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12:00 AM
Then I prove that if $C⊂X$ is closed $\iff f(C)⊂Y$ is closed then $g$ is continuous.
 
Then you showed $g$ is continuous in the reverse implication
You should see instantly that $f$ is continuous
Ahh I'm typing with my pinkies
 
@HenryT.Horton Well, the problem is I cannot.
 
@HenryT.Horton only your pinkies?
 
I'll go back to peeling apples-
 
@BillDubuque The trouble is that internetizens are an ungrateful lot. If the site gets hosed somehow, somebody needs to be working on getting it fixed within 5 minutes, or the users will eventually migrate elsewhere. That means either a round-the-clock operations staff or someone on call -- and neither is something you get from volunteer labor, or from some university volunteering the service of their computing dept.
 
12:01 AM
Tell me what it means for $f$ to be continuous in terms of closed sets
yes I'm eating right now
 
@HenryT.Horton $f$ is continuous $\iff$ for every closed subset $F$ in $Y$, $f^{-1}(F)$ is a closed subset of $X$.
 
@HenningMakholm "The trouble is that internetizens are an ungrateful lot." - hell, yes. "Waah, it's down, please fix it now plzzzz!!!1!"
 
It is pretty funny if the government has to hire an external IT company because their own people are too incompetent 8-).
 
bbl (30 mins)
 
People with PhDs and such in CS. While that company even has people without a high school degree.
 
12:03 AM
@PeterTamaroff And based on your assumption for this implication, what can we say about $F \subset Y$ if it is closed?
 
Long haired barefoot walkers. With beard sometimes!
 
@JonasTeuwen There is some law of nature that government IT is always either totally fucked up or at least ten times expensive as it ought to be.
 
@HenryT.Horton You mean $F\subset X$?
 
@HenryT.Horton That it is a $f^{-1}(N)$ for some $N$ in $Y$?
 
12:06 AM
Well
 
@HenningMakholm I know many people at that company... The government their department with >200 people fuqed up their security certificates completely (blacklisted...) and they "fixed" it with only a couple of people 8-).
 
$f(f^{-1}(F)) = F$ is closed in $Y$
So what does that mean for $f^{-1}(F)$ in $X$
 
Apparently the secret service does not trust their own computer guys as they seem to be a large customer... (if you work there you have to through their security survey which can be the same as the one for the prime minister...).
 
@HenryT.Horton Right. I was writing it the other way around. I was writing $f^{-1}(f(F))$
Perfect.
 
@HenningMakholm Ten, huh?
 
12:08 AM
@JonasTeuwen I think it's because the decisionmakers don't understand what they are paying for, or exactly why having it is a good thing, and consequently can't make proper decisions about what needs to work, what would be nice to have, and what is not needed at all.
@J.M. Sorry, not ten, 10. In some appropriate base.
3
 
:D
 
@HenningMakholm Yea, well, they did this after Diginotar. Was quite funny. The certificates were revoked and blacklisted and the minister was stating on the television: "just click ignore, and it will work!"
And then you have the FoxIT guy in the media who basically said: wa da fuq?
 
@HenryT.Horton Thanks.
 
@JonasTeuwen Well, web PKI is irredeemably fucked up anyway. Too many strange little CAs that everyone trust. But no, it doesn't sound good.
 
@HenningMakholm Yep, but that still makes the response quite bad. About two hours later the response was completely different.
What are the "transferable skills" you gain by doing a math PhD other than "I know how to type"?
 
12:16 AM
"I know how not to panic when I'm faced with a problem that there are no known solutions to".
 
Hmm. "I don't have a mental breakdown anymore when I'm faced with a problem <...snip...>" would be more appropriate.
 
@HenryT.Horton Let $F=f(f^{-1}(F))=\subset Y$ be closed. Then $f^{-1}(F)$ is closed, so that $f$ is continuous.
 
"I know how to resolve hopeless dilemmas by redefining the boundary conditions such that the stupid approach we tried to improve for months actually works as-is".
 
Hmm "I don't mind feeling useless and doing boring repetitive tasks". Yeah, sounds good.
 
@PeterTamaroff Feels like a weird way to say it, homie
 
12:19 AM
@HenryT.Horton Just spell it out for me. I give up.
 
"I know how to spend weeks and months understanding something horribly complex that mere mortals would run away from screaming -- and then go in and do something about it".
 
I am still surprised when I see a guy with a math PhD... unzip like ~100 .zip documents which are in the same folder and have a common part in their name manually one-by-one. Why does it not occur to somebody like this that that... could be automated?
@HenningMakholm You mean "cylindrically archive it"?
 
