On Tue, 22 Jul 2008 16:41:10 +0200 (CEST), Daniel Chicayban
Bastos <dbast0s@[EMAIL PROTECTED]
> wrote in
<news:slrng8bsia.6tq.dbast0s@[EMAIL PROTECTED]
> in
alt.algebra.help:
> 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:
[...]
>>> 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?
No: if that space has more than one point, it isn't
metrizable, i.e., there is no metric on X that generates the
given topology.
A metric d on a set X is a function d : X x X --> R such
that:
(0) for all x, y in X, d(x, y) >= 0;
(1) for all x, y in X, d(x, y) = d(y, x);
(2) for all x, y in X, d(x, y) = 0 if and only if x = y;
(3) for all x, y, z in X, d(x, y) <= d(x, z) + d(z, y).
For x in X and e > 0 define
N(x, e) = {y in X : d(x, y) < e},
the open e-ball around x, and let B be the collection of all
such balls:
B = {N(x, e) : x in X & e > 0}.
Then B is a base for a topology T on X, the topology
generated by the metric d.
If X has more than one point, this T cannot be the
indiscrete topology. To see this, suppose that x and y are
distinct points of X, and let e = d(x, y); by (0) and (2)
above, e > 0. I claim that N(x, e/2) and N(y, e/2) are
disjoint open neighborhoods of x and y, respectively.
First, it's clear that x in N(x, e/2) and y in N(y, e/2).
Also, N(x, e/2) and N(y, e/2) belong to B and hence to T, so
they really are open neighborhoods of x and y, respectively.
Suppose that z is in their intersection. Then
e = d(x, y) <= d(x, z) + d(z, y) < e/2 + e/2 = e,
which is obviously impossible. Thus, no such z exists, and
N(x, e/2) and N(y, e/2) are disjoint. In the indiscrete
topology there are no disjoint non-empty open sets, so T
isn't indiscrete. (In fact I've just shown here that every
metric space is Hausdorff.)
> If this true, then the topology of a space describes at
> least the nature of its metric, yes?
If (X, T) is a topological space, there need not be any
metric that generates T; in that case we say that (X, T) is
not metrizable. If (X, T) *is* metrizable, i.e., there is a
metric that generates T, then there are infinitely many
other metrics that also generate T.
> So the metric d(a,b) = |a - b| in R makes R our intuitive
> topological space.
If by this you mean that the metric d(x, y) = |x - y|
generates the usual topology on R, you're correct.
> Now, not every topological space is a metric space, right?
A topological space isn't a metric space at all: a topology
isn't a metric. If you mean that not every topological
space is metrizable, however, you're right, as noted above.
For example, if X = {0, 1}, and T = {Ø, X}, then (X, T) is
not metrizable, because it's not Hausdorff.
> What's the use of non-metric spaces? What sort of things
> do they describe?
What's the use of negative numbers? What's the use of
non-Abelian groups? I really don't know how to answer
questions like these. Oh, in each case I could doubtless
come up with an application of some sort, though you might
not have the background to follow it, but my own reaction is
'Who cares? They're *interesting*'.
[...]
>> 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?
More or less, yes.
> 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?
Not at all.
> Otherwise when I say let x be in B, what's x exactly? Is
> it an object of type B1 or B2?
It could be any member of B. If B contains elements of
different types (whatever that may mean), then x could be an
object of any of those types.
> 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.
If a, b, and p are in R, and a < b, then (a, b) and (p, oo]
are both intervals in R*, but they're different kinds of
intervals: (p, oo] contains its right endpoint, while (a, b)
doesn't.
> 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?
Yes.
[...]
>> 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?
Yes.
> 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.
Yes. For example, for each x in R let N(x) = (x-1, x+1) =
{y in R : |y - x| < 1}. These are open sets in R, because
they're open intervals. Let C = {N(x) : x in Z}; I claim
that C is an open cover of R. Certainly it's a family of
open sets, so we need only show that UC = R, where 'U' here
is the union symbol. Let y in R, and let n = floor(y).
Then n <= y < n + 1, so y in N(n) in C, and therefore y in
UC. Since y was arbitrary, this shows that UC = R.
>> 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?
No.
> Otherwise we always have a trivial subcollection --- C
> itself.
And indeed C is a subcover of 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.
It isn't. In fact, the open cover C that I constructed
above has no finite subcover. One way to see this is to
note that if n is an integer, N(n) is the *only* member of C
that contains n. Thus, if we throw away any member of C, we
no longer have a cover of R. That is, C has no proper
subcovers at all, and hence certainly no finite subcover.
[...]
> Let's look for a trivial example: (X = {0,1}, T =
> {{},{0},{1},{0,1}}).
More generally, any finite space is compact. Let (X, T) be
a topological space, and assume that X is finite, say X =
{x_1, x_2, ..., x_n}. Let C be an open cover of X. Then
for each x_k in X there is a V_k in C such that x_k in V_k.
But then {V_1, V_2, ..., V_n} is a finite subcollection of C
that still covers X, i.e., a finite subcover of C. Since C
was arbitrary, (X, T) is compact.
[...]
> What does this mean intuitively? Wikipedia[*] is saying
> that closed and bounded is equivalent to compact, [...]
Only in metric spaces. In fact, the definition of
boundedness requires having a metric. There are compact
Hausdorff spaces that are not metrizable, though even the
simplest examples that come to mind would take a fair bit of
explaining.
> But what does bounded mean for a topological space? [...]
It doesn't. If (X, d) is a metric space, a subset A of X is
said to be bounded (with respect to the metric d) if there
is some x in X and some e > 0 such that A is a subset of
N(x, e). (In other words, every point of A is within e
units of the point x.)
Brian
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