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Re: which linearly ordered sets are sets of reals?
Posted:
Aug 2, 2003 3:34 PM
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> > A theorem of Cantor says any two densely ordered sets
> > without endpoints that are countable are order-isomorphic.
> > "Densely ordered" means ordered in such a way that between
> > any two elements there is another.
> > A corollary is that having a countable dense subset
> > suffices for your purpose. -- Mike Hardy
William Elliot (mars@agora.rdrop.com) answered:
> Then I've counterexample.
>
> Let S = Rx{0,1} ordered lexicographically
> (r,a) <= (s,b) when r < s or r = s, a <= b
> S is the double pointed line. Write r_a for (r,a).
>
> A countable dense subset, that's also densely ordered, is Qx{0}.
I answered:
> Is that dense? Let's see ... between (0, 0.1) and (0, 0.2)
> there is no member of Qx{0}, so Qx{0} is not dense according to
> the definition I stated above. It's also not "topologically" dense,
> since the interval from (0, 0.1) to (0, 0.2) is an open set that
> does not intersect Qx{0}. So this is not actually a counterexample.
Oh .... I see that you had curly braces: "{0,1}", denoting
a set with two members. I had read this as "[0,1]", with square
brackets, denoting the closed unit interval. It is not the case
that strictly between any two members of your linarly ordered set
Qx{0,1} (with curly braces) there is a member of the subset Qx{0},
so in that sense the subset is not dense in the larger set.
I didn't have in mind topological denseness necessarily.
Mike Hardy
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