Mathematics - 2015-02-02 (page 4 of 4)

Ted Shifrin

11:00 PM

I will give them prizes :P

Studentmath

11 weeks is a lot of time from my perspective

Ted Shifrin

well, @Studentmath, think about how many weeks are in 40 years.

Karlo

Why 2^N is finite set

Studentmath

A bit more

Ted Shifrin

Huh @Karlo? That looks like a number, not a set.

Studentmath

11:01 PM

Wait what, 2^N is a finite set?

Karlo

I mean if it is Decart

:D

gives up

it is wrong yes

What's decart?

but the statement

2^A is not countable is true

2^A means guys if A={0,1} then { 0,1,01,empy set}

Studentmath

11:03 PM

Oh. $P(A)$.

user112495

@TedShifrin So a subset of a metric space is bounded if it can be contained within a ball of radius r (>0) centred at a, where $a \in M$. If we equip M with the discrete metric (M could be $\mathbb{R}$), then I think this gives the same topology?

Studentmath

Well, as you already seem to know $P(A)$ has $2^A$ elements...

Or to be precise

$2^{|A|}$

Karlo

@Studentmath mhm

Mike Miller

@Studentmath One also might denote the set of maps $B \to A$ as $A^B$, hence the notation $2^A$ instead of $\mathcal P(A).

Ramanewbie

the left side of the screen is so green now...

Studentmath

11:05 PM

@MikeM Hah, I know the definition but never saw it like that, nice

Karlo

Can someone elabore why A x B (decart's products) is the same as f:A->B.Because decart's product will make the result and send it there?

Studentmath

@Karlo well, if $|A|$ is not finite, then $2^{|A|}$ is not finite..

Karlo

@Studentmath yes

evinda

@DanielFischer I will think about it... :/

Hallo @Alessandro

Alessandro

Hallo @evinda!

evinda

11:07 PM

Wie geht es dir?

Studentmath

That should answer your question regarding $2^N$. And what do you mean the same?

Ramanewbie

@evinda mir geht sehr gut -_- is that correct ?

evinda

@Ramanewbie Mir geht es sehr gut should it be :)

Karlo

"We often write instead f is a subset of A x B f:X->Y and we say a function is given"

Alessandro

Mir geht es auch sehr gut ^^ und dir?

evinda

11:09 PM

@Alessandro Ganz ok.. Wie läuft der Deutschunterricht?

Ilya_Gazman

@robjohn I made a progress and found a mathematical proof to my claim.

user112495

@TedShifrin The other example that's confusing me is the following.

We are given a list of properties of metric spaces, and for each one, we have to decide whether or not it is a topological property. One of them is:

$\forall x \in M \exists y \in M$ such that $d(x,y)=-1$. Our lecturer says that this is a topological property. But by the definition of a metric space, don't we have to have $d(x,y)\geq 0$?

Mike Miller

@user112495 This property is false for all $M$. Thus it being true is preserved under homeomorphism... something false for all spaces (or true for all spaces) is automatically a topological property. but this seems like an odd point to bring up.

Ramanewbie

@evinda ok... I haven't spoken German for a year

Alessandro

@evinda gut, vielleicht ein bisschen zu schnell, aber ich muss am 20. Mai die B1 Prüfung schreiben :)

evinda

11:13 PM

@Ramanewbie Did you learn German in school?

Ramanewbie

@evinda just a bit, yes

evinda

@Alessandro Bis welchen Level muss du die Prüfungen schreiben? Bis C2?

@Ramanewbie Gehst du noch zur Schule?

user112495

@MikeMiller Oh, okay :p. Thanks. If instead we had $d(x, y)=1$, then how would you go about showing that this isn't a topological property?

Ramanewbie

@evinda ya, ich bin 14 Yahre alt

Mike Miller

@user112495 Find a space for which it's true and a homeomorphic space for which it's false.

Alessandro

11:18 PM

@evinda nein, C1 (oder testDaF oder DHS) ist genug

Ted Shifrin

@user112495, do you think the discrete metric gives the usual topology on $\Bbb R$? If two points are close in one metric, are they close in the other — and vice versa?

user112495

@MikeMiller Oh, so I could take $(0, 1)$ and $(0, \infty)$ with the euclidean metric?

