Re: Cantor's "diagonal argument". My Objection.

george wrote:
On Dec 3, 7:31 am, John Jones <jonescard...@xxxxxxx> wrote:
How about using a circular list of numbers in Cantor's square?

THERE IS NO SUCH THING
as a "circular list"!! A *list* HAS to be ordered linearly like the
natural
numbers. But THAT IS NOT EVEN THE POINT! The POINT is,
YOU INHERENTLY HAVE *TWO DIMENSIONS* In this problem!
You INHERENTLY HAVE a COLLECTION OF SUBSETS OF A GIVEN SET,
and you are INHERENTLY COMPARING them for size! SO SQUARE
IS PREDEFINED into the question! "circle" ISN'T EVEN MEANINGFUL
in this context!

You see,
it's a bit odd that we have to continue the diagonal, zigzag, whatever,
to a place where we lose sight of it.

You're being and idiot: WE NEVER lose sight of it! EVERY position in
it
IS FINITE!

I say let's keep sight of our
operations and do a big circle that's in full view.

You say that YOU ARE AN IDIOT.
You are CONSTRUCTING A SUBSET of 1 set FROM A LIST
of subsets OF THE SAME SET. THE LIST IS INHERENTLY rectangular
if the set has a natural ordering. The vertical dimension of the list
(along
the rows) IS INHERENTLY linearly and connectedly ordered BECAUSE IT'S
A LIST, and that's PART OF THE DEFINITION of list. The horizontal
dimension
does NOT HAVE to put the elements of the underlying set in any
particular
order, but IF YOU ARE DOING THIS WITH NATNUMS, then the order IS THERE
WHETHER YOU LIKE IT OR NOT. More to the point, you ONLY NEED ONE
element-of-disagreement between the constructed subset and each
subset-
on-the-list, but any line you could draw across a circle cuts it in
TWO places
unless it's tangent -- a circle is just OVERKILL. But more to the
point, since
THE LIST IS INFINITE IN BOTH DIMENSIONS, a circle CAN'T COVER it!
A circle COULD NOT REACH some rows/subsets.

In other words,
YOU'RE STUPID.

> But then Cantor's
demonstration wouldn't work.

You're also a damn liar. The fact that you chose to draw a circle
COULD NOT STOP the diagonal subset FROM EXISTING, or from getting
complemented. Certain sets HAVE to exist, if certain others do.

And the point is, no matter how big we make the circle, Cantor's proof
will never work for it.

NO proof will work for it!
No matter how big you make the circle, IT DOESN'T INTERSECT SOME ROWS!
So it OBVIOUSLY IS NOT RELEVANT to the question!

In which case, we must ask ourselves what we are
doing different for a diagonal.

YOU must ask because YOU ARE SO STUPID that you don't already know.
What is different about the diagonal is that it DOES intersect every
row,
in ONE place (and one is all you need). The circle intersects (at
most) 2
rows in 1 place, a finitely few more in 2 places, AND INFINITELY MANY
IN NONE AT ALL. So it is just not relevant to the issue. It doesn't
describe
what happens with the subset regarding elements-corresponding-to-the-
rows
that it (the circle) DOES NOT INTERSECT!


It's shouting george. Hello, shouting george. I'll have a look at your effort and pick out anything that looks promising. OK by you?

Let's see. Well, I can make the circle intersect the row or the column at the tangent. Didn't you think of that?

A circle can be as ordered as a straight line. You must know that, so I am reminding you. All I have to do is bend a line.

YOu argued that a circle gives too much, that it re-crosses rows or columns. But that's no disadvantage. Re-crossing isn't significant. I can make any shape I like in Cantor's square.

I can make a circle big enough to intersect every row and column. It still wouldn't satisfy Cantor.

Goodbye, shouting george.



.

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