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9:04 PM
@leo $f(x)=\dfrac{e^{\frac1{d(K,x)-m}}}{e^{\frac1{d(K,x)-m}}+e^{-\frac1{d(K,x)}}}$ when $x\not\in K$ and $d(K,x)< m$ where $m=d\left(K,U^C\right)$
@robjohn I see. But then the previous parts of the exercise are useless here
@leo Oops, did I do this too quickly?
@robjohn no. I think I should post this on main so you can understand
@leo What am I missing?
@robjohn no nothing. This works just find, the problem is that it seems like if the actual exercise I'm trying to solve is designed to proof the $C^\infty$ version of Urysohn Lemma by other approach
9:11 PM
This only does it when one of the sets is compact...
@robjohn yes
I have that
@leo But it can be extended by looking at bigger and bigger compact sets in $\mathbb{R}^n$
@AymanHourieh Done!
Wanna see it?
@PeterTamaroff Sure.
@AymanHourieh Take $x=0$ and $y=x+h$, but $f(y)\neq 0$, that is, $h\neq 2\pi k$, $k\in \Bbb N$. Let $n\in \Bbb N$, and write $x'=x+2\pi n$ and $y=y+2\pi n$. Then $|x'-y'|=|x-y|=h$ and $|f(x')-f(y')|=|2\pi n\sin h+h\sin h|>2\pi n|\sin h|$. Then, given any $\epsilon >0$ we can take $h\neq 0$ but sufficiently small such that $|x'-y'|=h<\delta$ but take $n$ large enough such that $2\pi n|\sin h|>\epsilon$ since $|\sin h|>0$ for $h$ was taken to be $\neq 0$ and $\neq 2\pi k$.
@AymanHourieh What is the veredict?
9:23 PM
$h = \pi \neq 2\pi k$ can also be problematic. Looks good otherwise.
@AymanHourieh Hmm, yes, I mean that =D
@AymanHourieh I meant such that the function doesn't evaluate to zero.
So I should've said $h\neq k\pi$
Yeah, you can even restrict it to $h < \frac{\pi}{2}$.
@AymanHourieh I have another exercise now, which I suspect is about $\limsup$, but the author uses other words.
@PeterTamaroff Upper limit? Outer limit?
He says "a number $x$ is called almost an upper bound of $A$ if there exists only a finite number of elements of $A$ for which $y\geq x$. Similarily we define almost a lower bound".
Suppose $A$ is an infinite bounded set. Prove the set $B$ of all almost upper bounds of $A$ is nonempty and bounded below.
We then define $\overline{\lim}A=\inf B$, and call it the limit superior of $A$.
9:29 PM
That's not really equivalent to lim sup
@PeterTamaroff Seems applicable to any set, not just sequences. First time I see it.
@EdGorcenski Sure?
This is Spivak talking!
Actually, I said that before your second line showed up. You closed your quotes so I thought you were done.
@EdGorcenski Oh, OK.
I still think that's strange... at first glance it suggests that the supremum of all subsequential limits of $A$ is less than some elements in $A$.
If... by limit superior he means $\limsup$
9:32 PM
@PeterTamaroff Actually, yes, you can prove that $\limsup$ is the only number with this property.
This is theorem 3.17 in Baby Rudin
@AymanHourieh With what property? The least of all the almost upper bounds of $A$?
@AymanHourieh Let me see.
Though, I'm thinking of increasing sequences... it makes much more sense for decreasing sequences
@AymanHourieh It is only the very worded definition that is a little baffling.
If you have a monotonically decreasing sequence, then every $a_n$ is an almost upper bound... the limit of any subsequence is smaller than that $a_n$
Of course, the sequence need not be monotonically decreasing... but I believe it may not be monotonically increasing.
@PeterTamaroff Baby Rudin also uses a non-constructive approach to $\limsup$. One can prove that both definitions are equivalent.
9:35 PM
I like Rudin's approach to limsup. I find his exclusive use of limsup in the ratio/root test definitions infuriating.
I mean it makes sense, it just seems to introduce an extra step
Rudin does not even call it limit superior. He calls it upper limit!
Rudin's a strange guy...
In Big Rudin, he uses the constructive approach to $\limsup$!
I like his treatment of it though; it makes more sense to look at $\limsup$ in a "nested" way
I really enjoy his books though. Still studying through Big Rudin.
