Re: notation: sup, inf, limit, sequences

On Mon, 21 Jul 2008 17:40:45 +0200 (CEST), Daniel Chicayban
Bastos <dbast0s@[EMAIL PROTECTED]
> wrote in
<news:slrng89bll.5c2.dbast0s@[EMAIL PROTECTED]
> in
alt.algebra.help:

> In article <150720082137136894%plsperry@[EMAIL PROTECTED]
>,
> Paul Sperry wrote:

>> In article <slrng7jspg.6um.dbast0s@[EMAIL PROTECTED]
>, Daniel Chicayban
>> Bastos <dbast0s@[EMAIL PROTECTED]
> wrote:

>>> In article <150720081458151297%plsperry@[EMAIL PROTECTED]
>,
>>> Paul Sperry wrote:

[...]

>>>> Well, let's take a look at the usual "Calculus"
>>>> definition of continuity, in the extended Reals, of a
>>>> function f at oo. It would say lim(f(x) : x -> oo) = L
>>>> and f(oo) = L. Hmmm, well, OK. What does _that_ say?
>>>> For every e > 0 there is a d > 0 such that if 0 < |x -
>>>> oo ..... Oh, oh. What's x - oo? Turns out that any
>>>> attempt to define arithmetic in the extended Reals
>>>> really messes up the algebra. That's why the extended
>>>> Reals, although much beloved by analysts and
>>>> topologists, is much despised by algebraists.

>>>> (There _is_ a metric for the extended Reals but it is
>>>> _not_ an extension of the usual metric on the Reals
>>>> --- I think.)

>>> I think I see the problem. If the metric on the extended
>>> reals is not an extension of the usual metric, then we
>>> have a distinguished system, yes? I mean, what's valid
>>> in one is not necessarily valid on the other.

>> Correct. A sequence may converge according to one metric
>> but not converge according to the other. See Brian's
>> most recent response to me in this thread.

> Yes, I begin to notice that it all depends on the
> topology. I remember seeing a non-Hausdorff space in
> which a sequence would converge to any point in its
> neighborhood of convergence. It was trivial example. If I
> recall correctly about this, then the space X would have
> a topology ({}, X). So if s(n) converges in X, then it
> must converge to any point in X, after all there aren't
> any other choices. True?

Yes: in that space every sequence converges to every point.
(That topology is called the indiscrete topology on X.)

[...]

>> It is not as bad as it seems - you are trying to "extend"
>> the wrong thing.. As far as I know, no one is
>> particularly interested in a metric on R* (= the
>> extended reals).

Though it's not hard to provide one.  Define a function 
f : R* --> [-pi/2, pi/2] as follows: f(x) = Arctan(x) if x
is in R, f(oo) = pi/2, and f(-oo) = -pi/2.  Now for any x
and y in R* define d(x, y) = |f(x) - f(y)|; it's easy to
check that d is a metric on R* that generates the right
topology.

>> However, all is not lost - far from it. We endow R* with
>> the order topology.

>> I don't know how much you know about topology so this
>> will be sketchy and vague.  [...]

>> Anyway, consider the left and right rays in R* : 
>> {x : x < a} and  {x : x > a} for any a in R* (of course if 
>> a = oo the right ray is empty, etc.). 

>> Consider the set consisting of all rays and all open
>> intervals, (a, b)  for a and b in the usual Reals. Call
>> this collection B. We can say a  sequence a_n converges
>> to L in R* if, whenever L belongs to an  element U of B,
>> the sequence eventually lies in B - that is, there is 
>> an N such that if n > N, a_n is in U.

>> If you think about it that says that if L is an ordinary
>> Real convergence means just what it always has but for
>> example,  1, 2, ..., n, ... really _does_ converge to
>> +oo according to this definition.

> Hm. Let's see. We allow that in the topology defined
> above, yes? You said a can be +oo, so there is an
> interval such as (p, oo) in B 

No, because B was defined to contain only rays and intervals
of the form (a, b) with a and b in R, not R*.  What you want
is the right ray at p, i.e., {x in R* : x > p}.  Note that
this *does* include the point oo; if one were to use
interval notation in R* to represent it, it would be 
(p, oo], not (p, oo).

> which as long as we find an
> N such that for all n > N, a_n is in (x, oo), 

Correction: a_n is in the right ray at p, i.e., in 
{x in R* : x > p}.  Of course this amounts to saying that
for all n > N, a_n > p.

> then it converges there. So, for 1, 2, ..., n, we find N =
> 2 and U = (1, oo) which works. Does this make sense?

To show that <1, 2, 3, ...> --> oo, you must show that for
*each* U in B that contains oo there is an n(U) such that
for all n > n(U), a_n is in U; it's not enough to do it for
a single U.

The only members of B that contain oo are the right rays, so
let U = (p, oo] be any right ray.  Let n(U) = p; clearly if
n > p, then n is in U.

>> For your amusement and edification look up "one point
>> compactification" - Wikipedia will do. There, oo is used
>> as _both_ ends of R!

> Don't know intuitively what a compact space is, and that
> seems to be a requirement for reading this article.

Let (X, T) be a topological space (i.e., X is the underlying
set, and T is the family of open sets).  A subset C of T is
an open cover of X provided that the union of C is all of X;
equivalently, provided that every x in X belongs to at least
one member of C.  If C is an open cover of X, U is a
subcollection of C, and U is also an open cover of X, we say
that U is a subcover of C; if in addition U has only
finitely many members, U is a finite subcover of C.

(X, T) is compact if and only if every open cover of X has a
finite subcover.

Brian