Re: Real Number Paradox?
- From: "Proginoskes" <proginoskes@xxxxxxxxxxxxx>
- Date: 21 Jun 2005 20:28:57 -0700
stush@xxxxxxxxxxxxxx wrote:
> Consider the set of real numbers A = { 0.3, 0.33, 0.333, ... }. Note
> that the elements of A all have a finite decimal expansion.
>
> Now consider the set X = the set of all real numbers less than 1/3 but
> greater than all a in A.
>
> Now traditionally I think you would prove that X is empty using, in
> some way, decimal expansions to show that such x in X can not be
> constructed.
>
> But then I consider:
>
> Let a_i mean the element of a with i three digits in its decimal
> expansion, i.e. A = { a_1, a_2, a_3, ...}.
>
> For any a_i in A, there are uncountably many real numbers between a_i
> and 1/3. Call this set X_i. Obviously all X_i are uncountable.
>
> So you end up with (countably) infinite many sets X_1, X_2, X_3, ...
> all of which are uncountable.
>
> Then is it not true lim i->inf |X_i| = 0? Then I do not see how an
> infinite sequence of uncountable sets can diverge to 0. [...]
>
> Please enlighten.
(1) Consider this: Let Y_n be [n,infinity); i.e., all real numbers >=
n.
Every Y_n is uncountable, and every finite intersection of Y_n's is
uncountable, but there's nothing in the intersection of ALL the Y_i's.
(No real number is greater than ALL integers.) Do you have a problem
with this?
(2) Consider this: lim (x-> a) of f(x) need not be the same as
f(lim (x->a) x).*
For instance, let f(x) = 1 if x =/= 0 and 0 otherwise.
Then lim (x -> 0) f(x) = lim (x -> 0) 0 = 0.
But f(lim (x -> 0) x) = f (0) = 1.
*It _is_, if f is continuous at x = a. But not all functions are
continuous.
--- Christopher Heckman
.
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- Real Number Paradox?
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