9:40 PM
Grandfather Rudin is good too!
I didn't understand $\limsup$ at all the first time I encountered it
Hi @meand!
@WillHunting Hello Willie!!!
I saw that again you were the victim of the moving target!
@WillHunting yeah...
9:42 PM
@MeAndMath There are usually three victims. A wants to ping B but pings C instead. So the victims are ABC.
@WillHunting Wanted help in something,but it is so simple that I'm embarrassed...
@MeAndMath I only know very simple stuff for now. I forgot everything!
@WillHunting funny!
@MeAndMath You spelled "embarrass" correctly with double r and double s, very good!
@MeAndMath Hit me with your best shot.
9:45 PM
@WillHunting A star for me!
@MeAndMath A smiley only. Don't get pretentious.
@MeAndMath Did you see the Musical (or Movie) "Rock of Ages"?
Those are both interpreted there.
@PeterTamaroff not yet
@meandmath You know, I usually don't know slang terms, but I just looked up Urban Dictionary and discovered the meaning of "willie"!
9:49 PM
@MeAndMath You should. I was so ecstatic I hugged the main character when the play finished.
@PeterTamaroff Pedro is so emo.
@WillHunting That's why we all crack up when we see "Willy Wong"!
@WillHunting Wait, what?
@PeterTamaroff You mean Willie.
@WillHunting I don't know these terms either.That's why sometimes I may offend people without inttention
@WillHunting Potato Pohtato
9:51 PM
@PeterTamaroff Nothing, hug=emo.
Hi everyone, I am a new user on the site. I would like advice on whether to post the link of a question I had asked some time back. It has not received any answers yet. I have read the chat room etiquette post about that. So pls advise.
@WillHunting so,what's Willie?
@WillHunting Huh?
@PeterTamaroff Well, just short for "emotional", not a bad thing.
@WillHunting I know that it means, but I don't see how it relates here.
9:52 PM
@MeAndMath Erm, are you trolling me?
@PeterTamaroff I just make random associations all the time. Someone who hugs is emotional. QED.
@rrampage go ahead; it's unlikely that chatters will be interested in your particular question however
@WillHunting No,I am not.Something people must know:I'm not funny at all.
@rrampage Just ask.
@MeAndMath Erm, I am scared to write the meaning here. You may look it up yourself...
9:53 PM
@MeAndMath It means "dong".
@PeterTamaroff HAHAHA
@WillHunting Oh dear...
@WillHunting such kind name...people are twisted...
9:55 PM
@MeAndMath Well, nothing bad, it's natural to think about such things...
@rrampage What does it mean for a positive integer to measure other numbers?
@anon , imagine measuring 10 kg with 2 weights of 5 and 15 kg. Its analogous to that. Basically, if a and b are weights, we can measure a+b and a-b as well
@anon think of it as a physical weighing balance.
@WillHunting people only think about that...
@MeAndMath Well, and math too of course!
@rrampage So basically "weighing" means "writing as the sum of other integers."?
10:03 PM
@anon yes
I am not Peter..
@PeterTamaroff Solved the drinking problem.
@anon sorry.
@JonasTeuwen Do tell.
10:04 PM
It is solved.
@JonasTeuwen What is the solution?
Must be a whisky!
@JonasTeuwen Glad you're back!!!How are you?
@JonasTeuwen You should get some Artie Shaw with that.
Carioca, maybe.
10:11 PM
@MeAndMath Uh, well. Fuqdup?
@robjohn, this is my problem
@JonasTeuwen Why?
@leo Yes, easy.
Even drunk.
@JonasTeuwen excuse me,what's fuqdup?
@MeAndMath I don't know.
10:14 PM
@JonasTeuwen 1.-4. are easy. prov 5. by using the others: don't know
@MeAndMath "Fuq'd up"
It is probably Klingon for "In trouble."
@PeterTamaroff yeah ...probably..
Haha, I saw a cool letter from the tax service to one of my colleagues... Apparently somebody before him registered as "Master Yoda".
I wonder how he did that...
@JonasTeuwen He clearly used the force.
@PeterTamaroff jIyajbe'
10:18 PM
"We need Master Yoda to sign this otherwise we cannot... wait wtf?"
