[FOM] refuting one particular argument which claims continuum's size cannot be any \aleph_n
Tom Dunion
tom.dunion at gmail.com
Wed Sep 21 15:24:25 EDT 2011
Probabilistic intuitions can be notoriously unreliable. (Recall the
"Monty Hall" game show conundrum, highly publicized in the popular
press some years back, as one example.) Now, in Chris Freiling's
Axiom of Symmetry paper, after his first argument (against the CH) the
author proposes a further "natural" proposition, which is used to
construct a proof, in ZFC, that the size of the continuum must be
larger than any \aleph_n, for finite n. His claim:
(*) Given any function f ({x,y}) which assigns a countable set of reals to
each unordered pair of reals from [0,1], there exist 3 points, x,y,z, such
that z \notin f({x,y}) AND y \notin f({x,z}) AND x \notin f({y,z}).
Is there a function which might serve as a counterexample to that intuition?
Suppose c = \aleph_2, and let's try to find such a special function
refuting (*). Start with a well-ordering of [0,1], of length \omega_2.
(We are working in ZFC, so we can identify cardinals with initial ordinals.)
Now let us suppose there to be a fixed collection of functions, one
for each ordinal \kappa strictly between \omega_1 and \omega_2, and
such that f_\kappa bijects \kappa to \omega_1. How might we define f
to produce a countable set?
First, let \alpha_1 < \alpha_2 be the ordinals corresponding via the well-
ordering to reals we may denote as x and y. We then right-add each of the
alpha's in turn, to the ordinal (\omega_1 + 1), to obtain \beta_1 < \beta_2.
(Note that \omega_1 < \beta_1 < \beta_2 < \omega_2)
Finally, consider f_\beta_2 (\beta_1), which is some countable ordinal,
call it \gamma. The countable set we seek can be taken to be
those reals t in [0,1] with f_\beta_2 (\beta((t)) (defined, and)
giving a countable ordinal less than or equal to \gamma. (We may
"throw in" some countably infinite set such as the rationals, to avoid
all possibility of ending with "merely" a finite set, if so desired.)
Back to (*) -- note that for any unordered triple {x,y,z}, one of them
(say z) must give the largest of the three Beta's, then we see that
either x \in f (y,z) or y \in f (x,z), contradicting (*).
-- TD
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