Mathematics - 2017-02-15 (page 4 of 5)

1:01 PM

its a question about whether the logic of 1+1=2 is true. Whenever the logic fails (such as in the case where 1+1 is true but 2 is not) the statement would not be true

DHMO

@user400188 hmm...

user400188

as far as I can tell; things like this can always be done.

and its one of the reasons logic was concived; becuase even if the things we were talking about were false; we could still say (ask) things "about" them.

DHMO

@user400188 is 1 true?

user400188

Depends on how1 is defined. if it is a tautology then yes.

DHMO

@user400188 is {} true?

user400188

1:06 PM

{} would have the truth values of how we defined it earlier

the statement of "is {} what we defined would be true"

or is it true that {} = that definition

DHMO

you have said it yourself

only a statement can be true or false

user400188

hmm?

yes

{} is a statement

DHMO

oh god

user400188

becuase its definition is a statement

and it is its definition

DHMO

so everything is true because everything is defined as it is defined?

user400188

1:09 PM

no. But the question: is everything defined as it is define true? Is true

anyway I think we are getting nowhere. So best drop the topic

DHMO

how is {} a statement?

user400188

also it appears that I am driving you insane

we defined {} earlier using a statement.

becuase we take {} to mean its definition it will be a statement too

DHMO

but {} itself is not that statement

that statement is a statement such that if we replace X by {}, it will be true

user400188

ah I see.

still it just means that you have given a broader definition to {}. To mean whatever you want it to be.

When you wish it to mean the definition of the statement we had earlier you just substitute it for X

DHMO

you can't substitute a statement by an object...

user400188

1:14 PM

what do you mean?

so if I have an object A, I cannot substitute A as B+C ?

DHMO

The empty set is an object X such that there is nothing in X

@user400188 B+C is not a statement

user400188

by + i meant OR

DHMO

B and C are not statements

BAYMAX

Sorry to interupt...Guys i am reading a research paper mainly on differential equations and their applications , its a MSc thesis , i want to make it to a research paper so that it would help me in admission to PhD programmes abroad...Please advise me in ths regard...i would be very much grateful to you for this ... i find many high level concepts or theory used...which sometimes demotivate me ... any help/guidance guys...

2

user400188

actualy I think I have a definition now for an object

Ill just call them atomic statements. Atomic statements are those which can't be broken down any further.

this way I wont be able to substitute things in place of them

DHMO

1:18 PM

*facepalm*

why do you still call them statements?

user400188

Ill just call them objects and drop the issue :/

hi guys

@ShaVuklia hello

@user400188 lol

if $\gcd (a,b)=\gcd(a,c)=1$, how can I show that $\gcd (a,bc)=1$?

or.. maybe just a hint is the best

DHMO

1:21 PM

35 mins ago, by DHMO

@user400188 now, if X is a set such that "$\forall x: x \in X \equiv x = \{\}$", what can X be?

Sha Vuklia

because i would like to solve it at least partly

DHMO

@ShaVuklia bezout's lemma?

Sha Vuklia

okay I'll have a try

Fawad

@DHMO so a,b,c are all primes?

DHMO

@Fawad no

user400188

1:23 PM

@DHMO X must be {}

DHMO

@user400188 why?

Sha Vuklia

ah, it's the contrapositive right

Can't I just say the following:

user400188

sorry it should be X is {{}}

becuase due to x been in X; X must have at least 1 pair of brackets. And becuase X$\equiv$ x; they must have same truth value. Which is fine becuase X contains x and nothing else. Now becuase {}=x, X must be {{}}

@DHMO

DHMO

@user400188 and we can go back to the question what is Y if X={{{}}}

1. u is an element of Y
2. there exists a z such that [(u is in z) and (z is in X)]
3. [u is an element of Y] if and only if (there exists a z such that [(u is in z) and (z is in X)])
4. for all u ([u is an element of Y] if and only if (there exists a z such that [(u is in z) and (z is in X)]))

Sha Vuklia

If a number divides $bc$, then it either divides $b$ or $c$ (or both). So the greatest divisor of $bc$ will be the $\max \{\text{divisor of }b, \text{divisor of }c\}$. In either case, we're basically looking at either $\gcd (a,b)$ or $\gcd (a,c)$, which equals 1.