Mike Miller

@user112495 Yup!

Alessandro

@evinda you can actually apply to a university with a B2 certificate and do a DHS exam as part of the admittance process

Hippalectryon

@Ramanewbie ins Bett gehen jetzt !!!!

Ted Shifrin

11:20 PM

@Mike @user112495: Be careful, you were asking about the discrete metric ($d(x,y)=1$ when $x\ne y$).

evinda

@Ramanewbie Achso,bist ja noch jung!!!

@Hippalectryon @Ramanewbie Seid ihr Geschwister?

Ramanewbie

@hippa jajaja, don't get mad, I just came mad from a shower !

Ted Shifrin

LOL !!

Hippalectryon

@evinda Brüder

evinda

@Alessandro And will you do it like that?

Ramanewbie

11:21 PM

@evinda ja, Brüder

Ted Shifrin

tous les deux tout à fait méchants!!

evinda

@Hippalectryon Bist du der ältere?

Ramanewbie

@ted lol

user112495

@TedShifrin Yeah, this was a slightly different question.

Ramanewbie

@evinda er ist

Mike Miller

11:22 PM

Huh, @Ted? The question was "How do you show that ($\forall x \in M \exist y \in M$ s.t. $d(x,y) = 1$) is not a topological property?"

Hippalectryon

@evinda tatsächlich

evinda

@Hippalectryon Wie alt bist du?

Ted Shifrin

Das fängt noch an ...

Hippalectryon

@evinda 17

Ted Shifrin

oh, I thought he was still on the discrete topology, @Mike.

evinda

11:22 PM

@Hippalectryon Also gehst du auch noch zur Schule?

Alessandro

@evinda yes, I'm supposed to get my B1 the 24th of April and a B2 around the end of july (woops, I mixed up the dates, the 20th of May I'll begin the B2 course)

Hippalectryon

@evinda de.wikipedia.org/wiki/Classe_pr%C3%A9paratoire

evinda

@Alessandro So will you start studying at the university the next year?

@Hippalectryon Interessant!

Hippalectryon

@evinda En effet :D

user112495

@TedShifrin What is the usual topology on $\mathbb{R}$?

Ted Shifrin

11:25 PM

coming from the usual metric $d(x,y) = |x-y|$, @user112495

evinda

@Hippalectryon @Ramanewbie Wohnt ihr noch zusammen? :D

Hippalectryon

@Ramanewbie Sie sind müde ... müde ...

@evinda Wir sehen uns am Wochenende

Ted Shifrin

Du sagst deinem Brüder nicht "Sie" :P

Ramanewbie

@hippa ich bin müde und go to ned now

evinda

@Hippalectryon Hilfst du ihm in Mathe?

Ted Shifrin

11:26 PM

smacks @Ramenwb good night

user112495

@TedShifrin No, it won't then.

evinda

@Ramanewbie Gute Nacht :)

Karlo

If in a unconnected directed graph there are 2 roads which are not loops that have maximum length I have to prove they have common verticie.Solution:If they don't have common edge then we connect the both roads and we get new bigger path contradiction?

Hippalectryon

@evinda manchmal

Chris's sis

@robjohn How about starting from here $$k^2 \log(k-1)+k^2 \log(k+1)-2 k^2 \log(k)-1$$? The second term can be arranged such that we can get telescoping terms.

evinda

11:27 PM

@Hippalectryon Aha..

Ted Shifrin

LOL, I've forgotten which question you were answering, @user112495

Hippalectryon

amnesia

Ted Shifrin

too many twists and turns @Hippa, even for you

user112495

@TedShifrin haha :p. This is the 'do you think the discrete metric gives the usual topology on R? If two points are close in one metric, are they close in the other — and vice versa?' one.

Alessandro

@evinda wenn ich alle die Prüfungen bestehe dann werde ich am Oktober (Wintersemester) zu studieren anfangen (that sentence required a bit of improvisation, i'm not very sure about its correctness :P)

Ted Shifrin

11:28 PM

ah, good, @user112495

Hippalectryon

@TedShifrin Don't worry, I'm dead tired. I'm off.

