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What is a basis for the order topology on $[0,\Omega)$ ?

$\Omega$ is the initial ordinal of cardinality $\aleph_1$

My knowledge of ordinals is very limited

I know the union of countable sets is countable

Since $\Omega$ is uncountable ,I don’t think a union can represent [$0$,$\Omega$) and if that is the case no basis can be found,but I am sure I am wrong.

I get this idea from https://math.stackexchange.com/a/340091/919044

Source:A first course in Topology:Conover

bof
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Plotinus
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    What sort of basis do you want? The order topology by definition is given by a certain basis... – Eric Wofsey Jun 30 '21 at 04:14
  • Any will do. The question is not specific and does not ask you to prove it. – Plotinus Jun 30 '21 at 04:17
  • Well do you know the definition of the order topology? It by definition provides you with a basis. – Eric Wofsey Jun 30 '21 at 04:18
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    The collection of all nonempty open sets is a base. – bof Jun 30 '21 at 04:56
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    Union of a countable family of countable sets is countable. Obviously, for example, any set is union of the family of the singletons of its elements. –  Jun 30 '21 at 05:25
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    It's not a countable union of base elements, but $[0,\Omega)$ is a union of base elements (pick one for each element), just an uncountable one. Topologies are closed under all unions... – Henno Brandsma Jun 30 '21 at 07:40

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First you need to realise how ordinals are built up from smaller ordinals, maybe read more on that on their Wikipedia page; I'm not sure how well your text by Conover covers this (or does he just assume this as required prerequisite knowledge?).

You basically start with the smallest ordinal $0$ (as a set, $\emptyset$) and for each ordinal $\alpha$ we already have we can form its successor, denoted by $\alpha+1$, which as a set is just $\alpha \cup \{\alpha\}$. As expected $1$ is defined as the successor of $0$, $2$ as that of $1$ etc, and so on for all natural numbers that are familiar. The first "new" ordinal number comes at the so-called limit stage: we have the ordinals $0$, $1$, $2$, $3$, $4$, etc. and we collect them (or union them, as sets, which is the same here, this justifies its existence from the union axiom) into $\{0,1,2,3,4,5,\ldots\}$ (the minimal infinite ordinal) and call it $\omega$. The new ordinal $\omega$ has a successsor $\omega+1$ which is just the set $\omega$ together with a new maximal element $\omega$, and this has a successor we denote by $\omega+2 = (\omega+1)+1$ (ordinal addition is defined to be consistent with this so that indeed $\omega$ plus $2$ becomes $\omega+2$ in this way), this has two maximal elements above the natural number ordinals in $\omega$; this goes on till we have a new limit stage when we have all ordinals $0,1, \ldots, \omega, \omega+1, \omega+2, \ldots, \omega+10000, \ldots $ and we form a union again to make an ordinal $\omega+\omega$ which has all natural numbers $n$ and all ordinals $\omega+n$ as elements; it's essentially two copies of $\omega$ one places after the other in order. But it's still a countable ordinal of course. $\omega+\omega$ is also denoted $\omega \cdot 2$ (which in ordinal multiplication is not the same as $2 \cdot \omega$!) which is a handy shortcut to denote the next countable sequence of ordinal successors $\omega \cdot 2 +1, \omega \cdot 2+2, \ldots, \omega \cdot 2+n, \ldots$ for all $n$ a natural number again. A new limit stage then gives us $\omega \cdot 3$, essentially $3$ copies of $\omega$ placed in consecutive order.

Of course this goes on and on and we get $\omega \cdot n$ for all $n =1,2,3,4,$ etc. (all made by limit steps; these ordinals do not have a predecessor $\alpha $ so that e.g. $\alpha+1 = \omega \cdot 2$, but are made from taking the union of all previous ordinals when we have formed all successors we can. This is a basic distinction that will come back in the order topology.

The limit step after all $\omega\cdot n$ is denoted $\omega \cdot \omega$ and this starts a whole new sequence again. It gets dizzying to visualise them after a while (as orders) but we still get countable well-ordered sets using these steps. It is part of the standard theory in sets and orders that every well-order has a unique "order type" and that the ordinals are sort of canonical representatives of them: every ordinal is well-ordered and every well-order is order isomorphic to a unique ordinal number (which we can define quite generically in ZFC as a transitive set, linearly ordered by $\in$ (or $\subseteq$).

There are also uncountable sets that are all-ordered so the theory tells us that there are not only countable ordinals (at all countable stages of what we did before, even if we get to $\omega^n$ and $\omega^\omega$ and beyond, we only still have countable ordinals and these get really big to visualise, as your linked post aptly tried to describe). But at some stage (by minimality) we have the smallest uncountable ordinal, usuallty denoted $\omega_1$ nowadays (older texts follow old Cantor in naming it $\Omega$, but it's only the first uncountable ordinal, hence te subscript $1$).

But now to the order topology: we have an uncountable set $X=[0,\omega_1)$, with a minimal element $0$ and no maximum ($\omega_1$ a limit ordinal). If $\alpha$ is in the set, so is the next larger $\alpha+1$ (just $\alpha$ with a maximum added), and if $\alpha_n, n=1,2,3,$ is a sequence in $X$, their union is also an ordinal and larger than all of them, denoted by$\sup_n \alpha_n$ and also a member of $X$ as it still is a countable ordinal (a countable union of ocountable sets is countable).

All elemts are either $0$, a successor of the form $\alpha+1$ or a limit (having no predecessor). Remember how the base for the order topology works: a minimal element (here $0$) has a local base of the form $[0,x)$ where $x \in X$. In particular, $[0,1) = \{0\}$ is an open set and $0$ is an isolated point of $X$. If $x=\alpha+1$ is a successor ordinal, then $\{x\} = (\alpha, x+1)$ is also open. (open intervals are a local base, but this is clearly the smallest open set around $x$ that is possible; it's an open interval hence open.) So $0$ and all successors are open singletons and isolated. Limit ordinals are more interesting and indeed are limti poits of their predecessors: e.g look at $\omega \in X$. A basic neighbourhood is an open interval, so pick $n < \omega$ and $y=\omega+1 > \omega$, then $(n, \omega+1)=\{n+1, n+2, \ldots, \omega\}$, so a "tail" of the natural numbers plus $\omega$ itself. So basic neighbourhoods for limits "look the left" : we pick some tail from the predecessors (by picking a left endpoint for the interval) and use the successor of the limit as the right hand end point of the open interval and the result is a "tail plus limit ordinal" set. This is basically (pun intended) what open neighbourhoods of limit ordinals look like. So it's un uncountable set, with loads of isolated points (as may as $|X|$) and scattered throughout them limits that have infinite left looking neighbourhoods and are not isolated, but sort of glue them together (also $|X|=\aleph_1$ many). If we take any incresing sequence in it, its limit is also in $X$; we cannot reach the "end" of the set merely by countable sequences; it's one of the main reasons (besides beauty, of course) to bring it into topology courses, to show that sequences are not always quite enough to do all things in topology.

Read up on ordinals and to help visualising I found the book "Infinity" (and the mind?, not quite sure of the full title) by Rudy Rucker very interesting as a high school kid and it tickled my imagination in these respects. It helped motivate me to study pure maths later.

Henno Brandsma
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