I did not use Bezout's lemma though

DHMO

1:29 PM

@ShaVuklia the first line is incorrect; 10 divides 2x5 but it neither divides 2 nor 5

user400188

∀x:x∈X≡x={}", what can X be?
1. x is an element of X
2. x equals {}
3. x is an element of X if and only if x equals {}
4. for all x, x is an element of X if and only if x equals {}

Sha Vuklia

ah

DHMO

@user400188 what is this?

user400188

I was trying to apply the same method to the new question.

sorry I thought that was what you meant when you asked to go back to the Y question.

@Fawad what do you mean by 4?

Fawad

*sorry,@DHMO what's is this discussion in which topic?(topology?)

DHMO

1:33 PM

@Fawad set theory and logic

@user400188 did we confuse X with Y?

Sha Vuklia

I don't know how to use Bezout's lemma then.. It says if $a$ and $b$ are relatively prime, then $a\mid bc\implies a\mid c$. So here we have that $a$ and $b$ are relatively prime, and $a$ and $c$ are relatively prime. So $a\mid bc\implies a\mid c$, and $a\mid bc\implies a\mid b$. I don't see the next step though... :(

user400188

you asked a question ∀x:x∈X≡x={}", what can X be? I answered it but when you typed "@user400188 and we can go back to the question what is Y if X={{{}}}"
I thought you wanted me to use the same method I used to solve that old question to solve the new one that involved X

DHMO

@ShaVuklia I also don't. Let's use another method

Sha Vuklia

HahaXD

I've also learned the following

DHMO

@user400188 did we confuse {{}} with {{{}}}?

user400188

1:37 PM

I think we might have

what the question you want me to answer at this point?

Is it:
what is Y if X={{{}}}?

Sha Vuklia

$\gcd(a,b)=\Pi_pp^{\{ord_p(a),ord_p(b)\}} $

where $p$ is prime

could I use that? I already tried some things, but it didn't work for me

DHMO

> For all X (there exists a Y such that (for all u ([u is an element of Y] if and only if (there exists a z such that [(u is in z) and (z is in X)]))))

@user400188 remember the above?

user400188

The Axiom of Union

Yes

DHMO

so I am asking you, what is Y when X = {{{}}}

@user400188 sorry, be back later

user400188

1. u is an element of y so y will have at least 1 B (where B stands for a pair of brackets)
2. u is in z and z is in x. This means z will have at least 1 B and X will have 2.
Now since X has 3 pairs of brackets, this means u will have at least 1 pair to start with.
Which means that Y will have 2 pairs of brackets.

Thus Y is {{}}

@DHMO
Ok. Ill wait till your back

This will probaly have to be the last question however. I plan to go to sleep soon.

Semiclassical

1:48 PM

hi chat

skill patrol

hi

Fawad

Hi

Alessandro Codenotti

@ShaVuklia suppose $\gcd(a,bc)=n$ with $n>1$, that means $n|a$ and $n|bc$. Clearly $n\not|b$ and $n\not|c$ since $\gcd(a,b)=\gcd(a,c)=1$. Now work on the prime factorization of $n$ to get a contradiction

Semiclassical

@user400188 Something to think about if you've an interest in tautologies/paradoxes/etc:

user400188

@Semiclassical Hmm?

Semiclassical

1:50 PM

Consider the following infinite list of statements:

"S1: Every statement after this one is false.
S2: Every statement after this one is false.
..."

How would you assign truth values to that?

I saw some of your conversation from earlier and thought you might like a taste of the Yablo paradox :)

Fawad

@Semiclassical can I know what it means "how would you assign truth values to that" ?

