Re: Definition of functions.

On Mar 6, 9:31 pm, "MoeBlee" <jazzm...@xxxxxxxxxxx> wrote:
On Mar 6, 5:32 pm, "zuhair" <zaljo...@xxxxxxxxx> wrote:

On Mar 6, 3:04 pm, "MoeBlee" <jazzm...@xxxxxxxxxxx> wrote:
take any non empty set x.

Now f:x->0 ,

Nope. It is a theorem:

~Efx(~x=0 & f:x->0).

I.e., there is no f that maps a nonempty x into 0.

is as you defined is a relation from x to 0 ( i.e a
subset of xx0 ( the cartesian product of x and 0)

Let's use 'y' instead of 'x' since we need 'x' here for 'cross'.

So, for any y, we have yx0 = 0.

Now in general for any A and B, AxB is defined as

AxB={<x y>| xeA & yeB}.

Now if A=x and B=0 were x is a non empty set we have
AxB=0 , i.e we have xx0= 0

Right.

Now Cantor defines equality of cardinality of x and 0 in the following
manner.

Do we need to mention Cantor? Why not just mention the usual results
of ZFC?

|x|=|0|<-> (Ef(f:x->0 & f is injective) & Eg(g:0->x & g is
injective)).

In ZFC that's not usually the definition, but it is a theorem, yes.

The definition of f:x->0 & f is injective is

f:x->0 & f is injective <-> (f is a subset of xx0 &
Axyz(<x y>ef & <x z>ef)-> z=y) & dom(f)=x & range(f) subset_of 0 &
Axyz((<x y>ef & <z y>ef)-> z=x)).

We don't need all that for a DEFINITION. Please just go back the
definitions I gave. However, we do get all that as a THEOREM.

since f is a subset of xx0, then f=0.

In this way there exist f that is injective from x to 0.
since Axyz((<x y>e0 & <x z>e0)-> z=y) is trivially true
and since Axyz((<x y>e0 & <z y>e0)-> z=x) is trivially true
and dom(f)=x and range(f) subset_of 0 , which is 0.

No, your fallacy is in taking f to be 0 and also claiming the dom(f) =
x. That holds only if x = 0.

ah,i.c. here is the thing I have been missing.
but just in order to get a clearer grasp of it
how do you define dom(f),in the standard way.

I just thought that dom(f) would be something like

x=dom(f)<->Azy(<z y>ef->zex), which is incorrect

since this definition defines a subset of dom(f) , which

is not necessarily dom(f).

Perhaps it should be

x=dom(f)<->Azy(<z y>ef<->zex).

If this is correct then you are right since if f=0

Then dom(0) should be 0.

So what we are getting here is that for every non empty set x
There cannot exist a function f such that f:x->0.

as you put it: ~Efx(~x=0 & f:x->0) is a theorum.

ah i c.

That's OK.

while the codom(f) is defined as:

y=codom(f)<->Axz(<x z>ef->zey).

range(f) is defined as

y=range(f)<->Axz(<x z>ef<->zey).

so range(f) subset_of codom(f).

so we can have functions from 0 to x even if x is non emtpy
because range(f) is a subset of codom(f), and if f=0
then rang(f)=0 , and 0 is a subset of any set.
So actually we always have a function from 0 to x. i.e

AxEf(f:0->x) is a theorum.

and of course this f is injective, surjective, bijective.

Actually I knew the theorum you've mentioned.
But I didn't know how to prove them. Now I saw
the standard definitions , I think I got a more
grasp on them

Thank you.

Zuhair




LOGIC, zuhair, LOGIC. Please, please, please find a way to get that
textbook we've talked about.

Yea, actually the new news is that I found that way!
But it would be a matter of weeks to months till they ship it.

and of course for the same reason we also have Eg(g:0->x & g is
injective).

According to the definition you gave, then we have
Ax(Ez(zex)-> |x|=|0|) which is incorrect.

Yes, that formula contradicts set theory. You got to that formula by
way of the fallacy I mentioned.

Perhaps I am missing something. NO doubt I am, but I want to know what
it is.

What it is is: 'Logic: Techniques of Formal Reasoning' by Kalish,
Montague, and Mar.

what I want to say is that this definition you've mentioned in which
it is not required that range(f) should be not empty
,this definition will lead to any non empty set having injection from
it to empty , which shouldn't be the case.

And it is not the case. The definition is correct. Furthermore, any
CORRECTLY FORMED defintion will not lead to contradictions in a
theory. The fallacy is what I mentioned: You argued about 0 as if it
were the function while not recognizing that 0 can be the function
only if x itself is also 0.

While the version of definition of f that requires range(f) to be non
empty avoids this, since there would be no function
from non empty x to 0, but on the other hand there is always
exist an injection from 0 to x, justifying Cantors defintion that |x|>|
0|.

Forget that. And learn that definitions that obey certain forms do not
ever permit contradictions in a theory. My definition obeys one of the
forms and your argument toward a contradiction was incorrect in the
step I mentioned.

Now the problem with the definition that requires range(f) to be non
empty is that it cannot define injections from 0 to 0.,because the
range of f:0->0 is always a subset of 0
and accordingly violating this definition, i.e there cannot exist a
function from 0 to 0 and accordingly no injections can exist between
them , nor bijections.

It's nugatory. The standard defintion is just fine as it is.

I don't know perhaps I am missing something.But just in case I am
right, I have question why not the following definition of f:x->y.

f:x->y <-> (f is a function & dom(f)=x & range(f) subset_of y &
(~dom(f)=0 -> ~range(f)=0) ).

This would solve the confusion.

Just drop this because adding (~dom(f)=0 -> ~range(f)=0)
is redundant.

You've got to find a way to get hold of that book to start, then move
on to a good set theory textbook. I really don't think your confusions
will cease until that happens.

Yes, you are right. I need to study this in a systematic way, it is
more delicate than amateur handeling of it.


By the way, just an idea: When you need to use 'x' or 'X' for the
Cartesian cross, choose another letter other than 'x' for a variable.

OK.

MoeBlee


.

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