[FOM] 696: Refuting the Continuum Hypothesis?/7
Mario Carneiro
di.gama at gmail.com
Tue Jul 5 22:37:35 EDT 2016
On Tue, Jul 5, 2016 at 7:41 PM, Harvey Friedman <hmflogic at gmail.com> wrote:
> I now want to look at
>
> $[k]. For all f:R into R, there exist k (distinct) reals, none being f
> at an integral shift of the sum of the others.
>
> Thus $ = $[2]. $[1] makes sense, and is equivalent to: for all f:R
> into R, there exist a real which is not 0. Hence $[1] is provable.
>
>From my reading, it seems that this should be $[1] = for all f: R into R,
there exist a real which is not f at an integral shift of 0, or in other
words, there is a real not in f(Z). Since f(Z) is countable, this is still
provable, though.
Mario
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