Semiclassical

Well, the first statement asserts that all the statements after it are false. Can that statement be true?

skill patrol

not "all"

Fawad

First statement will be true (I think)

Semiclassical

Well, suppose the first statement is true.

user400188

1:53 PM

its not a question of if the first is true or not; but if it were true what would happen

and if it were false what would happen

Sha Vuklia

@AlessandroCodenotti Hi, thanks. I'll try it out!

Semiclassical

Heheheh. Follow me down the rabbit hole.

7

If the first statement is true, then every statement after it (S2,S3,...) is false. In particular, S2 is false.

But S2 being false means that at least one of the statements after it is not false.

Alessandro Codenotti

@ShaVuklia think about how you'd get a contradiction is $n$ were prime and then adapt it to a generic $n$

user400188

is there a last statement ?

Semiclassical

So therefore one of S3,S4,... would have to be true. But we already asserted that S1 being true means that every one of these statements is false.

Good question. In the paradoxical version, no.

user400188

1:56 PM

if there is a last statement the first one could be false and everyone after it would be flase also except for the last

Semiclassical

That doesn't quite work, in fact. But if there's a last statement then you can make it work if you have the second-to-last statement be true and the last statement (the only one the second-to-last one cares about) be false.

So yeah, having a finite number of statements prevents this from being paradoxical. But what if there is no last statement, i.e. we have a countably infinite list of such statements?

user400188

its still works actualy. Just becuase you do not have confirmation does not mean you cannot assign a true to it.

Semiclassical

What?

user400188

anyway: i think the point of this paradox is to avoid contradiction? Correct?

Semiclassical

To try to, at any rate.

What I argued above is that if the first statement is true, then you'd get a contradiction.

So it seems that the first statement would have to be false.

It's not hard to see that if you look at the second statement, then the same approach (assume it's true and derive a contradiction) shows that it's false as well. And so forth.

user400188

2:01 PM

well if its about contradiction; you may actualy make a mistake right at the end and beucase you do not reference anything you would avoid contradiction.

Semiclassical

By that do you mean: If P implies a contradiction, then P is false?

user400188

Im not using an if here

but I might be; Ill need to rhink about it

Semiclassical

Then I really have no idea what you're saying. But it sounds I'm not catching you at a good moment.

user400188

nope

Its pretty late and I should be in bed by now

Semiclassical

Mmkay.

Try it another time, then.

user400188

2:05 PM

anyway reading If P implies a contradiction, then P is false? I got:
($p\rightarrow C)\rightarrow (p=0)$ where C is a contradiction arises and p is the statement.

which simplifies to (p and no contradiction) or (p is false)

Semiclassical

Right. So if you assume P and find that it leads to a contradiction, then P can't be true.

user400188

what I really meant was that a contradiction only arises when a true statement is AND'ed with a false one

but if there is nothing to and it with; it will never be contradicted

you avoid the contradiction by never meantioning (anding it with) the thing that will contradict it

Semiclassical

Uh.

user400188

if there is a last statement; then this will in fact be the case

Fawad

Now there is no last statement,so what could be correct statement? @Semiclassical ?

user400188

2:09 PM

I was talking about the case where the first is false

Semiclassical

Ah.

user400188

f the first where false then 2 scenarios could occur where the whole thing is true

one is what you mentioned; the second last could be true and the last flase

Semiclassical

I'm fine with saying the first one is false, I just wanted to make it clear that the first one should not be true.

user400188

and the other is the last is true and it simply avoids mentioning the statement aftet it

DHMO

@user400188 let's examine the statement "there exists a z such that [(u is in z) and (z is in X)]", where X={{{}}}, and find out what it means for u

X has only one element

so (z is in X) is equivalent to (z={{}})

agreed? @user400188

user400188

2:11 PM

well; if we are going to have a last statement. then concivably the first statement may be the last. So in this case the first could be true without contradiction

Oh Hi again

Semiclassical

In the interest of this conversation not getting too confusing, I'll let this conversation be for now.

user400188

fair enough

@DHMO Agreed

DHMO

@user400188 so we can transform the statement to "there exists a z such that [(u is in z) and (z={{}})]"

Semiclassical

by "too confusing" I really mean that I don't want to interrupt further what you and DHMO were talking about

