Topology with first countable.
mina_...@hanmail.net
I know that
Let S have the cofinite or the cocountable topology. Show
S is countable iff S is 2nd countable iff S is 1st countable.
- William Elliot -
I have a question.
If X is a countable set, then (X, T) is first countable.
is this true ?
José Carlos Santos
No. Just take the one-point compactification of Q, that is, Q U {oo}
with the topology T for which a set A is open when:
1) if oo does not belong to A, then A is an open subset of Q;
2) otherwise, then the complement of Q is compact.
Best regards,
Jose Carlos Santos
David C. Ullrich
This seems like a very good question - seemed trivially
true at first, but I suspect that it's false.
Of course one could look this up in a book on topology
(it's equivalent to saying that any separable space must
be first-countable).
Here's something that I _think_ is true, which explains
why I _suspect_ that the answer to your question is
no: Say F is a family of subsets of the natural numbers
N. Say F is "incomparable" if for every A, B in F with
A <> B, A is not a subset of B and B is not a subset of A.
I believe that there exist uncountable incomparable
familes F...
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Dave L. Renfro
>> I have a question.
>> If X is a countable set, then (X, T) is first countable.
>>
>> is this true ?
David C. Ullrich wrote (in part):
> This seems like a very good question - seemed trivially
> true at first, but I suspect that it's false.
I thought the same thing myself, and then realized that neighborhood
bases involve collections of subsets of the space and not just
subsets of the space. While subsets of the space must be countable,
collections of subsets can be uncountable (a less precise version
of your thoughts, not copied above, about the existence of a
collection of uncountably pairwise "incomparable" subsets a
countably infinite set).
After these thoughts I then realized that I've previously posted
some references on extreme examples of this kind . . .
-----------------------------------
A countable space need not be first countable. In fact, there
exist countable spaces that have no points of first countability.
For some examples, see:
Peter J. Harley, "A countable nowhere first countable Hausdorff
space", Canadian Mathematical Bulletin 16 (1973), 441-442.
http://tinyurl.com/5bdddb [.pdf file of Harley's paper]
Ronnie Levy, "Countable spaces without points of first countability",
Pacific Journal of Mathematics 70 (1977), 391-399.
http://tinyurl.com/6f9u24 [.pdf file of Levy's paper]
Richard Willmott, "Countable yet nowhere first countable",
Mathematics Magazine 52 (1979), 26-27.
In fact, Leslie Foged constructed 2^c nonhomeomorphic countable
spaces having no points of first countability in his 1979 Ph.D.
Dissertation "Weak Bases for Topological Spaces" under Ron
Freiwald at Washington University.
-----------------------------------
Dave L. Renfro
David C. Ullrich
I can't quite parse (2) as a definition of which sets are
open. If you mean the one-point compactification of
Q (with the discrete topology) then
(1) Any subset of Q is open
(2) Any subset of Q U {oo} with a finite complement is open.
(Ie, the complement (in Q U {oo}) of any compact subset of Q
is open.)
Except that seems second-countable to me: If F_n is a
sequence of finite subsets of Q with union equal to Q
then don't the complements of the F_n give a local base
at 00?
> Best regards,
>
> Jose Carlos Santos
--
David C. Ullrich
mina_world
- Arens-Fort Space -
Let (X, T) be the set of all ordered pairs of nonnegative integers
with each pair open except (0, 0).
Open neighborhoods U of (0, 0) are defined
so that for all but a finite number of integers m,
the sets S_m = {n | (m, n) in U} are each finite.
Thus each open neighborhood of the origin contains all but
a finite number of points in each of all but a finite number of columns.
(Hm... unintelligible to me)
(X, T) is not first countable, for it does not have a countable local basis
at (0, 0).
For suppose {U_i} was such a basis; then for each positive integer i,
there are integers m_i, n_i each greater than i,
such that (m_i , n_i) in U_i.
Then X - {(m_i , n_i) | i = 1, 2, 3, ...} is an open neighborhood of
the origin which contains none of the sets U_i.
---------------------------------------------------------
- Appert Space -
Let X be the set of positive integers.
Let N(n, E) denote the number of integers in a set E subset X
which are less than or equal to n.
We describe Appert's topology on X
by declaring open any set which excludes the integer 1,
or any set E containing 1 for which lim{n->oo} N(n, E) / n = 1.
(Hm...not easy.)
X is not first countable, since the point 1 does not have
a countable local basis.
Suppose {B_n} were a countable local basis at 1,
Then each B_n must be infinite,
so we can select an x_n in B_n such that x_n > 10^n
then U = X - {x_n} does not contain any of the sets B_n,
yet it is an open neighborhood of 1,
for N(n, U) > n - log_10 (n), and thus lim{n->oo} N(n, U) / n = 1.
- Counterexamples in topology - Steen, Seebach -
Anyway... difficult !
