# T1 topology

87 views

### pierre.c

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

### David C. Ullrich

Oct 4, 2004, 9:15:27 AM10/4/04
to
On Mon, 04 Oct 2004 12:37:55 GMT, pierre...@apx.fr (pierre.c)
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

### pierre.c

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

### 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:

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

### G. A. Edgar

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/

### William Elliot

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
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.

### David C. Ullrich

Oct 4, 2004, 5:08:52 PM10/4/04
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On Mon, 04 Oct 2004 13:57:53 GMT, pierre...@apx.fr (pierre.c)
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.

### David C. Ullrich

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:

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

### shedar

Oct 5, 2004, 12:45:14 AM10/5/04
to
"David C. Ullrich" <ull...@math.okstate.edu> wrote in message
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

### pierre.c

Oct 5, 2004, 5:43:39 AM10/5/04
to
On Mon, 04 Oct 2004 12:37:55 GMT, pierre...@apx.fr (pierre.c)
wrote:
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

### William Elliot

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