DHMO

and then to "u is in {{}}"

@Semiclassical let's resolve your paradox by resorting to intuitionistic school

@user400188 agreed?

user400188

2:14 PM

yes

had to think about it for a sec sorry. That was u in z just then

DHMO

and I substituted the z

but since {{}} has only one element, "u is in {{}}" is equivalent to "u={}"

user400188

from that u is {} and from that y must be {{}}

DHMO

summary: "u={}" is equivalent to the statement "there exists a z such that [(u is in z) and (z is in X)]"

Alessandro Codenotti

groans I barely wrote a page of this stuff I have to write for uni and I already understand why people always leave all the proofs to the reader

user400188

ok

DHMO

2:15 PM

since we require the first and second statement to be equivalent, we require "u is in Y" and "u={}" to be equivalent

therefore, Y must be {{}}

as you have correctly stated

user400188

I see

Semiclassical

ew, intuitionism.

user400188

At this point I feel I have the Axiom of Union down pretty well

DHMO

@user400188 heh, I haven't given you the hard questions

Semiclassical

You'll take the law of the excluded middle from my cold dead hands.

2

DHMO

2:16 PM

kills Semiclassical

user400188

@Semiclassical I agree; but this is all I have learned so far. So to me its the only way at the moment. Plus its quicker to learn late at night

DHMO

takes law of the excluded middle

Semiclassical

twitch

DHMO

@user400188 do we want to go to my next question, or do you want to explore the rabbit hole?

user400188

Whats the Rabbit hole?

in either case Ill probaly want to go to bed

DHMO

2:18 PM

24 mins ago, by Semiclassical

Heheheh. Follow me down the rabbit hole.

Alessandro Codenotti

$\neg\neg P\iff P$ is something every sane person should agree on.

DHMO

@AlessandroCodenotti is the continnum hypothesis true?

Alessandro Codenotti

(Not all computer scientist are sane according to my definition of sane)

user400188

oh no thanks. Thats too leasurely for this hour. If I am going to stay up I might as well learn something as opposed to waste time.

DHMO

@user400188 do you want to do my questions?

user400188

2:19 PM

yes

But to tell the truth Im a bit to tired to continue on either

DHMO

so what do you want

user400188

To go to bed and continue this later. If you are willing

alright

Goodnight everyone

Cya

2:28 PM

Hi there

I have a small problem

I have kind of a vector field, but with time as an additional paramenter

How's it called ? Something like : $\mathbb{R}^3,\mathbb{R} \to \mathbb{R}^3$

Semiclassical

$\mathbb{R}^3\times \mathbb{R}$ is probably what you want

But just plain $\mathbb{R}^4$ will do.

PearlSek

why not a comma ?

Semiclassical

Because that's not how you denote a combination of sets.

Specifically, $A\times B$ is the set of ordered pairs $(a,b)$ with $a\in A$, $b\in B$.

PearlSek

Yes, right

Semiclassical

Which in your case translates into $(\vec{x},t)\in \mathbb{R}^3\times \mathbb{R}$ where $\vec{x}\in \mathbb{R}^3$ and $t\in \mathbb{R}$.

DHMO

2:34 PM

I guess I asked this before

How can I prove that $A \times A = A \implies A = \varnothing$?

PearlSek

And I want to have an equation that means that this function evolution over time ($\frac{d\vec{x}}{dt}$ ?) is equal to $\vec{x}$

Semiclassical

Isn't it true that $\mathbb{Z}\times \mathbb{Z}$ has the same cardinality as $\mathbb{Z}$ itself?

DHMO

@Semiclassical it is

PearlSek

More concretely, the vector describes what is it's velocity over time

Semiclassical

In which case it would seem you can set up an isomorphism between $\mathbb{Z}$ and its Cartesian square.

DHMO

2:36 PM

@Semiclassical but they are still not equal

PearlSek

like a wave that has some energy that concretizes as speed

Semiclassical

How could they be? One is a set of ordered pairs, the other isn't.

