Conversation started Aug 4, 2012 at 0:49.
Aug 4, 2012 00:49
@HenryT.Horton Is there a non-metrizable Hausdorff space?
@JonasTeuwen I have a quick question
@BenjaLim How quick? I call my questions Usain.
@BenjaLim Hi.
@JonasTeuwen If I have $e^{ix} = e^{iy}$
@JonasTeuwen Then I should get that $x = y +2\pi k$ yes?
Aug 4, 2012 00:54
@BenjaLim Same as asking when you have $\sin(x) = \sin(y)$ and $\cos(x) = \cos(y)$ at the same time, right?
@JonasTeuwen yes
@PeterTamaroff Yes, some examples are $\Bbb R$ with basis open sets $[a, b)$, the long line, and $\Bbb R$ with basis open sets of the form $U - \text{countable set}$, where $U$ is open in the standard topology
@BenjaLim So, you say, if you have one value, this will also work if you just add integer multiples of $2\pi$, right?
@JonasTeuwen yes
@HenryT.Horton Good.
I'm away to eat.
BBL
Aug 4, 2012 00:57
@PeterTamaroff The Nagata-Smirnov metrization theorem gives necessary and sufficient conditions for a topology to be metrizable
@JonasTeuwen I have it don't worry.
@BenjaLim Good, so you have no question for me :-).
@HenryT.Horton Is it complicated?
@BenjaLim Finally... mailed Pierre.
@HenryT.Horton I have to prove that the subspace of a metrizable/Hausdorff space is metrizable/Hausdorff.
Aug 4, 2012 01:12
@PeterTamaroff Well... restrict the metric. DONE?
@JonasTeuwen Yes, yes. That's not the question.
Then why do you ask that question? 8-).
Isn't one a corollary of the other?
I have that $\rm metrizable \Rightarrow \rm Hausdorff$
But not conversely.
All metric spaces are Hausdorff.
@HenryT.Horton We just finished chapter 1 of Hall
Aug 4, 2012 01:13
@JonasTeuwen Yes, I know.
Yes, sure. Take a discrete topology on $\mathbf R$.
Singletons are open.
@HenryT.Horton And I just proved that every continuous functions from $\Bbb{R}$ to the circle is of the form $e^{iax}$ for some real number $a$
@HenryT.Horton Can I ask you something
We know that $\Bbb{R}, \times$ is not a group
@BenjaLim Sure...............
Not sure if that works. It is past 3 AM.
@HenryT.Horton But I think $\Bbb{R},+$ is a matrix lie group
Aug 4, 2012 01:14
@JonasTeuwen Hehehe nou
Take the discrete metric in $\Bbb R$
@HenryT.Horton $\Bbb{R} \cong \left\{ \left( \begin{array}{cc}1 & a \\ 0 & 1 \end{array}\right) : a \in \Bbb{R} \right\}$
@PeterTamaroff Modafuqa... Yeah.
@PeterTamaroff Fine. Weak topology (on a dual of a Banach space).
@HenryT.Horton And the latter is a matrix lie group
@JonasTeuwen Hahaha no idea what Banach spaces are!
@HenryT.Horton It is a closed subgroup of $GL_2(\Bbb{C})$
@HenryT.Horton Do you think it's right?
Aug 4, 2012 01:20
@PeterTamaroff Take the space of all real valued functions on $[0, 1]$ with the topology of pointwise convergence (that is, the smallest topology that makes all of the functions continuous), this is not metrizable.
@BenjaLim That should work
@PeterTamaroff Because in a metrizable space the singleton $\{0\}$ would be $G_\delta$. So it would be the intersection of a sequence of open sets.
@JonasTeuwen What is $G_\delta$?
@PeterTamaroff Second part.
@JonasTeuwen Pardon? =)
@JonasTeuwen I might have other words for that.
Aug 4, 2012 01:28
@JonasTeuwen Oh. Interesting notation!
@PeterTamaroff So if we have the topology of pointwise convergence on our space, then if $\{0\} = \bigcap_n G_n$ then this would mean that...
@JonasTeuwen I'm not used to function spaces =/
I'd have to think about it for a bit.
So for each $G_n$ there would be a set of functions in there that converges to $0$ right? Otherwise we would not have $\{0\}$ in the intersection.
@PeterTamaroff This is a very easy one! The easiest actually.
@JonasTeuwen You're considering $0$ to be a function, right?
Aug 4, 2012 01:36
Uh, wait, no. That would mean there is a set of functions that does not converge to $0$. Otherwise the bloody thing would not be open.
$\{f=0\}$
So, that means there is an $\epsilon_n > 0$ and a finite set $S_n$ such that $G_n \supset \{f : |f(x)| < \epsilon_n \text{ and } x \in S_n\}$.
For each $n$. So set $g(x) = 0$ for all $x$ in the $\cup S_n$ and $0$ otherwise. This will be in the intersection and is not $0$ hence not metrizable. Done.
@PeterTamaroff Yes.
@JonasTeuwen I need to assimilate a little about function spaces.
@PeterTamaroff I hope I did not mess up. Anyway, it is called "topology of pointwise convergence" if you look it up you will probably find this too. It is also called "weak topology".
Good night! It is 20 to 4 AM 8-).
@JonasTeuwen Dude, sleep!
Aug 4, 2012 01:40
Yeah, something like that. Bye.
@JonasTeuwen Bye.
 
Conversation ended Aug 4, 2012 at 1:41.