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.

[Your proof of (b)]

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.

[The successor case of (c)]

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 $[0,1)$.

[The limit case of (c)]

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 $[0,1)$.

[Your proof of (d)]

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.

Assume $L$ had a countable basis, call this basis $U_i$ where $i\in\mathbb{Z}^+$.

$U_i$ isn't the basis, it's a member of it. More precisely, $\{U_i:i\in\mathbb{Z}^+\}$ is a countable basis.

$x_\alpha$ be the collection of all points of the form $y\times 0$ where $y\in S_\Omega$. $x_\alpha$ is uncountable, being indexed by $S_\Omega$.

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 $x_\alpha$.

Then by (e), there is some uncountable collection $V_\alpha$ of neighborhoods around $x_\alpha$. Note that we can pick, more strongly, $V_\alpha$ such that it is pairwise-disjoint.

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.

[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.

## Best Answer

$f(x)$ is defined only when $x$ has a finite number of $1$s. Indeed, the set $$\bigcup_{n=1}^\infty\{0,1\}^n\times \{0\}^\omega=\{x\in \{0,1\}^\omega: x_i=0\text{ for } i\text{ sufficiently large}\}$$ is countable with $f$ as a bijection to $\mathbb{Z}^+$. However, you'll notice that when $x$ has infinitely many $1$s, then $f(x)=\infty \not\in \mathbb{Z}^+$.