David C. Ullrich
wrote:
You quoted the following from somewhere. When you
do that you should state where the quotation came from.
>Let (X, T) be the set of all ordered pairs of nonnegative integers
>with each pair open except (0, 0).
>Open neighborhoods U of (0, 0) are defined
>so that
(i)
> for all but a finite number of integers m,
>the sets S_m = {n | (m, n) in U} are each finite.
(ii)
>Thus each open neighborhood of the origin contains all but
>a finite number of points in each of all but a finite number of columns.
>
>(Hm... unintelligible to me)
Well, I think you must have typed something wrong, because
(i) and (ii) do not say the same thing, although (ii) is clearly
supposed to be just an explanation of what (i) means. What seems
likely to me is that (i) should have been
(i')
for all but a finite number of integers m,
the sets S_m = {n | (m, n) not in U} are each finite.
In other words:
If (n,m) <> (0,0) then {(n,m)} is open. That gives a local base
at (n,m).
On the other hand, U is a neighborhood of (0,0) if and only if
(0,0) is an element of U and the set of m such that
{n : (n,m) is not in U} is finite. Except for finitely many
m, the set S_m has finite complement.
>(X, T) is not first countable, for it does not have a countable local basis
>at (0, 0).
>For suppose {U_i} was such a basis; then for each positive integer i,
>there are integers m_i, n_i each greater than i,
>such that (m_i , n_i) in U_i.
This holds because, except for finitely many m, all but finitely
many n have the property that (m,n) is in U_i. So there
exists m_i > m such that all but finitely many (m_i, n) are
in U_i, and hence there exists n_i > i such that (m_i, n_i)
is in U_i.
Note that to make the rest of the argument easier to state
we can also assume that the sequence (m_i) is
strictly increasing. (If we've chosen m_i for 1 <= i < k
then we can choose m_k to be greater than all those
values, since all but finitely many values of m will
work for m_k.)
>Then X - {(m_i , n_i) | i = 1, 2, 3, ...} is an open neighborhood of
>the origin which contains none of the sets U_i.
Say S = X - {(m_i , n_i) | i = 1, 2, 3, ...}. Then S is a
neighborhood of (0,0), since in fact _every_ S_m
is finite! (If m = m_i then S_m = {n_i}, while
if m is not one of the m_i then S_m = N, the
set of all natural numbers).
And S_i is not a subset of S, since (m_i, n_i)
is an element of S_i but not an element of S.
>---------------------------------------------------------
>- Appert Space -
>
>Let X be the set of positive integers.
>Let N(n, E) denote the number of integers in a set E subset X
>which are less than or equal to n.
>We describe Appert's topology on X
>by declaring open any set which excludes the integer 1,
>or any set E containing 1 for which lim{n->oo} N(n, E) / n = 1.
>
>(Hm...not easy.)
>
>X is not first countable, since the point 1 does not have
>a countable local basis.
>Suppose {B_n} were a countable local basis at 1,
>Then each B_n must be infinite,
>so we can select an x_n in B_n such that x_n > 10^n
>then U = X - {x_n} does not contain any of the sets B_n,
>yet it is an open neighborhood of 1,
>for N(n, U) > n - log_10 (n), and thus lim{n->oo} N(n, U) / n = 1.
>
>- Counterexamples in topology - Steen, Seebach -
Ah, so that's where it came from. Ok.
>Anyway... difficult !
David C. Ullrich
<dull...@sprynet.com> wrote:
>In article <6deis0F...@mid.individual.net>,
> José Carlos Santos <jcsa...@fc.up.pt> wrote:
>
>> On 07-07-2008 12:20, mina_...@hanmail.net wrote:
>>
>> > I know that
>> > Let S have the cofinite or the cocountable topology. Show
>> > S is countable iff S is 2nd countable iff S is 1st countable.
>> > - William Elliot -
>> >
>> > I have a question.
>> > If X is a countable set, then (X, T) is first countable.
>> >
>> > is this true ?
>>
>> No. Just take the one-point compactification of Q, that is, Q U {oo}
>> with the topology T for which a set A is open when:
>>
>> 1) if oo does not belong to A, then A is an open subset of Q;
>>
>> 2) otherwise, then the complement of Q is compact.
>
>I can't quite parse (2) as a definition of which sets are
>open. If you mean the one-point compactification of
>Q (with the discrete topology) then
>
>(1) Any subset of Q is open
>(2) Any subset of Q U {oo} with a finite complement is open.
>
>(Ie, the complement (in Q U {oo}) of any compact subset of Q
>is open.)
>
>Except that seems second-countable to me:
I meant "first countable" here.
>If F_n is a
>sequence of finite subsets of Q with union equal to Q
>then don't the complements of the F_n give a local base
>at 00?