@MeAndMath LAWL Babylon has a Klingon to English dicc.
Babel fish could exist...
@MeAndMath ¿
@PeterTamaroff you never read hitchhker's guide to the galaxy?
@MeAndMath No. Highways are good enough for me.
10:23 PM
The Hitchhiker's Guide to the Galaxy is a science fiction comedy series created by Douglas Adams. Originally a radio comedy broadcast on BBC Radio 4 in 1978, it was later adapted to other formats, and over several years it gradually became an international multi-media phenomenon. Adaptations have included stage shows, a "trilogy" of five books published between 1979 and 1992, a sixth novel penned by Eoin Colfer in 2009, a 1981 TV series, a 1984 computer game, and three series of three-part comic book adaptations of the first three novels published by DC Comics between 1993 and 1996. T...
@MeAndMath Hitchhiker's Guide is the single best piece of sci-fi comedy I have read. If it weren't for the chance of getting banned, I would answer every question with 42
@rrampage AMAZING!
@rrampage what's the name of the poem no one likes to hear?
@MeAndMath : Oh freddled gruntbuggly/thy micturations are to me/As plurdled gabbleblotchits on a lurgid bee.
Groop I implore thee, my foonting turlingdromes. And hooptiously drangle me with crinkly bindlewurdles,
Or I will rend thee in the gobberwarts with my blurglecruncheon, see if I don't!
remembered Vogons. searched wiki for it
@rrampage hahahahahahahahahha!
@MeAndMath have you read all the books in the series?
10:29 PM
first only..:(
@MeAndMath oh.. you must the read 2,3,4. Have read them. The 3rd book is especially good.
@rrampage I will!
I did not have the time yet...
10:50 PM
@JonasTeuwen Could you help me?
11:25 PM
Wow, someone copied a comment of mine and tried to post it as an answer.
@EdGorcenski Dude.
@EdGorcenski I'm trying to prove something with a seemingly difficult idea.
What's that?
@EdGorcenski Well.
@EdGorcenski Recall what we just talked about the "almost upper bound".
Now, let $A$ be a bounded infinite subset of the reals.
Let $B$ be the set of all almost upper bounds of $A$. I want to prove that $B$ is nonempty and bounded below. The first one is easy.
Since $A$ is bounded, then there is a number $\mu$ such that $x\leq \mu$ when $x\in A$.
So $\mu$ will be in $B$ either because the last inequality holds vacuously, or because $\mu \in A$.
It is beer o clock and I have not yet had a beer. This might take a second
$A$ is bounded
@EdGorcenski Yes.
11:36 PM
$B \subset A$. There is a $y \in \Bbb R$ such that $y \le a$ for all $a \in A$. Therefore, $y \le b$ for all $b \in B \subset A$.
actually, we have to refine that $B \subset A$ comment, because it's not true
@EdGorcenski I just proved $B$ contains all upper bounds of $A$.
but, it is true that by definition for all $b \in B$ there are infinitely many $a \in A$ such that $a < b$.
@EdGorcenski Yes, yes.
Right, so $B$ is bounded below.
There are infinitely many $a \in A$ that are lower bounds of $B$.
@EdGorcenski Well, I need to prove that there is at least one. How do you prove that?
11:39 PM
By the definition of an almost upper bound. $b$ is an almost upper bound if there are only finitely many $a$ such that $a \ge b$. Therefore, there are infinitely many $a'$ such that $a' < b$.
So if $B$ nonempty, then it is necessarily bounded from below.
I love when the MathJax preview crashes my edit
@EdGorcenski How long is this particular edit?
@EdGorcenski I think I can just use the lower bounds of $A$ to bound $B$ from below.
It was about 60 seconds worth of work
@PeterTamaroff That's what I was getting at.
No element of $B$ can be smaller than the $\inf A$
@EdGorcenski Tsk, tsk. I've taken to typing in a text editor and pasting into the text box at regular intervals, when typing an Arturo-length answer.
Thankfully, it was not that much work.
Blegh, I don't want to do homework tonight. I want to take a painkiller and rest my knee.
Sadly, I have no good painkillers and a surplus of homework.
11:56 PM
@EdGorcenski The nearest drugstore is far, far away?
The best stuff I have reduces the sharp pain to a dull throb. Anything better requires a prescription, which I don't have.
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