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Oct 4, 2004, 8:37:55 AM10/4/04

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I read that a T1 topology is metrizable . I do not understand because

:

:

- a metric gives a separate topology according to the T2 separation

condition:

For each couple of points x and y, there is an open set U which

contains x , an open set V which contains y and U. V is empty.

The triangular inequality forbids that U.V be non empty.

- a T1 topology is not separated this way but rather :

For each couple of points x and y, there is an open set U which

contains x and not y , an open set V which contains y and not x .

This property says nothing about U.V which can be non empty.

So a T1 topology + a metric should be a T2 topology

did i read properly? is it a mistake?

pierre.c

Oct 4, 2004, 9:15:27 AM10/4/04

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On Mon, 04 Oct 2004 12:37:55 GMT, pierre...@apx.fr (pierre.c)

wrote:

wrote:

At first it sounded like you'd read that _any_ T1 topology

was metrizable - that's certainly wrong. But it seems like

what you read was that some particular T1 topology was

metrizable. There's no problem with that. Yes, it follows

that the topology is actually T2. So what? Any T2 topology

is also T1.

>pierre.c

************************

David C. Ullrich

Oct 4, 2004, 9:52:12 AM10/4/04

to

So i did'nt read properly. I thought of having read that any T1

topology was metrizable.

thanks

pierre.c

Oct 4, 2004, 9:57:53 AM10/4/04

to

On Mon, 04 Oct 2004 08:15:27 -0500, David C. Ullrich

<ull...@math.okstate.edu> wrote:

<ull...@math.okstate.edu> wrote:

I found this

http://mathworld.wolfram.com/T1-Space.html

Wher it is said that T1 spaces are complete and metrizable but with

the restriction that the space be locally convex.

Do you know where i can find a clue?

pierre.c

Oct 4, 2004, 10:38:16 AM10/4/04

to

>

> http://mathworld.wolfram.com/T1-Space.html

>

A confused web page. A T_1 space may be called a "Frechet space".

There is something else, in another branch of mathematics, also

known as a "Frechet space" (complete metrizable locally convex

topological vector space). The web page mixes the two.

--

G. A. Edgar http://www.math.ohio-state.edu/~edgar/

Oct 4, 2004, 10:39:29 AM10/4/04

to

On Mon, 4 Oct 2004, pierre.c wrote:

> <ull...@math.okstate.edu> wrote:

>

> I found this

>

> http://mathworld.wolfram.com/T1-Space.html

>

> Wher it is said that T1 spaces are complete and metrizable but with

> the restriction that the space be locally convex.

>

> Do you know where i can find a clue?

>

Not a clue what. A space is complete when every Cauchy sequence> <ull...@math.okstate.edu> wrote:

>

> I found this

>

> http://mathworld.wolfram.com/T1-Space.html

>

> Wher it is said that T1 spaces are complete and metrizable but with

> the restriction that the space be locally convex.

>

> Do you know where i can find a clue?

>

converges. For Cauchy sequences, one needs a uniform space, which

includes topological groups and metric spaces. Nor do I know what

is meant by locally convex. From this and from what I read, it seems

he's rambling and thinking about something else more than T0 spaces.

topological spaces. The big bad Wolfram is reputed for errors.

As for a T1 counter example, the cofinite topology of an infinite space is

not metrizable for not being T2, ie Hausdorff.

There is Urysohn's reknown metrization theorem, that a 2nd countable

regular T0 space is metrizable. However do note, regular with T0 or T1

does imply Hausdorff which all metrizable spaces must be, as well as

1st countable and (every variety of) normal, just all metric spaces are.

Even so, what a locally convex metric may space be, I know not. Perhaps

another will elucidate us.

Oct 4, 2004, 5:08:52 PM10/4/04

to

On Mon, 04 Oct 2004 13:57:53 GMT, pierre...@apx.fr (pierre.c)

wrote:

wrote:

>On Mon, 04 Oct 2004 08:15:27 -0500, David C. Ullrich

><ull...@math.okstate.edu> wrote:

>

>>On Mon, 04 Oct 2004 12:37:55 GMT, pierre...@apx.fr (pierre.c)

>>wrote:

>>

>>>I read that a T1 topology is metrizable . I do not understand because

>>>:

>>>[...]

