Re: notation: sup, inf, limit, sequences

In article <zz6onkvqsfra.15szu6gae8q2t.dlg@[EMAIL PROTECTED]
>,
Brian M. Scott wrote:

> 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.)

And may I say the metric in that space would cause the distance between
any two points to be always the same? If this true, then the topology of
a space describes at least the nature of its metric, yes?

So the metric d(a,b) = |a - b| in R makes R our intuitive topological
space. Now, not every topological space is a metric space, right? What's
the use of non-metric spaces? What sort of things do they describe?

>>> 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).

Okay.

>> 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.

Okay.

>> 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.

Hm. I'm just noticing that a ray is one thing and an open interval is
another. A ray is an interval that doesn't end, right?

So yes, the only members of B that contain oo are the right rays ---
that is, the ``intervals'' that begin at p and ``end'' in oo. So okay.

Okay, except I have a set theory objection here. Is it okay to have rays
and intervals in the same set? Don't objects need the same, say, nature
to belong in the same set? Otherwise when I say let x be in B, what's x
exactly? Is it an object of type B1 or B2?

But I'm thinking that what's going on here is that in such set B, 
(p, oo] is just an interval as (a, b) is. 

Anyway, I followed your proof above. The problem I still have is in the
definition itself. As given by Paul we have

      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.

Let me rewrite that in my words to see if I get something equivalent. 

The sequence a_n converges to L in R* if for all U in B such that L is
in U, there is an N such that for all n > N, a_n is in U.

That is, given any neighborhood U which contains the limit L, we must
find an index N such that the sequence a_n from that point on is all in
U. True?

>>> 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;

What's the ``union of C''? Answered attempted below.

> equivalently, provided that every x in X belongs to at least
> one member of C.  

So C must be a collection of sets like T. Then perhaps the union of C is
the union of all of its members? 

So if I choose enough sets in T which contain all points of X I have an
open cover of X. That's my C here.

> 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.

Alright. U is a smaller open cover of X. Does ``subcollection'' imply a
proper subcollection? Otherwise we always have a trivial subcollection
--- C itself.

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

I see. An example seems challenging. :-) Let's take R and its usual
topology which is all open intervals, yes? Hopefully R is compact. Hm,
no, we will never be able to finitely (sub)cover R.

Let's look for a trivial example: (X = {0,1}, T = {{},{0},{1},{0,1}}).
What are the possible open covers of X? They are C(1) = {{0},{1}} and
C(2) = {{0,1}}. Do they have finite open subcovers? It depends on
whether the subcollection must be proper. Let's assume so and change the
the covers to C(1) = {{}, {0}, {1}} and C(2) = {{}, {0,1}}.

Now let U(1) = { {0}, {1} } and U(2) = { {0,1} }. So every cover of X
has a finite subcover. So (X,T) is compact.

What does this mean intuitively? Wikipedia[*] is saying that closed and
bounded is equivalent to compact, although Bourbaki wanted to leave the
word compact for Hausdorff spaces only.

[*] http://en.wikipedia.org/wiki/Compact_space

I don't know if (X, T) above is closed and bounded. I must say it is,
since I think it is compact. 

A set is closed if it's complement is open. X is closed, since {} is in
T. But what does bounded mean for a topological space? I've no idea what
a lower and upper bound here would be.