Re: Cantor Confusion

David Marcus writes

The formula that has to be true for all |x| < 1 is

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

In other words, for all x

if |x| < 1 then lim_{n->oo} |x|^n = 0.

So, the formula only has to be true for each x individually. Not, for
all x uniformly (to use the technical jargon).

How would you phrase it if you meant each x in the interval
individually?

You are quite correct that

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

is not zero. In fact, for any n > 0, sup_{|x|<1}|x|^n equals 1. So,

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


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 ?


--
Andy Smith

.

Quantcast