Re: Question: What conditions are sufficient to prove a subset does not exist?

On Jul 30, 11:52 am, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
I would like to explore this a bit more to make sure I understand. If
the existence of A is my sole assumption and the existence of f is
such that it exists iff A exists, and the existence of A generates a
contradiction, then why must I assume f does not exist as opposed to
A; eg, exist(A) -> f and f -> ~exist(A)? Or in otherwords, exist(A) ->
( f -> ~exists(A) ).

You keep adding more conditions, which naturally change the questions
you are asking. By changing the question, you are sometimes taking
a valid answer to question A, and wondering why it is that this answer
makes so little sense with respect to question A'. Well, because they
are DIFFERENT questions.

Sorry - I have a tendency to want to explore new subjects/conditions
when people mention them.

From 5 and not(exists(f)), you can CONCLUDE (by contrapositive)that
not(exists(A)) also holds. So from 1 and 2 (as opposed to 1' and 2),
you can CONCLUDE (not "assume") that BOTH not(exists(f)) and
not(exists(A)) would hold.

Ah, thank you. Very succinct.


As a follow-up question, if I reject the existence of f am I not also
rejecting the existence of A; eg, ~f -> ~exist(A)?

Only if you are assuming 1, as opposed to 1'. In the proof of Cantor's
Theorem, you do NOT have 1, you have 1' which is weaker.

Agreed.

.

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