DHMO

@Semiclassical We know nothing about $A$; it may as well be a set of ordered pairs

Semiclassical

Sure, but then $A\times A$ would be ordered pairs of ordered pairs.

DHMO

I don't consider it a proof...

Alessandro Codenotti

2:38 PM

@Semi you can have $A\times A\subseteq A$ with $A$ nonempty

DHMO

@AlessandroCodenotti wait, what? how?

Semiclassical

Huh.

I could understand it being isomorphic to a subset of $A$, but equal?

Though I also don't really know how one defines $A\times \emptyset$, come to think of it.

DHMO

@Semiclassical the same way as to defining $A \times B$

$A \times \varnothing = \{(a,b) | a \in A \land b \in \varnothing\} = \varnothing$

Semiclassical

Hmm, okay.

I guess the simplest way to put it is whether there's a non-empty set which satisfies $A\times A=A$.

Alessandro Codenotti

define $V_0=\varnothing$, $V_{\alpha+1}=\mathcal{P}(V_\alpha)$ for successor ordinals and $V_\lambda=\bigcup\limits_{\beta<\lambda}V_\beta$ for limit ordinals (that's called the Von Neumann hierarchy)

then you have $V_\omega\times V_\omega\subseteq V_\omega$

Semiclassical

2:41 PM

ow.

in short, ordinals are f***ing weird.

DHMO

@AlessandroCodenotti but $V_\omega$ doesn't have ordered pairs

Alessandro Codenotti

$V_\omega$ is the set of hereditarily finite set, that means finite sets whose elements are finite sets whose elements are finite sets whose...

also the set of sets you can write with a finite number of brackets

@DHMO it does, for a reasonable encoding of ordered pairs as sets

DHMO

@AlessandroCodenotti which?

Alessandro Codenotti

like $(a,b)=\{\{a\},\{a,b\}\}$

Semiclassical

Ah. And a Cartesian product of sets with a finite number of brackets is another set with with finitely many brackets.

DHMO

2:44 PM

@AlessandroCodenotti you win

Semiclassical

That be weird.

At the same time, you had to go that far just to get $A\times A\subseteq A$ with $A\neq \varnothing$. So that still leaves $A\times A\neq A$ as plausible.

No clue how to prove it, though.

Alessandro Codenotti

$A\times A=A$ is impossible

2

DHMO

is $V_\omega \times V_\omega = V_\omega$?

Alessandro Codenotti

that was discussed before in chat, starting about here

Semiclassical

It mentions at the end of the post that $A\times A=A$ is true if one assumes the negation of the axiom of foundation.

Which I guess makes me wonder if "nonempty $A\neq A\times A$" is in fact equivalent to the axiom of foundation.

DHMO

2:50 PM

@AlessandroCodenotti eh, is it yes or no?

mercio

lol at the $\Omega = \{\Omega\} \implies \Omega = \Omega \times \Omega$ remark

Alessandro Codenotti

@DHMO look at my message shortly above, I linked the answer to a question showing that $A\times A\neq A$

DHMO

@AlessandroCodenotti so exactly which element of $V_\omega$ is not in $V_\omega \times V_\omega$?

Semiclassical

They also give a recursive construction. Start with a set $A_0$, iterate $A_{n+1}=A_n\cup (A_n \times A_n)$, and define $A_{\omega}=\cup_n A_n$.

mercio

@DHMO the empty set ?

DHMO

2:53 PM

@AlessandroCodenotti does this need the axiom schema of replacement?

@mercio mind = blown

you're right, there is no empty set in $A \times A$

@Semiclassical does this need the axiom schema of replacement?

Semiclassical

shrug

Alessandro Codenotti

@DHMO no idea. Doesn't look like it tho

DHMO

@AlessandroCodenotti I think it needs

or else how are you going to construct the infinite set $\{V_0,V_1,\cdots\}$

Fawad

Z²×Z =Z³ means? @Semiclassical or @anyone

Semiclassical

It means there's not much difference between writing ((1,2),3) and (1,2,3).

