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12:07 AM
1

Let $\displaystyle f:[ a,b]\rightarrow \mathbb{R}$ be a continuous function, that is $\displaystyle f\in \mathcal{C}[ a,b]$. It is to be proven that $\displaystyle m\in \mathcal{C}[ a,b]$, where $\displaystyle m( x) =\min_{a\leq t\leq x} \ f( t)$. Claim: $\displaystyle m$ is a monotonically decre...

Can anybody please review only the part where $m(c-)-m(c+)$ has been shown to be zero?
$m(c-), m(c+)$ are one sided limits. $m(c-)=\lim_{x\to c^{-}}m(x)$.
In particular, I request review of only relation no. $(4)$.

why did copper get suspended?

12:49 AM
wirey personality
disregarding code of conductivity
...i'll see myself out

he didn't did he?
oh from chat.se ?

1:14 AM
I thought it would spin

1:37 AM
anyone know how to write this in a way that doesn't make my eyes bleed (without factoring yet)...I can't seem to get \split to render...
$$V =\\ e^{a\left[(mx^2/\hbar)+it\right]} \left(i\hbar -aie^{âˆ’a(mx^2/\hbar)}\cdot e^{-ait} \\ + \frac{\hbar^2}{2m}\left[{\frac{âˆ’2am}{\hbar}}e^{(âˆ’am/\hbar)x^2}\cdot e^{-ait} \right] (\frac{âˆ’2am}{\hbar}x^2 +1)\right)$$
(although I need to go get my finger stitched together now, so no hurry :P)

best of luck with the finger

2:01 AM
Thx. My partner used to study sepsis, so any time I get wounded I think I might err on the side of over reacting :P

2:20 AM
if possible, it's good to have all of your fingers, and even better, the use of all of your fingers

3:03 AM
@leslietownes The professional medical opinion was: "it was a good thing you cook with a sharp knife."

3:23 AM
ha! nice

3:36 AM
dis iz ncie

3:59 AM
And I got chewed out in comments for my **language in here. While I now have a second downvote on that question, nary a comment.
To be chewed out in comments for chatroom behavior is totally inappropriate.
I might have been short-tempered after my colonoscopy. So maybe I get suspended next.

downvoting without commenting is bizarre to me. unless there is something like a top voted post that contains an obvious mistake that is already pointed out in comments.
i wasn't aware that anybody had been suspended. where do people keep track of this stuff?

4:24 AM
Da onli tim i sau someon beinj suspendid iz wehn i got suspendid

I don't know, I just find out when I see someone has 1 rep but it seems not all suspensions are like that, or who knows.

4:41 AM
I want to prove Wilson's theorem but I get stuck at one point. $(p-1)!=-1(mod p)$
Here, $p$ is an odd prime.

ok. lots of proofs out there. it is sometimes an introductory exercise in finite group theory. (p-1)! is the product of all of the elements in the group of units mod p. even without the group theory language, if you know about multiplicative inverses modulo p, that idea can go through.
i think there are also other elementary ways of getting it via arithmetic but that is the main path i can think of right now.

For this, I use this lemma: For every finite abelina group $G=\{a_1,a_2,\cdots,a_n\}$, if $x=a_1a_2\cdots a_n$ then $x^2=e_G$, the identity in G. Its obvious as $x^2=(a_1a_2\cdots a_n) (\text{their inverse})=e_G$
Hi @leslietownes
Then I took multiplication modulo p group that is $U_p$ which has $p-1$ elements viz. 1,2,3,..., p-1

But you can do better than that proof. How many elements here can be their own inverse?

So using this, it follows that $((p-1)!)^2=1 (mod p)$. It means that $p|(((p-1)!)^2-1)\implies p| ((p-1)!-1)((p-1)!+1)\implies$ either $p|((p-1)!+1)$ or $p|((p-1)!-1)$. How do I contradict this second possibility?

4:52 AM
Hi Ted. I think that p-1 elements can be their own inverse.
because p is odd.

We have a specific group here.
Think about the proof you just did, in a much simpler context.

So since $p$ is prime,$U_p$ must be cyclic. That means that there is $\phi (2)=1$ element which is of order 2 that is its own inverse

Way too complicated.

don't forget to also count the element that isn't of order 2 that is its own inverse.
you basically did the proof above, but it would help to translate it into simpler language

the problem is that I don't understand how to contradict $p|((p-1)!-1)$

4:58 AM
Your proof is the wrong proof, but you can fix it by being more direct. Surprise, surprise.