@JonasTeuwen Oh, that's because it's a math Ph.D. and not a CS one.
 
Now I have to prove that given a metric space $(X,d)$ and a subspace $(Y,d')$, $F'/O'\subset Y$ is closed (open) $\iff$ there exist a closed (open) subset $F \;(O)$ in $(X,d)$ such that $F' (\; O')=Y\cap F(\;O)$.
 
Mmm... I do know some CS PhD guys who I could expect of doing such a thing. But they study stuff like regex.
 
12:22 AM
Let $F \subseteq Y$ be closed. In order to show that $f$ is continuous, we need to show that $f^{-1}(F)$ is closed in $X$. But $f(f^{-1}(F))$ is closed in $Y$, so by hypothesis $f^{-1}(F)$ is closed in $X$.
 
Well... then it is even worse. Maybe *.zip is too simple a regex.
That math guy also logs in as a root on his Linux machine. The sysadmin quarantined his machine 8-). So now he uses windows instead, where he can just do this.
 
12:41 AM
@HenryT.Horton Hey shawty rite there
 
Shake dat thang gurl, get dirrty
Well this is an interesting ad
 
@HenryT.Horton I told peter he is over complicating this new problem
 
Which new problem
 
@HenryT.Horton The one above on open and closed sets and continuous functions and inverse fucntions
@HenryT.Horton And why are we talking like rappers
Ok let us reassign
You be P. Diddy
and I'll be ludacris
@HenryT.Horton Or you wanna be Nelly
 
I just wanna be who you want me to be
 
12:48 AM
@HenryT.Horton I need to go take a shower then I'm going shopping :D :D :D :D
 
I always need to go take a shower
2
 
@HenryT.Horton :D
@HenryT.Horton do you like comm algebra?
 
I saw
We have our very own edition in Kings Cross
@RagibZaman I'm going to Paddys today
and then back to canberra tomorrow
 
To compare what happens in Kings Cross to that could be considered uninformed and insulted.
insulting*
 
12:56 AM
@BenjaLim I don't know
I like homological algebra, does that count
 
@BenjaLim Capitalist swine.
@RagibZaman I think he was testing some psychotics on himself.
 
Peter your new question doesn't seem clear
Are $F$ and $F'$ maps? What do you mean with $F'/O'$?
 
@PeterTamaroff Well he must have been doing so continually for quite a long time (as opposed to a one off trial like the guy who ask another mans face in florida who was on 'bath salts') because this is definitely premeditated - he purchased his materials over a span of 6 months and booby trapped his house for police.
 
@RagibZaman "Bath salts"... I sure haven't heard of that euphemism before.
 
@J.M. I hadn't either, its quite new I think, the drugs are called 'designer drugs' because they are made carefully from chemical procedures (not extracted from anything natural).
 
1:04 AM
I only do organic drugs, none of this bath salt stuff for me
 
@RagibZaman We've always had designer drugs... :) There's always been an entire community dedicated to molecular tweaking. Here, just a simple substitution can radically affect how it behaves in the brain.
The new part for me is that they have new euphemisms for these things. :D
 
@J.M. Oh I didn't know about that. I read an article where it said the abuse of designer drugs was relatively recent.
And that it was hard to legislate against them because if they ban a certain chemical, you can just slightly change the molecular structure to something that isn't illegal.
 
@RagibZaman "if they ban a certain chemical, you can just slightly change the molecular structure to something that isn't illegal." - precisely the impetus. So, you have some of these countries that write laws saying "molecules with this or that group cannot be synthesized without a license". Which can sometimes be overly broad as to be inconvenient for those who have less clandestine motives.
@RagibZaman Yeah, the things going mainstream is relatively new. Ecstasy is the classical example.
(I'm a chemist, so I keep abreast of these things...)
 
@J.M. Ahh I see. Wow it was hard to guess what you did! You know a lot from so many different areas.
 
@JM Ah a chemist! Of what?
 
1:13 AM
@JonasTeuwen At one point I was an (X)Emacs developer.
 
@RagibZaman The math is just a hobby... :)
 
@BillDubuque Kickass!
 
@JonasTeuwen Didn't I tell you before? :)
 
@J.M. Can't remember... my memory fails me (usually not).
 
@J.M. Tis true at this point! (for me)
 
1:16 AM
@JonasTeuwen I work with pharmaceuticals, currently. So, I always hear drug-chatter (conventional or clandestine)...
 
