Re: The strange set in R^2
- From: David Libert <libert.david@xxxxxxxxx>
- Date: Sat, 04 Aug 2007 05:55:17 -0000
On Aug 2, 9:18 pm, The World Wide Wade <aderamey.a...@xxxxxxxxxxx>
wrote:
In article <f8th9b$m3...@xxxxxxxxxxxxxxxxxxxx>,
Ben Rudiak-Gould <br276delet...@xxxxxxxxx> wrote:
Virgil wrote:
This set, being countable, projects vertically onto a countable subset
of the x axis, but the whole of the x axis is not countable, A so does
not contain a point on every vertical line.
Which is fine, since the original problem didn't say that it should. But I
would be interested to know if you can solve the "exactly one" variant
without AC.
-- Ben
Sure. Let D be any countable dense subset of the plane that contains
at most one point on each horizontal and each vertical line, e.g. one
of the several posted in this thread. Let D_x, D_y be the projections
of D into the x and y axes resp. D is then the graph of a bijection f
: D_x -> D_y. Because D_x, D_y are countable, R \ D_x, R \ D_y have
the cardinality of R. Let g be any bijection of R \ D_x onto R \ D_y.
If we define h = f on D_x, h = g on R \ D_x, then h is a bijection of
R onto R. The graph of h has the "exactly one" property as well as
being dense in the plane.
So the above settles the "exactly one" variant without AC, as
noted.
As in the comments above, the "exactly one" variant can be seen from
its statement, not thinking directly about actual constructed
solutions but just inferring from the statement of the question, that
any such set as required must be the graph of a bijection of R >->> R.
In
[1] Robert Sheskey Aug 2 '07 "Re: The strange set in R^2"
http://groups.google.com/group/sci.math/msg/bfa4927136081278
there is the elegant solution
For any irrational t, let A = { (p+qt, pt+q) : p,q rational }.
Not having read [1] carefully after it was posted I wrote
[2] David Libert Aug 4 '07 "Re: The strange set in R^2"
http://groups.google.com/group/sci.math/msg/d505e638a313e80e
giving a longer solution.
On the other hand, [2] gets, without AC, all the points in AS to
have both coordinates rational, answered a question of Stephen
Herschkorn.
Also, we can modify [2] in combination with Wade's solution from he
parent article above. Make an omega list of the rationals, and
interleave with the steps in [2] steps to add a pairing, avoiding
previously used values as in [2], which in turn throw the n'th
rational (according tho the omega listing) as first a vertical value
and then separately a horizontal value, if that n'th rational was not
previously so used, else do nothing for that step.
This has the effect of making the contructed A be not merely a
bijection between 2 subsets of Q = the rationals, but a full
bijection of Q to itself, ie all of Q got thrown in.
Next take the identity function on R-Q, and union that with A the
graph of the bijection of Q. This is like Wade's solution, but
knowing Range A = Domain A, we can use the identity function on the
other part.
So we end up in ZF with a graph dense in R^2, identiy function off
Q. So on a comeager set and on a full Lebesgue measure set the
function is just the identity.
We can also replace Q by any other countable dense set, for example
"sparse" subsets of Q.
--
David Libert
.
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- Re: The strange set in R^2
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- Re: The strange set in R^2
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