[FOM] Re: Friedman's characterization of R
Ali Enayat
enayat at american.edu
Sat Feb 12 17:44:54 EST 2005
This is a reply to Jesse Alama's message (Fri Feb 11 23:51:37 EST 2005 )
regarding Salehi's proposed counterexample to Friedman's characterization
of the real line.
Friedman (Sat Feb 12 03:00:35 EST 2005) has already pointed out that the
function F in Salehi's example is not continuous, by the way. It is curious
to note, however, that Salehi's function F becomes continuous if its domain
is changed to R x R [but then the glb property is lost, e.g., the "y-axis"
is bounded but has no glb or lub].
In his mesage, Alma wrote:
>Your ordered set X has the least upper bound property (this is
>inherited from the least upper bound property of R) and a countable
>dense subset (namely, Q x ([0,1] & Q)). If the standard
>characterization of R (an ordered set that has the least upper bound
>property and a countable dense subset) is accurate , then it would seem
>that your X is order-isomorphic to the real line. Perhaps I'm missing
>something; I would be grateful if you would discuss your counterexample
>in more detail.
Contrary to the above assertion, the set Q x (Q intersection [0,1]) is not
dense in R x [0,1] (with the topology given by the lexicographic order, as
opposed to the familiar Euclidean topology), e.g.,look at the open set
whose end points are (alpha,0) and (alpha, 1), where alpha is irrational.
Moreover, R x [0,1] has no countable dense set since the open sets of the
above form are uncountable and pairwise disjoint [in the language of
topology, the space is not c.c.c.]. This implies that all dense sets of
Salehi's space are uncountable (and even better: of power continuum).
Regards,
Ali Enayat
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