Indeed, the format definition of the ordered pair (a,b) is the set {{a},{a,b}}

In which case the first one means {{{1},{1,2}},{{{1},{1,2}},3}}

DHMO

3:02 PM

@Semiclassical and the formal definition of (1,2,3) can be ((1,2),3)

Semiclassical

Yeah.

I'm going based on the Wiki page, and I'm not seeing the definition of an ordered triplet there.

mercio

then we shuold just take $X^3 = X^2 \times X$ as a definition and move on

Semiclassical

Heh.

That, or $X\times X^2$ :P

Alessandro Codenotti

I think there isn't the definition of an ordered triplet because nobody ever cared to write down one since it doesn't matter which convention we pick

DHMO

^

N3buchadnezzar

3:23 PM

Heya

DHMO

@N3buchadnezzar god dag

(Hey, it's the same as Swedish!)

N3buchadnezzar

You speak swedish?

DHMO

no, I just know "god dag" in Swedish lol

N3buchadnezzar

DHMO

@KasmirKhaan god dag

N3buchadnezzar

3:25 PM

Hi, btw. Anyone have any clues on proving the above equality?

Kasmir Khaan

Hi @DHMO

DHMO

@KasmirKhaan in Norwegian it is also "god dag" lol

N3buchadnezzar

$$ g(re^{i\theta}) = \sum_{k=1}^\infty \hat{g}(k) r^k e^{ik\theta} $$

Kasmir Khaan

Yes

Fawad

3:49 PM

0

Q: Question concerning the domain of a function

In the function $f:\mathbb{N}\rightarrow\mathbb{N}$ defined by $f(x)= 5x$ why is range not equal to co domain? I don't understand why this is not a surjective function while the same function is surjective if $f:\mathbb{R}\rightarrow\mathbb{R}$.

special-functions

mercio

?

Fawad

I don't understand why this is not a surjective function while the same function is surjective if $f:\mathbb{R}\rightarrow\mathbb{R}$.

Hi @mercio

mercio

what's $f^{-1}(1)$ ?

Fawad

$\frac 15$

mercio

is $\frac 15$ an element of $\Bbb N$ ?

Fawad

3:52 PM

Oh,it is not natural number,thanks

What does $\mathbb{Z}\times \mathbb{Z}$ mean? What is $\mathbb{Z}$ ? Is under geometry part?

Semiclassical

$\mathbb{Z}$ is the set of integers i.e. $\Bbb Z=\{0,1,-1,2,-2,\cdots\}$

so therefore $\Bbb Z\times \Bbb Z$ is the set of ordered pairs of integers.

In other words, it's any point in the plane with integer coordinates.

Fawad

Something like all possible values ({0,0},{0,1}…,{1,0}…{-1,0}…,{-1,1}……?)

Semiclassical

Right.

Except you'd use () not {}

Fawad

$\Bbb Z$ can be taken as one axis , so ${\Bbb Z}^2\times \Bbb Z = $\Bbb Z\times{ \Bbb Z}^2={\Bbb Z}^3$ ?

Semiclassical

Right: Any 3D point (x,y,z) with integer coordinates has integer coordinates for (x,y) and integer z

Fawad

4:05 PM

Also $\times$ means axis needs to be perpendicular to which is is multiplied?

@Semiclassical why not say integer coordinates (x,y,z) ?

Semiclassical

because you said Z^2 times Z.

I'm just emphasizing that one can split it up that way or you can leave it as just (x,y,z).

Fawad

And about this?

Semiclassical

Oh, perpendicular? Nah, that's not necessary.

As an example, suppose you had $z=x+y.$

Then the set of points $((x,y),x+y)\in\Bbb Z^2 \times \Bbb Z$ would correspond to the integer coordinates on the plane $x+y=z$ and there'd be no inherent sense of perpendicular.

(In applications it often is the case that there's some notion of orthogonality for different coordinates, but it's not built into the Cartesian product. You have to incorporate that additionally.)

