First-Countable, Second-Countable, HUH?
Euclid's Dog
it is second-countable. I put together what looked like a convincing enough
proof, but I ended up not needing the fact that the set X was
first-countable (kind of like putting together a bicycle and having a single
nut left over). Why then do I need that for the proof? All I could figure is
that it is a roundabout way to say that the set X has a local base at each
point x in X.
My proof goes something like this:
Let X be a countable space and U an open set in X. For each x in U, U is a
neighborhood of x. Since X is first-countable, x has a local base, so we can
find an open set V containing x, such that x in V in U. We can find such a V
for every x in U, and the union of all such open sets V is also open and
"equal" to U. Since there are only countable many x, there are countable
many V and X is second-countable.
See? I used the fact that X was first-countable only to claim that each x
had a local base. I really didn't need that it was (or wasn't)
first-countable. The fact that the number of elements in X (and any subset
thereof) is countable, gave me the second-countable part.
I know I missed something. Can you help? Thanks in advance...
-jk
Fred Galvin
You must have missed something, because there are topological spaces
which are neither first countable nor second countable. I can't
pinpoint the mistake, because I can't understand any part of your
proof. Instead, I will give you an example of a countable space, and
let you find a countable base for it.
Let X be the set of all positive integers. A subset U of X is open if
and only if it satisfies at least one of the conditions: (i) 1 is not
an element of U; (ii) the limit of |{x in U: x <= n}|/n, as n
approaches infinity, exists and is equal to 1.
Johann Zwischenkieferknochen
> Given a countable set X that is also first-countable, I'd like to show that
> it is second-countable. I put together what looked like a convincing enough
> proof, but I ended up not needing the fact that the set X was
> first-countable (kind of like putting together a bicycle and having a single
> nut left over). Why then do I need that for the proof? All I could figure is
> that it is a roundabout way to say that the set X has a local base at each
> point x in X.
>
> My proof goes something like this:
>
> Let X be a countable space and U an open set in X. For each x in U, U is a
> neighborhood of x. Since X is first-countable, x has a local base, so we can
> find an open set V containing x, such that x in V in U. We can find such a V
> for every x in U, and the union of all such open sets V is also open and
> "equal" to U. Since there are only countable many x, there are countable
> many V and X is second-countable.
You proved that the set U is a countable union of sets like V. Now, if you
start with another set U, you will get _another_ countable union of sets
like V. You did not prove that _any_ set U is the countable union of sets
like V.
The family of all sets U is not countable in general (even if X is), so you
end up with a non-countable collection of sets like V. "Sets like V" means
the sets you would like to put in your basis.
To prove the statement, define a countable family of open sets at the
beginning (guess how!) and show that it is a basis.
> See? I used the fact that X was first-countable only to claim that each x
> had a local base. I really didn't need that it was (or wasn't)
> first-countable. The fact that the number of elements in X (and any subset
> thereof) is countable, gave me the second-countable part.
>
> I know I missed something. Can you help? Thanks in advance...
>
> -jk
--
johann
Fred Galvin
> Given a countable set X that is also first-countable, I'd like to
> show that it is second-countable. I put together what looked like
> a convincing enough proof, but I ended up not needing the fact
> that the set X was first-countable (kind of like putting together
> a bicycle and having a single nut left over).
All right, I think I know what's wrong here.
> Why then do I need that for the proof? All I could figure is that
> it is a roundabout way to say that the set X has a local base at
> each point x in X.
>
> My proof goes something like this:
>
> Let X be a countable space and U an open set in X. For each x in
> U, U is a neighborhood of x. Since X is first-countable, x has a
> local base, so we can find an open set V containing x, such that x
> in V in U. We can find such a V for every x in U, and the union of
> all such open sets V is also open and "equal" to U. Since there
> are only countable many x, there are countable many V and X is
> second-countable.
The mistake seems to be where you say "and X is second-countable". The
argument up to that point seems to be more or less OK, but you haven't
proved that X is second-countable. All you've proved is that:
(A) given an open set U, you can find a countable collection of open
sets whose union is equal to U.
That's fine (and I could suggest a much simpler proof), but that's not
what "second-countable" means. To prove that X is second-countable,
you have to show that X has a *countable base*, in other words:
(B) there is a countable collection B of open sets such that *every*
open set [not just one that you've singled out for special attention]
is the union of some subcollection of B.
You didn't say how you thought you could get from (A) to (B). I'm just
guessing, but maybe you meant to repeat the process for each open set
U. The collection of all V's you get that way will be a *base* all
right, but it won't necessarily be *countable*: you are picking
countably many V's for each U, but there may be uncountably many U's.
Dave L. Renfro
[sci.math Sun, 15 Apr 2001 03:27:26 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/singnendspex>
wrote
> Given a countable set X that is also first-countable, I'd like
I assume "set" means "topological space"?
> to show that it is second-countable. I put together what looked
> like a convincing enough proof, but I ended up not needing the
> fact that the set X was first-countable (kind of like putting
> together a bicycle and having a single nut left over). Why then
> do I need that for the proof? All I could figure is that it is
> a roundabout way to say that the set X has a local base at each
> point x in X.
The fact that every point has a local base should be something
proved right after the concepts were introduced. This is always
true. But this isn't what first countable means. First countable
means that every point has a COUNTABLE local base.
> My proof goes something like this:
>
> Let X be a countable space and U an open set in X. For each x
> in U, U is a neighborhood of x. Since X is first-countable,
> x has a local base,
As I mentioned above, this doesn't require first-countability.
For each x in X let U(x) be the collection of all neighborhoods
of x. Then U(x) will satisfy what I assume you mean by local
base at x. [Certain subcollections of U(x) will also be a local
base at x, unless you're dealing with a really trivial space.]
> so we can find an open set V containing x,
> such that x in V in U. We can find such a V for every x in U,
> and the union of all such open sets V is also open and "equal"
> to U. Since there are only countable many x, there are countable
> many V and X is second-countable.
Here's an analogy that might help to show where your
argument breaks down:
Each semi-open interval (-oo, r) in the real line for any real
number r is the union of all sets V of the form (-oo, x) for x
a rational number less than r. Since there are only countably
many x's, there are countably many V's, and hence countably many
sets (-oo, r). [Don't tell Ross A. Finlayson about this.]
> See? I used the fact that X was first-countable only to claim
> that each x had a local base. I really didn't need that it was
> (or wasn't) first-countable. The fact that the number of
> elements in X (and any subset thereof) is countable, gave me
> the second-countable part.
>
> I know I missed something. Can you help? Thanks in advance...
A first countable space need not be second countable --- consider
any uncountable discrete space.
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 Math. Bull. 16 (1973), 441-442.
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.
As for what you want to prove, namely that for countable topological
spaces, first countability implies second countability, you
want to be a bit more explicit in your proof.
First, tell your reader exactly which countable collection of open
sets you're going to show is a base. [Each point has a countable
local base (these bases may not be unique, so make sure you don't
refer to any of them as "the local base at x" until you've fixed
a choice of them in your proof), there are countably many points,
a countable union of countable sets is countable, . . .]
Having done this, now prove this collection is a base for the
topology on X. Look up the definition of a base and make sure
that you cover each part of it (pun originally accidental, but
I'll take credit for it).
If you really want to impress sci.math readers, write an essay
explaining how this result can be incorporated into logic based
probability using the proximity function. [See David C. Ullrich's
April 13, 2001 post at
<http://forum.swarthmore.edu/epigone/sci.math/bydwehkheh> for an
excellent example along these lines.]
Dave L. Renfro