@J.M. Ahhh! Well, I thought you were just a very good "amateur" (as in mathematics). I tend to know a bit about different thingies too.
 
@BillDubuque Yeah, it's been a pretty rewarding hobby... :) I'm sure it's the same for you.
@JonasTeuwen It's the unfortunate consequence of going through so many jobs... :)
 
@J.M. At first I thought your reply was a joking followup to Jonas' kickass comment. For me math has always been part pro and part hobby and everywhere in-between. The diversity is nice.
 
@JM Computing integrals by cutting paper eh?
 
@JonasTeuwen I was way less mathematically enlightened in those days. :D (I think I said as much in that question, too.)
 
1:19 AM
@JM If your are a pharmacist, does the "magnifying" effect of stuff like -xetine together with -idone have to do with the fact that both are metabolized with CYP2D6?
 
@JonasTeuwen (My interest in numerics started a number of months after that.)
 
@JM So you are a self-sufficient intellectual :-).
 
@JonasTeuwen Sure; if you knock out that particular liver enzyme, there's nothing left to process the other one... so, don't take both at once.
 
@HenryT.Horton It is just two cases written out in one.
 
@JM Yes, so, that is the interaction, and not something on neuron level or whatever. Might not be too selective stuff but selective enough not to do that? :-).
 
1:21 AM
@HenryT.Horton Given a metric space $(X,d)$ and a subspace $(Y,d')$, $F'\subset Y$ is closed (open) $\iff$ there exist a closed (open) subset $F$ in $(X,d)$ such that $F'=Y\cap F$.
 
@PeterTamaroff Oh, $F$ and $F'$ are closed sets
 
@JM But it makes whisky work better.
 
Horrible notation :)
 
@HenryT.Horton Come at me bro.
 
@JonasTeuwen Maybe they also interact in the brain, but I'm not updated on the cutting-edge research with SSRIs. (I have a number of papers on my to-read list...)
@JonasTeuwen Heh.
 
1:22 AM
They are extremely good at "FOKITOL".
FOKITAL = Haloperidol & Clomipramine?
 
@BillDubuque Not at all. The nice thing about this site is that in RL, I don't really have anybody else to talk about this hobby with, so this site serves me splendidly.
 
@PeterTamaroff What are your ideas here, son? And $d'$ is just $d$ restricted to $Y$, right?
 
@JonasTeuwen Oh, this? The formulation is proprietary, I'm told... :D
@JonasTeuwen but the Haldol is certainly a knockout.
 
Would probably make a pretty good substitute. Add some biperiden (take >20mg of the first).
 
@HenryT.Horton Yeah. No ideas so far, pops. But I will try and think about it.
 
1:25 AM
@JM You can actually spot the Haldol zombies from meters away!
 
@JonasTeuwen It dries your mouth, lips, and eyes, which is why a few people aren't fond of it...
 
Not the extrapiramidal stuff? That sucks the most.
Hmm, or wait, that is part of it right?
 
The shakes are worse. Thankfully, not that common.
 
Prof. Scott just answeres a question of mine.
How I feel:
 
@PeterTamaroff I'm still wondering how ponies and rainbows ever got associated...
 
1:27 AM
@JM Akathisia. That. Sucks. The. Most. Ever.
 
@J.M. In case you didn't know, ponies eat rainbows.
 
@JonasTeuwen Common SSRI problem. Drives people nuts, or off a ledge.
@PeterTamaroff Really? My source said that they excrete rainbows, not eat them... :)
 
@JM SSRI? Really? Not AP? I thought it was a good idea for a shrink to give every whiner some nice atypical to experience this and know what real misery is.
 
@J.M. That's the most wonderful part of it all: the rainbows are preserved.
 
@JonasTeuwen Most of the cases nowadays are SSRI-triggered, but atypicals can do that too. That's why you have some of these Americans suing the companies after somebody they know went kaput from trying to get rid of the akathisia. For them, it was sucky to trade depression for akathisia...
@PeterTamaroff Hmm, I guess you're right.
@JonasTeuwen ...and it really is messy to clean up after a guy that went *splat*.
 
1:31 AM
@JM Oh. I knew that it was also caused by SSRIs, but thought it was more common with our other friends. Perhaps because SSRIs are more commonly used?
 
@JonasTeuwen Yes to your last question.
They get pushed a lot to docs, see...
 
I have seen people with much worse akathisia from AP than from SSRIs.
You feel the uncontrollable urge to move. But moving does not help to get rid of the urge.
 
@JonasTeuwen Oh, certainly. Atypicals are even more nasty in the grand scheme, but thankfully people on atypicals aren't that many, relative to SSRI users.
 