N3buchadnezzar

4:28 PM

Is it possible to write

$$
\sum_{k=0}^\infty \sum_{n=0}^\infty a_k a_n = \sum_{k=0}^\infty \sum_{n=0}^k a_{k} a_{k-n}
$$ ?

mercio

probably not

PearlSek

5:06 PM

Hi back

Let's say I have a function $f: \mathbb{R}^3 \times \mathbb{R} \to \mathbb{R}^3$

(a vector field with time)

I want the vectors to "move" over time at a fixed speed (1) but the direction of the vectors describe the direction of this movement

mick

5:34 PM

0

Q: Zero's and fixpoints of $ f (\exp(b z)) $

Let $ b > 1$. Let $f(z)$ be -meromorphic over the entire complex plane. Also $f(z)$ maps the real line to the extended real line. And $f(z)$ is non-linear. Consider the strip $A$ such that $ 0 =< \Im(z) < \frac{2 \pi}{\ln(b)}$. 1) Is it true that there is always a complex number $q$ in $A$ suc...

complex-analysis exponential-function roots fixed-point-theorems periodic-functions

Notice the similarity to analytic fourrier series

Any ideas welcome

Astyx

5:50 PM

@N3buchadnezzar Yes

Vrouvrou

Please i need help for this exercise: math.stackexchange.com/questions/2138610/…

if i suppose that $f_A$ is continuous on $x_0\in A$

How to continue ?

SteamyRoot

@Vrouvrou You should address the comments about your function. It's a function $E \to \{0,1\}$

Vrouvrou

@SteamyRoot i don't understand

SteamyRoot

The image of some $x$ is either $0$ or $1$, not $(1,\{1\})$ or so

Vrouvrou

???

SteamyRoot

6:01 PM

You write $f_A: E \to (\{0,1\},\mathcal{P}(\{0,1\}))$

Vrouvrou

$\{0,1\}$ is given with that topology $\mathcal{P}(\{0,1\}$

SteamyRoot

You may want to write that in the question

It's rather common not to mention the topology in the definition of a function

(and you're missing a bracket ) at the end)

Also, I'm guessing Fr(A) comes from frontière. The more-or-less accepted international symbol is $\partial$, e.g. $\partial A = \bar{A}\setminus \mathring{A}$

Vrouvrou

i edited the question

N3buchadnezzar

@Astyx Thanks!

Any ideas about this one? It says it is an application of the Cauchy-Schwartz inequality. However I am having a hard time seeing how n+1 vanishes.

SteamyRoot

@Vrouvrou looks good.

Don't have time to answer it, but maybe math.stackexchange.com/questions/284432/… or math.stackexchange.com/questions/1424344/… are helpful

Vrouvrou

6:19 PM

this not help me

Alessandro Codenotti

6:56 PM

I'm trying to prove that in a complete metric space if I have a family $\{A_n\}_{n\in\Bbb N}$ of dense open sets then $\bigcap A_n$ is dense too (that's the Baire category theorem)

Actually the theorem is that complete metric spaces are Baire spaces, the open set stuff is one of many equivalent things to being Baire, but my (notoriously bad) intuition suggested to pick that characterization for the proof

Balarka Sen

That's a good choice

Alessandro Codenotti

So I pick an arbitrary open set $U$ and want to show that $U\cap(\bigcap A_n)$ is nonempty

Balarka Sen

I think you should work with open balls, WLOG

Alessandro Codenotti

@BalarkaSen I'm glad to hear that because the other characterizations don't seem as nice to work with

Balarka Sen

Because that's where you get to use the fact that you're working with metric spaces

Alessandro Codenotti

7:02 PM

@BalarkaSen sure, the metric must be useful for something

Maks

Hey, can someone help me find the answer to the system of equation
$ x^2 - y^2 = 2$ and $y^2 - x^2 = 2$ ?

Could it be that $(0,\pm\sqrt{2})$ is the only answer?

Balarka Sen

7:20 PM

@Alessandro did you figure it out

Alessandro Codenotti

Nope

Balarka Sen

well, you know $U \cap A_n$ are nonempty for all $n$. any ideas how to use that?