:'(

I want you to figure out $x$, not $x^2$.

Ahh. x=e_G
$x=e_G$ because inverses will come in pairs :)

Oh really?

Excluding the identity, we have an even number of elements left in G, so pairing of $a\in G$ with $a^{-1}\in G$

5:01 AM
Clearly wrong.

:'(
Yes. I made a mistake I realized it now.
$U_p$ has $p-1$ no. of elements (even) so it must have at least one order 2 element.

is there any way to rigorously understand optimization problems? i'm doing a calculus for economics class, and the optimization problems we're doing assume a tonload of stuff
any sort of book I could look into for this stuff?

Think of arithmetic mod p. What elements are their own mult inverse?
Assuming a tonload? You mean Lagrange multipliers?

well, we have to maximize f(a, b) constrained with g(a,b), and I'd just like to know if there's something like analysis of this procedure

Yes, Lagrange multipliers. Watch my YouTube lecture.

5:08 AM
great, I'll look into that. thank you!!
wow you have lectures on every proof we've just skipped over
neat

If you want proofs, you're in the wrong course!

@TedShifrin Thatâ€™s an amazing line to say. Want answer , watch my lecture

y dsoe mathgemtics ned proophs

I am afraid I'm new to group theory, I don't understand.

da mathgematikians cood simpli se, "bro truzt mi dis iz troo"

5:16 AM
About $U_n$, I know that it's a group whose elements are equivalence classes of integers congruence modulo n binary operation is $[a][b]=[ab]$ $[x]=$equivalence class of $x$ under the relation congruence modulo n.

da reeman hipodesis iz troo

So work out some examples, @Koro. You should always do that if you all you can do is see symbols and not have intuition.
@RussianBot You are truly annoying.
@Srijan: It's a lot easier than giving a lecture (with no pictures) here.

@TedShifrin that's what bots are for

That's what ignore is for.
9

Haha

5:21 AM
yes

@TedShifrin Yes. Also, amazed by your achievements sir.

Well, @Srijan, this truly is not why I gave that advice. But the videos have been helpful for lots of people, so might as well tell people.

My question was completely different though: simply how to get a contradiction to $p|((p-1)!-1)$ and intuitively it seems false. For example: by taking $p=3$, it's false. $p=5$, again false. It's just that I don't know how to contradict this.

@TedShifrin ofc I understood thatâ€™s not why. Just great achievements still

But I understand that to get alternative approach to proving Wilson's theorem, I need to understand U_p better.

5:25 AM
Why do I always feel like you don't read what I write/say, @Koro?
The way to understand is to work some examples.
Modular arithmetic is very concrete and should be something you master.

Ted, Please note that I do take a note of your suggestions. I do understand and I have accepted already that "I need to understand $U_p$ better."

But you keep going back to your wrong approach. Over and over.
I think I might need to take a break from chat. My patience isn't optimal these days.

that was only one of a series of copper-related puns. i forget the other ones.

Oh, if it was intentional, I'm deleting.
I am too addled to have thought of that, and I didn't read all the context.

5

5:31 AM
enthusiasm for puns is a sign of a misspent youth.

LOL, @shin, no need for that, but thanks.
Well, I am an inveterate punster, so don't analyze my youth.

It's just probably I didn't make the background of the question clear. It's an exercise problem in section 4 of Herstein's Abstract algebra. This exercise follows two exercises -1) For a finite abelian group, if $x=a_1a_2a_3...a_n$ where $a_1,a_2,a_3,...,a_n$ are all elements of G. then $x^2=e_G$, where $e_G$ is identity in G. 2) $x^2=1 (mod p)$ has only these solutions $x=1 (mod p)$ , $x=-1(mod p)$.

Oh, funny how you ignore 2) and concentrate on 1) despite my telling you to ignore 1).

2) is what ted was asking you above, if i'm not mistaken.

By the way, my absolute least favorite algebra book.
Too much algebra as symbol manipulation. One of the worst books in that regard.

5:33 AM
the ordering of 1) and 2) is this in microcosm.

I actually became a big fan of teaching rings first, starting with modular arithmetic galore.
But Artin's book still wins hands-down for the strongest students.

