Re: Is one-to-one mapping valid for comparing infinite-sized sets?

In article
<7232ab09-275e-43b7-9f36-3f0646430bcd@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
venkat.6123@xxxxxxxxx wrote:

On Sep 17, 12:32 pm, Virgil <Vir...@xxxxxxxxx> wrote:
In article
<e8a07277-af3d-4eb6-9748-08259d9aa...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

 venkat.6...@xxxxxxxxx wrote:
Consider the mapping f:N -> N: x -> (x+1). It is an injection, so that  
by your arguments N cannot be of the same size as N

Yes, for me, this mapping is as much correct as f:N -> N: x -> (x) if
we are using these mappings to determine the size of this set. The
weird result of N not being equal to N is the result of applying
mapping for infinite sets. However, it is as much correct as saying
size of N is equal to its own size. Can we say an infinite is equal to
itself in size? Then we need to define "equal" and arithmetic that
goes with it as applicable to infinite (if x = x, then x-x = 0).

But for infinite sets, as for finite sets, we already have an adequate
definition of "equal in size" (i.e., equal in cardinality), namely the
existence of a bijection between the sets.

It is a perfectly consistent definition of size which works equally well
for both finite and infinite sets.

At least as long as one does not assume that the existence of a
non-bijection between two sets prohibits the existence of a bijection.

It doesn't prohibit, but indicates that bijection is not the *only*
truth in this case.



It is not the only mapping, but one rarely has only one possible
mapping between two sets. The issue is whether among the set of all
mappings from A to B there exists one which is an
injection/surjection/bijection

You got bijection because you looked for it.

One is supposed to look for it. The issue is whether it is there to be
found. For instance, the total absence of any bijections between two
sets shows them to be of different cardinalities, but the existence of
even one, however difficult it may be to find, shows the sets to have
the same cardinality. That is true by definition of cardinality.

doesn't mean anything special enough to celebrate.

It means the sets are of the same cardinality, whereas lack of any
bijection means they are of different cardinalities.

This is not the
case with finite sets.

Yes it is! Even for finite sets, finding any bijection means the sets
are of the same cardinality, whereas lack of any bijection means they
are of different cardinalities.


Even if you look for it, you wont get an
injection while there is some other mapping that is a non-injection.

WRONG! For every set of more than one element, there are constant
functions from it to other sets which prove you wrong.

Example: Let A = {0,1} and B = {3,4} and f:A -> B: x|-> 3, be the
constant function whose value is 3 for each argument, and is a
non-injection.

According to your statement, the existence of the above function
prohibits existence of g:A -> B: x |-> x + 3, which is an injectin and
surjection and bijection from {0,1} to {3,4}






Our little logical reasoning using 1-1 mapping makes perfect sense
here.

Mine logical reasoning does! Yours doesn't!
.

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