Oskar Tegby

I can't continue from the answer in this question, and I'm looking at the same problem. I'm a little confused by the radii. Any ideas? math.stackexchange.com/questions/1321208/…

Alessandro Codenotti

I'm trying to get a contradiction by assuming that there's an open ball not intersecting $\bigcap A_n$

Balarka Sen

You don't need to do a proof by contradiction for this, actually

Alessandro Codenotti

7:21 PM

In particular I'm trying to construct a "bad" sequence by intersecting $U$ with $A_i$

Balarka Sen

What do you mean by bad?

Alessandro Codenotti

Either Cauchy but not converging or Cauchy but with a limit that leads to a contradiction

the latter seems more likely

Balarka Sen

Your plan is good but you should really forget about that contradiction. You know $U \cap A_k$ are nonempty, so pick $x_k$ from them. All you want to do is to show you can pick $x_k$ in a way that the sequence is Cauchy, right?

Alessandro Codenotti

ok, that's not a problem. I pick an arbitrary $x_0$ in $U\cap A_0$, then I pick $x_1$ from $U\cap A_1$ in a ball around $x_0$ small enough to be in $U$, let's assume of radius $<1$, $x_2$ in a ball of radius $<\frac 12$ around $x_1$ and so on

that can always be done because each $A_i$ is dense

Balarka Sen

That sounds about right

So I guess the question is why does the limit belong in $\cap A_i$

Alessandro Codenotti

7:37 PM

oh, wait, it's possible to do better than this. I start with an open ball $U_0$ and $x_0$ in $U_0\cap A_0$, pick $x_1$ in a ball $U_1$ such that $U_1\subseteq U_0\cap A_1$ ... and $x_n$ in a ball $U_n$ such that $U_n\subseteq U_{n-1}\cap A_n$ so that all those balls are nested

that's always possible because $U_i\cap A_{i+1}$ is open and the $A_i$ are dense

Balarka Sen

That's the picture I had, yeah

Alessandro Codenotti

I also choose them so that the radius goes to 0

Balarka Sen

Right, so the sequence is obvious Cauchy

ah, and the limit also belongs in $U \cap \bigcap A_n$ because the sequence fits inside a closed ball in the space

Alessandro Codenotti

Ah, I got it, rather than $U_0$ I should do the construction inside a closed ball contained in $U_0$ then

Balarka Sen

7:54 PM

right

Alessandro Codenotti

then the limit belongs to all the $U_i$ so to all of the $A_i$ so it's also in $U\cap\bigcap A_n$, neat

I guess that for the second version of the Baire category theorem (compact and Hausdorff implies Baire) the construction is very similar but I pick closed balls rather than open ones at every step and use the fact they have the finite intersection property so they must have nonempty intersection being in a compact space

Alessandro Codenotti

8:10 PM

ah, wait, I dropped the metric that doesn't work. But compact Hausdorff spaces are normal so I should be able to construct a decreasing sequence of closed sets. hm, I'll think about the details

Daniel Guldberg Aaes

Hi. i'm sure you guys know about the second degree polynomial, where f(0)=c and f'(0)=b there is showing us the tangentline in x=0 on the graph.

Do you know if there is anyting equal to the cubic polynomial?

mick

9:03 PM

0

Q: Zero's and fixpoints of $ f (\exp(b z)) $

Let $ b > 1$. Let $f(z)$ be -meromorphic over the entire complex plane. Also $f(z)$ maps the real line to the extended real line. And $f(z)$ is non-linear. Consider the strip $A$ such that $ 0 =< \Im(z) < \frac{2 \pi}{\ln(b)}$. 1) Is it true that there is always a complex number $q$ in $A$ suc...

complex-analysis exponential-function roots fixed-point-theorems periodic-functions

Oskar Tegby

I can't continue from the answer in this question, and I'm looking at the same problem. I'm a little confused by the radii. Any ideas math.stackexchange.com/questions/1321208/…

10 Replies

10:45 PM

Help? math.stackexchange.com/questions/2145707/…

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