Ted Shifrin

bonne nuit, @Hippa

Hippalectryon

:-)

evinda

@Alessandro Shall I improve it? :p

Karlo

can someone loot at what I asked :D

Ramanewbie

11:28 PM

@evinda danke

Alessandro

@evinda that'd be very nice :)

Chris's sis

@robjohn $$k^2 \log(k-1)+((k+2)^2-4k-4) \log(k+1)-2 k^2 \log(k)-1$$

evinda

Wenn ich alle Prüfungen bestehe, dann werde ich im Oktober (im Wintersemester ) anfangen zu studieren. @Alessandro

Alessandro

@evinda it wasn't too bad! I'll try to unravel the misteries of german grammar and understand why is anfangen before zu studieren tomorrow though, I'm too tired for that now!

evinda

@Alessandro Ok.. Sind die Prüfungen vom Goethe Institut?

Chris's sis

11:33 PM

@robjohn hmmm, here is an optimized idea $$((k-1)^2+2k-1) \log(k-1)+((k+1)^2-2k-1 )\log(k+1)-2 k^2 \log(k)-1$$ This is because of the terms $2k^2 \log(k)$ that we introduce it into 2 telescoping sums.

Karlo

Someone going to participate in Harvard-MIT tournament

?

Mike Miller

It seems likely that someone is going to participate in such a thing.

robjohn

@Chris'ssis Yes... I was looking at the telescoping sum. I should finish soon if it works.

Karlo

:D :D :D

Chris's sis

@robjohn OK. The really bad term there was that one. After taking care of it, all becomes an easy job.

user112495

11:35 PM

@TedShifrin I can't think of a different metric that would make $(0, \infty)$ bounded.

anon

can you think of a bijection between (0,inf) and (0,1)?

Chris's sis

@robjohn It works 100% (in a great way).

user112495

@anon I know that they're homeomorphic. But that doesn't mean (0, inf) can be bounded though does it? As boundedness isn't a topological property. Or am I understanding it incorrectly?

anon

what do you think "boundedness isn't a topological property" implies? is it relevant? does it say something about this?

Alessandro

@evinda Ich besuche einen Kurs in der ISL Sprachschule, sie ist auch ein lizenziertes B1/B2 Prüfungen Zentrum, aber ich glaube sie sind ähnlich wie die Prüfungen vom Goethe institut

anon

11:39 PM

if (0,1) and (0,inf) are in bijection, then just define the distance between things in the second interval to be the usual distance between the corresponding things in the first interval.

user112495

@anon If boundedness were a topological property, then if M was bounded, then every space homeomorphic to M would also be bounded

anon

correct

so if boundedness is not a topological property, then ...

user112495

@anon Oh, so it just means that it isn't the case that every space homeomorphic to it has the boundedness property?

anon

right

Alessandro

@evinda anyway, I'm going to sleep now, gute Nacht!

user112495

11:41 PM

@anon But that doesn't remove the possibility that some of the spaces homeomorphic to it are bounded.

evinda

@Alessandro Gute Nacht!!! :)

anon

@user112495 correct

did you read what I wrote?

3 mins ago, by anon

if (0,1) and (0,inf) are in bijection, then just define the distance between things in the second interval to be the usual distance between the corresponding things in the first interval.

for instance, (0,inf)->(0,1) via e^-x, so define d(x,y)=|e^(-x) - e^(-y)| on (0,inf). that makes (0,inf) bounded.

user112495

@anon Oh! Yep. That makes sense :D.

@anon What is an example of a set homeomorhpic to $(0, \infty)$ that isn't bounded then?

Mike Miller

$(0,1)$?

Axoren

@user112495 $(0, \infty)$ is :P

user112495

11:56 PM

@Axoren But haven't we just shown that we can find a metric under which it is bounded?

@MikeMiller Oh, so if we use a metric under which (0, inf) is bounded, then (0, 1) isn't and vice versa? At least for this metric anyway?

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