>

>I found this

>

>http://mathworld.wolfram.com/T1-Space.html

>

>Wher it is said that T1 spaces are complete and metrizable but with

>the restriction that the space be locally convex.

>

>Do you know where i can find a clue?

Right here. One clue: Don't believe what you read at mathworld.

It's full of errors - that page is one of the worst I've seen.

As Edgar said, the authors are confusing two totally different

notions that just happen to be described by the same word.

Oct 4, 2004, 5:10:31 PM10/4/04

to

On Mon, 4 Oct 2004 07:39:29 -0700, William Elliot <ma...@privacy.net>

wrote:

wrote:

That page is one of the worst I've seen at mathworld, which is saying

something. Hald of the comments are about what you know as T1 spaces,

which evidently are also called Frechet spaces. The other half is

about a totally different topic, in topological vector spaces.

************************

David C. Ullrich

Oct 5, 2004, 12:45:14 AM10/5/04

to

"David C. Ullrich" <ull...@math.okstate.edu> wrote in message

news:4se3m0h5he9v7u1m2...@4ax.com...

news:4se3m0h5he9v7u1m2...@4ax.com...

Yes, that is a terribly worded page! It uses the word "Frechet" in two

different contexts without being explicit about it (assuming the author is

even aware of it). In a "pure" topological setting, "Frechet" is often used

as a synonym for "T1", but in functional analysis, a "Frechet" space is an

example of a "special" kind of complete topological vector space (such as

one whose topology is induced by some countable family of semi-norms).

Indeed, being "T2" alone is not sufficient to guarantee metrizability, let

alone being just "T1". A well-known example of a T2 (in fact, completely

normal) space which is not pseudometrizable is the Sorgenfrey line <R,T>

(aka, Right Half-Open Interval Topology), where R is the set of all real

numbers, and T is the topology generated by the collection B = { [a,b) | a,

b in R and a < b} of basic open sets.

Shedar

Oct 5, 2004, 3:58:44 AM10/5/04

to

shedar wrote:

>>Right here. One clue: Don't believe what you read at mathworld.

>>It's full of errors - that page is one of the worst I've seen.

>>As Edgar said, the authors are confusing two totally different

>>notions that just happen to be described by the same word.

>>

> Yes, that is a terribly worded page! It uses the word "Frechet" in two

> different contexts without being explicit about it (assuming the author is

> even aware of it).

It seems that what we see here is mostly an editorial problem. The page

ackowledges two distinct authors. It is possibe that none of them is to

blame individually.

Best regards,

Jose Carlos Santos

Oct 5, 2004, 5:43:39 AM10/5/04

to

I have one more question about T1 topology

Is there a condition on a basis B so that the topology T induced by B

has the following properties

- T is T1

- for every open sets U, V , U/\V is non empty

(This topology should behave like the cofinite topology on an infinite

set)

or at least

- for every points x, y for wich there are open sets U and V so that

x lies in U and not y, y lies in V and not x

then U/\V is non empty

I'm looking for a condition on a basis not on the topology itself.

pierre.c

Oct 5, 2004, 7:12:24 AM10/5/04

to

On Tue, 5 Oct 2004, pierre.c wrote:

> On Mon, 04 Oct 2004 12:37:55 GMT, pierre...@apx.fr (pierre.c)

> wrote:

> Is there a condition on a basis B so that the topology T induced by B

> has the following properties

>

> - T is T1

> - for every open sets U, V , U/\V is non empty

>

Yes, it's called hyper-connected. Have I not answered that a few days

ago? The exact condition is

for all open nonnul U,V, U/\V nonnul

Thus to describe the space with those properties, would be to call it

hyper-connected T1.

> (This topology should behave like the cofinite topology on an infinite

> set)

>

> or at least

>

> - for every points x, y for wich there are open sets U and V so that

> x lies in U and not y, y lies in V and not x

> then U/\V is non empty

>

> I'm looking for a condition on a basis not on the topology itself.

>

Basis don't need empty sets, so by convention you could require basis not

to have them. Beyond that, it makes little difference where you put

the condition as

all nonnul open sets intersect

iff

all (nonnul) base sets intersect

Same with T1, it makes no differenct where you put the condition as

for all distinct x,y

some open U with x in U, y not in U

iff

some base set U with x in U, y not in U

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