>
>> Best regards,
>>
>> Jose Carlos Santos
David C. Ullrich
José Carlos Santos
>>> I know that
>>> Let S have the cofinite or the cocountable topology. Show
>>> S is countable iff S is 2nd countable iff S is 1st countable.
>>> - William Elliot -
>>>
>>> I have a question.
>>> If X is a countable set, then (X, T) is first countable.
>>>
>>> is this true ?
>> No. Just take the one-point compactification of Q, that is, Q U {oo}
>> with the topology T for which a set A is open when:
>>
>> 1) if oo does not belong to A, then A is an open subset of Q;
>>
>> 2) otherwise, then the complement of Q is compact.
>
> I can't quite parse (2) as a definition of which sets are
> open. If you mean the one-point compactification of
> Q (with the discrete topology) then
>
> (1) Any subset of Q is open
> (2) Any subset of Q U {oo} with a finite complement is open.
>
> (Ie, the complement (in Q U {oo}) of any compact subset of Q
> is open.)
>
> Except that seems second-countable to me: If F_n is a
> sequence of finite subsets of Q with union equal to Q
> then don't the complements of the F_n give a local base
> at 00?
But what I meant was the one-point compactification of Q with respect to
the usual topology.
David C. Ullrich
Ok. The reason I didn't realize that's what you meant is that if
that's what you meant then your version of (2) is simply wrong,
not just confusing. Because not every compact subset of Q
is finite.
Another reason I didn't guess that is that it's locally compact
spaces that _have_ a one-point compactification, and Q
is not locally compact. (If K is compact (hausorff) and oo
is a point of K then K \ {oo} is locally compact. Hence
Q has no one-point compactification.)
So right now I'm really not certain exactly what topological
space you were referring to. Whatever it is, I tend to doubt
that it can be an example of a non-first-countable countable
space, just because I doubt it's possible to give a description
of one that's _quite_ that simple.
>Best regards,
>
>Jose Carlos Santos
David C. Ullrich
José Carlos Santos
>>>>> I know that
>>>>> Let S have the cofinite or the cocountable topology. Show
>>>>> S is countable iff S is 2nd countable iff S is 1st countable.
>>>>> - William Elliot -
>>>>>
>>>>> I have a question.
>>>>> If X is a countable set, then (X, T) is first countable.
>>>>>
>>>>> is this true ?
>>>> No. Just take the one-point compactification of Q, that is, Q U {oo}
>>>> with the topology T for which a set A is open when:
>>>>
>>>> 1) if oo does not belong to A, then A is an open subset of Q;
>>>>
>>>> 2) otherwise, then the complement of Q is compact.
Oops! What I meant here was that "the complement of A is compact".
>>> I can't quite parse (2) as a definition of which sets are
>>> open. If you mean the one-point compactification of
>>> Q (with the discrete topology) then
>>>
>>> (1) Any subset of Q is open
>>> (2) Any subset of Q U {oo} with a finite complement is open.
>>>
>>> (Ie, the complement (in Q U {oo}) of any compact subset of Q
>>> is open.)
>>>
>>> Except that seems second-countable to me: If F_n is a
>>> sequence of finite subsets of Q with union equal to Q
>>> then don't the complements of the F_n give a local base
>>> at 00?
>> But what I meant was the one-point compactification of Q with respect to
>> the usual topology.
>
> Ok. The reason I didn't realize that's what you meant is that if
> that's what you meant then your version of (2) is simply wrong,
> not just confusing. Because not every compact subset of Q
> is finite.
>
> Another reason I didn't guess that is that it's locally compact
> spaces that _have_ a one-point compactification, and Q
> is not locally compact. (If K is compact (hausorff) and oo
> is a point of K then K \ {oo} is locally compact. Hence
> Q has no one-point compactification.)
I should have written: apply to Q the method which, when applied to
locally compact spaces, gives a one-point compactification. The
_definition_ of the topology makes sense, although, yes, Q U {oo} is not
compact.
> So right now I'm really not certain exactly what topological
> space you were referring to. Whatever it is, I tend to doubt
> that it can be an example of a non-first-countable countable
> space, just because I doubt it's possible to give a description
> of one that's _quite_ that simple.
Q U {oo} is not first countable. More precisely, there is no sequence
(A_n)_n of open neighborhoods of oo such that each neighborhood of oo
contains some A_n. An equivalent way of saying this is: there is no
sequence (K_n)_n of compact subsets of Q such that each compact subset
of Q is contained in some K_n. For each _n_ let q_n be a rational number
from ]-1/n,1/n[ which does not belong to K_n. Then, if you define
K = {0, q_1, q_2, q_3, q_4, ...}
the set K will be a compact of Q which will contain all the q_n's and
therefore cannot be a subset of some K_n.
David C. Ullrich
Ok. that's pretty simple.
>Best regards,
>
>Jose Carlos Santos
David C. Ullrich