Re: Cantor Confusion

On May 10, 10:15 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

Be sure, I studied every page very carefully, so carefully that I
typed every letter (some time ago).

And typing letters does not an understanding make!


2.   RELATIONS

Mathematicians often study relations between mathematical objects.
Relations between objects of two sorts occur most frequently; we call
them binary relations.
...
A binary relation is, therefore, determined by giving all ordered
pairs of objects in that relation; it dos not matter by what property
the set of these ordered pairs is described.

!!!!!!!!!!!! But it obviously does matter *that* it is
described. !!!!!!!!!!!!!!!

Yes, to DEFINE a PARTICULAR relation, we have to have a formula to
define that PARTICULAR relation. But the relation ITSELF is NOT the
formula that defines it and the formula is not even a component of the
relation. Morevover, there do exist relations that are not DEFINED.
For example, say it is a theorem:

Er(r is a relation & P)

(where the variable 'r'' occurs free in P).

So the theorem says that there exists at least one relation that has
the property that is specified by P. But it might be the case that
there is not a defining formula to DEFINE a PARTICULAR relation that
has the property P.

Now, if you don't like that, then that is understandable, and you may
reject such a non-constructive theory. However, you can't rationally
dispute the plain fact of what do and do not happen to be such
theorems of non-constructive theories such as Z set theory, since it
is objectively verifiable that there are proofs of such theorems,
notwithstanding anyone's acceptance or not of the truth of the axioms
or of a non-constructive theory. Moreover, it is not a contradiction
to have non-constructive theorems. It is quite understandable that a
person might prefer only constructive theories and reject non-
constructive ones, but that is different from the issue of
consistency, and a theory is not inconsistent merely for being non-
constructive.

2.3   Definition   Let R be a binary relation.
(a) The set of all x which are in relation R with some y is called the
domain of R and denoted by dom R = {x | there exists y such that xRy}.
dom R is the set of all first coordinates of ordered pairs in R.
(b) The set of all y such that, for some x, x is in relation R with y
is called the range of R, denoted by ran R. So ran R = {y | there
exists x such that xRy}

2.9   Lemma   The inverse image of B under R is equal to the image of
B under R-1.
2.11   Definition   The membership relation on A is defined by

                                A = {(a, b) | a  A, b  A, and a  b}

The identity relation on A is defined by

                                IdA = {(a, b) | a  A, b  A, and a = b}

2.12   Definition   The set of all ordered pairs whose first
coordinate is from A and whose second coordinate is from B is called
the cartesian product of A and B and denoted A  B. In other words,

                                A  B = {(a, b) | a  A, b  B}.

???You see a A and B here????

Yes, to define 'AXB' you have to mention A and B. So what? The
definition of the Cartesian product of A and B (AXB) is not the
defintion of the predicate 'is a relation'.

 >                                                   It is ridiculous to
 > follow this discussion. I can assure you, if one of my students would
 > not know that a function is a formula (or rule or whatever) together
 > with a domain where it is defined and a range, then he or she would
 > not pass the exame. And this is the same in the better math courses in
 > Germany.

If that is true, it tells us a lot about mathematics education in Germany.
Functions in set theory are *not* the same as functions in analysis.

I talked about mathematics, not about set theory. Nevertheless
functions *in set theory* consist of a prescription (formula, rule,
whatever), domain, and range.

No, that is not the DEFINITION of 'function'; and there exist
functions that have no defining formula; and every function has its
domain and its range, but those are not required to mention to DEFINE
'function' and a function is NOT and does not "consist of" a formula,
domain, and range, as the function ITSELF is:

A set of ordered pairs that is many-one.

THAT is what the function IS. Nothing more and nothing less. It is a
certain kind of set of ordered pairs and it is not a formula with a
domain and range.

Exercises

2.1 Let R be a binary relation; show that dom R c U(UR), ran R c
U(UR). Conclude from this that dom R and ran R exist.

The last exercise is a good homework for such "mathematicians" who
studied in Oxford or Harvard or at home and, therefore, do not know
that dom R and ran R do exist for any binary relation, or in other
words: without domain and range R would not be a binary relation.

We've done that exercise and many similar exercises. And we've done
the "double drill down" into the union of the union of a relation MANY
times, especially to talk about elements of the domain, range, or
field of a relation.

And for every new operation symbol (as 'domain of' and 'range of' are
locutions representing operation symbols) we prove the existence and
uniqueness clauses that support the definition. Doing the above
exercise is just one (two, actually) of the many proofs we perform to
prove the existence of sets having certain properties.

MoeBlee


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