Re: Mathematicians are in deep *** for 2 reasons

moeblee says

There is no contradiction in set theory here, as you will see:

(1) A countable set may have as a member a set that is uncountable.

(2) The universe of the model is countable, and it has as member
f('R'), where 'R' is defined in set theory as the set of real numbers
and f is the interpretation function for the model of the extended
language of set theory. R is uncountable, but it is not necessarily
the case that f('R') is uncountable.

(3) The universe of the model is countable, and it has as a non-empty
subset {x | f('U')x} where f('U') is the subset of the universe that
is the value of the interpretation function f for the argument 'U',
where 'U' is the symbol (of the extended language) that is defined in
Z set theory (extended by definitions) by 'Ux <-> ~x is countable'.
But it is not required that f('U') is the set of uncountable sets in
the universe of the model.




all moeblee has given us is
there are models where there is no contradiction
but there is A model where there is a contradiction


wiki

Using the Löwenheim-Skolem Theorem, WE CAN GET A MODEL of set theory
which only contains a countable number of objects. However, it must
contain the aforementioned uncountable sets, which appears to be a
contradiction


moeblee says

There is no contradiction in set theory here, as you will see:

(1) A countable set MAY have as a member a set that is uncountable.

note the word MAY

so if it can have countable and uncountable it is in contradiction
i say
thus p and ~p

all moeblee has given us is
there are models where there is no contradiction
but there is A model where there is a contradiction

(2) The universe of the model is countable, and it has as member
f('R'), where 'R' is defined in set theory as the set of real numbers
and f is the interpretation function for the model of the extended
language of set theory. R is uncountable, but it is not necessarily
the case that f('R') is uncountable.

i says
note
but it is not necessarily the case that f('R') is uncountable.

not necessarily these imply that there is a case where f('R')is
uncountable
all moeblee has given us is
there are models where there is no contradiction
but there is A model where there is a contradiction



moeblee says

(3) The universe of the model is countable, and it has as a non-empty
subset {x | f('U')x} where f('U') is the subset of the universe that
is the value of the interpretation function f for the argument 'U',
where 'U' is the symbol (of the extended language) that is defined in
Z set theory (extended by definitions) by 'Ux <-> ~x is countable'.



note moeblee says

But it is not required that f('U') is the set of uncountable sets in the
universe of the model


not required does not mean cant or IS NOT

it may be not required imply there is a case where f(U) is uncountable



all moeblee has given us is
there are models where there is no contradiction
but there is A model where there is a contradiction
so what have we
(1) A countable set may have as a member a set that is uncountable.


i say
thus p and ~p


2)imply that there is a case where f('R')is uncountable which agrees with
wiki which says there is A model where there is a contradiction


3)it may be not required imply there is a case where f(U) is uncountable


which agrees with wiki which says there is A model where there is a
contradiction

thus all moeblee has given us is
there are models where there is no contradiction
but there is A model where there is a contradiction

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