But soon everybody will be bipolar if we can follow the doom stories about DSM-V 8-))).
 
Hah.
 
1:36 AM
In any case: it is ridiculously late. Good night guys. Good night @JM. (or whatever time it is there). I can hear cute birds whistle.
 
See you! (It's a rainy morning where I am.)
 
2:07 AM
@PeterTamaroff Are you doing your work?
 
Hi folks
 
H... hello
 
Hi Henry
 
@HenryT.Horton Not really.
 
@M.B.M. How did you know my name!?
 
2:10 AM
@HenryT.Horton well...
 
@HenryT.Horton OK, I'll start now.
 
@HenryT.Horton your "How did you know my name!?" question brought a very distant memory into my mind.... i dunno if you've ever used AOL Intant Messenger.... but when I used it in college there were these spambots that would send you messages like "Hi <your AIM nickname>", etc.... Sometimes i had fun with them by replying with the question you've asked
 
@M.B.M. I assume you're Marvin the (Bad) Martian.
 
of course, most of the replies were boring... but there were a couple that were clever
 
I occasionally get a bot IMing me on AIM still
 
2:15 AM
@HenryT.Horton Clever Bot is clever.
 
@PeterTamaroff No, though that's a very creative interpretation
I like it
I no longer use AIM
 
@HenryT.Horton Could I argue by contradiction to make the proofs?
 
Go all out: split it into a million cases, each of which using contradiction-within-contradiction.
 
So, I have a really dumb question (I don't think it's worth writing it on the site)... Does unitary operation preserve positive-definiteness? That is, if U is a unitary matrix, and A is positive-definite, is UAU* positive-definite?
I'm pretty sure it does, but it's late and my brain is very tired...
the reason i think it does is: A is positive definite iff x*Ax > 0, where x is an arbitrary non-zero vector and * is conjugate transpose
x*U yields just another arbitrary vector, whose conjugate transpose is U*x
so x*UAU*x>0.... does that make sense in a correct way?
 
@M.B.M. Yes.
 
2:30 AM
Similarity transforms in general will preserve positive definiteness
Since they preserve eigenvalues
 
Oh -- I didn't know the term for that!
Now, is there a name for a transformation of the form PAP*, where P isn't unitary (that is P* is not an inverse of P)?
 
@M.B.M. It's a congruence transformation.
It doesn't preserve eigenvalues, but it preserves inertia.
 
seems like that would preserve positive-definiteness (except in a perverse case of P being all-zero matrix)
 
(and thus it also preserves definiteness)
 
Thanks, @J.M.! Is inertia the "distribution of the eigenvalues"?
 
2:35 AM
@M.B.M. Nope, it's a count of the number of positive, zero, and negative eigenvalues.
(Look up Sylvester's inertia theorem.)
 
aha!
yup, just found it
very interesting
 
@J.M. I have not received the postal mail from Don Knuth yet.
 
Corrected an error?
 
@FrankScience That takes time. He has to verify it, of course.
 
@FrankScience Just curious, which volume?
 
3:02 AM
@J.M. Oh, no. He finished because his assistant wrote to me, asked me about my postal mailing address and said that she would send Don Knuth's letter to me.
@M.B.M. Not The Art of Computer Programming.
@CLarue In fact, there's no error I found but my mistake.
 
@FrankScience I see. Then, it's just the postal system going at a turtle's pace, then.
 
@HenryT.Horton Oh, gawd, I just wasted so much time in 9GAG.
 
I have to at least prove one.
Do you recall the theorem I have to prove?
 
3:12 AM
I need to prove that an open subset in $Y$ is the interesction of an open subset of $X$ and $Y$ itself.
 
Yes you do
 
I really don't know where to start.
Because an open subset in $Y$ is also an open subset in $X$:
 
No it isn't
$X = \Bbb R$, $Y = [-1,1]$, $U = (0,1]$
$U$ is open in $Y$ but not $X$
But $U = Y \cap (0,2)$
 
Right. OK.
 
I'm going to go home though... goodbye, forever.
 
3:18 AM
@HenryT.Horton Oh, noes.
Bye.
 
Hi folks
 
@RajeshD Hi
@HenryT.Horton I think I got a bit of it
Assume $O'\subset Y$ and there is an open $O\subset X$ such that $O'=O\cap Y$.
Since both $Y$ and $O$ are open, so is $Y\cap O=O'$.
@RagibZaman Are you around?
 
3:35 AM
komusta
 
@BenjaLim Bennnnnnnn
 
yes
 
How are you today?
 
