Re: Cantor Confusion

Andy Smith wrote:
David Marcus writes


OK, here is a suggested alternative argument to show that
lim_{n->oo} |x|^n != 0 for all |x| <0

We can consider the mapping from x to y_n given by y_n = |x|^n. The
effect of the exponentiation is to 'stretch' [0,1] non linearly, so that
, depending on the size of n, progressively more points uniformly spaced
originally in [0,1] are concentrated in the bottom of [0,1] in y_n. But,
that doesn't matter - we have plenty of reals. There are as many reals
in any interval of y_n in [0,1] as that of x.

So it doesn't matter how large n becomes - y_n has the same "density" of
reals as x. And that must also apply at the limit n->oo - if it were
true that you could 'stretch' the reals by such a power law, so as to
map all the reals in 0,1 to either 0 or 1, this would be tantamount to
saying that the reals are not continuous on the line - the reals are
infinitely expandable?

So lim_{n->oo} |x|^n cannot be 0 for all |x|<1 ?

Did you read what I wrote? Doesn't seem like it. I'll try again. The
sentence

lim_{n->oo} |x|^n = 0, for |x| < 1

means

lim_{n->oo} |0|^n = 0,
lim_{n->oo} |0.1|^n = 0,
lim_{n->oo} |0.2|^n = 0,
lim_{n->oo} |-0.278|^n = 0,
lim_{n->oo} |0.78|^n = 0,
lim_{n->oo} |sqrt(2)/2|^n = 0,
etc.

That's all it means. It does not mean

lim_{n->oo} sup_{|x|<1} |x|^n.

When we want to say, "lim_{n->oo} sup_{|x|<1} |x|^n", we say, "lim_{n->
oo} sup_{|x|<1} |x|^n". We don't say, "lim_{n->oo} |x|^n = 0, for |x| <
1", because the latter means that lim_{n->oo} |x|^n = 0 for whatever
value of x we happen to try as long as |x| < 1. The only way that

lim_{n->oo} |x|^n = 0, for |x| < 1

can be false, is if there is a real number y such that |y| < 1 and

lim_{n->oo} |y|^n != 0.

Is there such a y?

I can see that I am irritating you.

No, you aren't.

It is not deliberate. It is a
consequence of a misunderstanding on my part of something fundamental.

I'm sure!

You say, is there such a y? Well, if we consider the mapping
w = Lim n->oo x^n, 0<=x<=1 , we can consider that as the limit of a
succession of mappings.

This is simply a language problem. The meaning of what is written is not
what you think. It is possible to write what you are thinking: see
below.

An analogy would be if I tell you what the word "cat" means and you ask
why can't a cat be the animal that barks. It could be, but it isn't. We
have another name for that animal.

So, for example we can think of some
representative points located in x

You can't say "in x". x is a number. We don't know exactly which, but it
is still a number. So, just as you can't say "in 7", you can't say "in
x". "x" is a name for a number, the number x. Similarly, "7" is a name
for the number 7.

e.g x - 0,1/4,1/2,1 on [0,1]. After
x^2 we get
0,1/16,1/4,1 and after x^n these points map to 0,1/(4^n),(1/2^n), 1.
But we still have a continuum of points in [0,1] after n iterations, and
each of these map back to an original location in x. And that has to be
true after n->oo

There is no "after n -> oo".

- the line is infinitely elastic. Otherwise you could
define a space between two reals in x?

There is then an inverse mapping from the points in w in [0,1] back to
their original locations in x? And these would all describe the y that
you request?

Name a number y with |y| < 1 that will work. You can't.

The following is from Chapter 24 of the book "Calculus" by Spivak.

"Let {f_n} be a sequence of functions defined on the set A. Let f be a
function which is also defined on A. Then f is called the uniform limit
of {f_n} on A if for every e > 0 there is some [natural number] N such
that for all x in A,

if n > N, then |f(x) - f_n(x)| < e.

"We also say that {f_n} converges uniformly to f on A, or that f_n
approaches f uniformly on A.

"As a contrast to this definition, if we only know that

f(x) = lim_{n->oo} f_n(x) for each x in A,

then we say that {f_n} converges pointwise to f on A. Clearly, uniform
convergence implies pointwise convergence (but not conversely!)."

Another way to write that {f_n} converges uniformly to f on A is

lim_{n->oo} sup_{x in A} |f_n(x) - f(x)| = 0.

Let f_n(x) = x^n. Let f(x) = 0. Then {f_n} converges pointwise to f on
[0,1), but not uniformly.

--
David Marcus
.

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