Re: Cantor Confusion

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

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

regards
--
Andy Smith
Phoenix Systems
Mobile: +44 780 33 97 216
Tel: +44 208 549 8878
Fax: +44 208 287 9968
60 St Albans Road
Kingston-upon-Thames
Surrey
KT2 5HH
United Kingdom
.

Relevant Pages

  • Re: Zenkins paper on Cantor
    ... > as least as many reals not in that list as are in the list to start with. ... A bijection between the naturals and reals is not accepted by ... My perception _was_ that you could toss out antidiagonals all day and after ... the mapping must be that way. ...
    (comp.theory)
  • Re: Zenkins paper on Cantor
    ... > as least as many reals not in that list as are in the list to start with. ... A bijection between the naturals and reals is not accepted by ... My perception _was_ that you could toss out antidiagonals all day and after ... the mapping must be that way. ...
    (sci.math)
  • Re: Complex Analysis - polar form notation!?!?!?!
    ... reals" idea. ... If the polar coordinates r and @ are used ... there need not be such an intermediate mapping. ... ordered pair to the element z in the complex plane. ...
    (sci.math)
  • Re: Calculus XOR Probability
    ... But this mapping denumerates only some of the reals (0,1] and omits ... What set of naturals is TO using as his domain for this function. ... and the number of unit intervals on the real line is Big'un, ...
    (sci.math)
  • Re: Cantor Confusion
    ... It is a mapping of one single edge on the path ... > representing 1/3, the next edge is mapped on the path representing pi, ... no one claims any well ordering of the reals. ... But the tree does. ...
    (sci.math)
Quantcast