@PeterTamaroff I just had yum cha
 
@BenjaLim Was it good?
 
3:37 AM
yes!!!!!!
my favourite
do you know what it is?
have you been to a yum cha restaurant in BA
 
woo dimsum
 
@BenjaLim Nope. No idea.
 
@PeterTamaroff you should try it
@RajeshD yes I love it
 
@BenjaLim I'll se if I can!
 
do you eat chinese food much?
 
3:39 AM
I'm trying to prove that if $Y$ is a subspace of the metric space $X$ then $O'\subseteq Y$ is open $\iff$ there exists an open subset $O\subseteq X$ such that $O'=Y\cap O$.
 
@BenjaLim Not really.
 
what do you usually eat? We'll tackle that problem afterwards
 
@BenjaLim As an example, today woke up at 12 am
Ate toasts and coffee.
 
ok
lunch what do you eat?
 
3:40 AM
Then at around 4 I ate some Milanesas.
 
you should try machaca
my friend ricardo sneaked it through customs
really good
 
then I drank more coffee and had some cereals, and then had a tuna and cheese sanwhich.
 
I just finished eating Pongal with mango pickle for BF
 
@PeterTamaroff By the way for your problem in topology actually that is the definition of the subspace topology
@RajeshD sweet yes?
 
@BenjaLim $$\Huge \text{ Metric spaces }$$
 
3:42 AM
no spicy
 
@RajeshD I have had it sweet
Ok suppose you have $U$ open in $X$
you want to prove $U \cap Y$ is open in $Y$ yes?
 
@BenjaLim I already proved that
 
Ok
 
The intersection of open sets is open, right?
 
be careful
now you are dealing with subspaces and stuff
you want to prove that if $U$ is open in $X$ then $U \cap Y$ is open in $Y$
 
3:44 AM
Right,
 
Take $x \in U \cap Y$
 
SO I have to prove $U\cap Y$ is a nbhd of all of its points,
 
there exists $\epsilon > 0$ such that $B_\epsilon(x) \subseteq U$ yes?
 
no $x \in U \cap Y$
 
3:47 AM
@BenjaLim Oh, sorry. Didn'tread that
 
but you see because $x \in U \cap Y \subseteq Y$
we can view $x$ as lying in $Y$
 
@BenjaLim Right,
 
actually when we chose $x \in U \cap Y$ we are supposing it is not empty
but if it is then there is no problem a priori :D
 
@BenjaLim Right, because $\varnothing$ is open.
 
in whatever metric space
Right how to proceed now ? @PeterTamaroff What do you think you should do now?
 
3:50 AM
@BenjaLim So I'm writing: Assume $O\subset X$ is open in $X$. We must prove $O'=O\cap Y$ is open in $Y$. If $O\cap Y=\varnothing$, we're done, so wlog assume the intersection is not empty.
 
yes
 
Since $O\cap Y\subseteq Y$, picking $x\in O\cap Y$ means $x\in Y$.
 
ok
 
(Now I must prove for any arbitrary picking of $x$, there is a nbhd of $x$ in $O\cap Y$,)
@BenjaLim Should I prove that $Y\cap O\subseteq Y$?
I mean, it is clear, but...
 
No
 
3:53 AM
@BenjaLim No to what?
 
@PeterTamaroff ok
let me simplify your problem
 
I mean $x\in O$ is just another constraint to $x\in Y$...
 
you know that in a metric space, the balls generate the open sets yes?
 
@BenjaLim Waittttt
 
So it suffices to prove your problem for balls
 
3:54 AM
@BenjaLim Yes,
 
So now $Y \cap B_\epsilon(x)$
is this open in $Y$?
$Y \cap B_\epsilon(x) = \{y \in Y : d(x,y)<\epsilon \}$
 
But I should say it: in a metric space, the open balls generate a basis for the systems of neighborhoods in $X$.
 
yes
stop the word systems of neighborhoods
and use "the topology on $X$"
 
I can't. I haven't that defined!
 
Here is the definition of a topology. Let $X$ be a set. We call a collection $\tau$ of subsets of $X$ a topology on $X$ if (1) $\emptyset, X \in \tau$, (2) $\tau$ is closed under arbitrary unions , (3) $\tau$ is closed under finite intersections
 
3:58 AM
OK.
 
@PeterTamaroff ok forget it
now looking at the definition of $Y \cap B_\epsilon(x)$
is it not clear that this is open in $Y$?
 
$B(x;\epsilon)$ is a ball in $X$, right?
 
yes
 

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