That's what I showed in very first comment. I probably didn't understand Ted's comment.
@leslie

You told us 1), never mentioned 2), and yet I asked explicitly about 2) twice.
I think Leslie is prepared to be far more patient than I, so I will bid the room adieu. Farewell, all.

goodnight ted.

Goodnight Ted
I didn't understand your comment, is my fault.

5:36 AM
i'm trying to think of how i would develop this as a series of exercises.
anyway koro it seems like you've proved 2? or at least you followed the reasoning that you'd need to prove 2 in the case that x = (p-1)! (although the argument does not depend on this)?

I have proven $2$ Leslie. It's just that for $U_p$ case, I don't understand how to contradict $p|((p-1)!-1)$

ok, so you've got this product of everything from 1 to p-1, inclusive.

If anyone can help me in a physics Q , In my textbook , centrifugal force is explained from NIF only and in which , they state that the box is at rest. This happens because we have a box inside the cabin and NIF is considered from the cabin. NIF=Non inertial frame of reference Then , N=mg means a normal force = mg. Ok. Then , they also say that there is friction force which acts towards centre when box is at rest
How is that

Right. So if I divide $(p-1)!$ by $p$, if I can somehow show that remainder is at least 2, then also I'm done! But that's what I have difficulty proving.

from what you proved, you know that 1 and -1 (aka p-1) are the only things in that list that are their own inverses. which means that none of the other p - 3 numbers in that product are their own inverses.
so each of those other p-3 elements can be paired with a different element, its inverse. those pairwise products will be 1. this simplifies the product of all of the elements.
ted might find this too abstract but i don't know why herstein focuses immediately on the square of the product of all of the elements of the group. there's a general idea here for any finite abelian group. the product of all of the elements in the group will be the same thing as the product of all elements of order at most 2 in the group. anything that is not its own inverse will cancel with something else in the product.
for any finite abelian G, prod_{g in G} g = prod_{g in G, g^2 = 1} g. when G is units mod p this simplifies further.

5:50 AM
@SrijanM.T try the physics stack exchange!

Leslie, simplifying $U_p$ for example, I should have either $x=1 (mod p)$ or $x=-1(mod p)$, where $x=(p-1)!$. But Wilson's theorem says $(p-1)!=-1 (mod p)$. That means $x=1(mod p)$ must be neglected somehow (This is what bothers me!)

(p-1)! = 1*2*3*4*...*(p-1). everything in this product cancels mod p except the terms that are their own inverses, who have nothing to cancel with.
so (p-1)! = 1*(p-1) mod p.

\pmod p $\pmod p$ looks nice.

pmod p is definitely the way to write it if one must.
i'm in denial.

I understand your point @leslietownes. That's actually very nice. So $(p-1)!=1.2.3....(p-2) (p-1)(mod p)$ and since $1$ and $p-1$ are the only elements which are their own inverses rest of all must cancel one another to give identity so in the end, we get $(p-1)!=(p-1) mod p$ which gives $(p-1)!=-1 (mod p)$.
Thanks a lot :)
I understand that now.

6:07 AM
similar idea proves that any finite group (abelian or not) of even order has an element of order 2. the elements of order > 2 can be paired with their inverses, so there's an even number of them. and there's the identity of order 1. gotta be at least one more element of the group.
of course there are other ways of proving that theorem that also provide generalizations. but that pairing idea is a simple and powerful idea.

I understand that @leslietownes. Indeed, if we have a group (finite) and of even order then excluding identity, we have odd no. of elements (consisting of pairs (a, a^{-1}) and then one will be left out. That's what I was trying to say earlier also but it created misunderstanding as I erroneously considered order of U_p as $p$.
Thanks a lot @leslietownes for the great help.

I'm getting hung up on the fact that sympy refers to $i$ as I and $e$ as E

hah. maple uses I by default, too. although i think you can change it

It seems question miss some information?

such as what $\Delta$ and $\Delta_1$ are? yes. if i had to guess, $\Delta = \det(S)$ and $\Delta_1 = \det(T)$. but i do have to guess.

6:22 AM
Ah yes, I just left this question in exam thinking it will be bonus, but you are correct, it's just that. If we consider that way I am getting one of the option.
Thanks @leslietownes

1 hour later…
7:37 AM
you saw it, i presume. :)

Yes lol
By the way, have any of you read through the proof of Riesz Representation Theorem in Rudin? It seems very long and convoluted. Is there an easier or better explained proof elsewhere?

mm, which rudin?

Real and Complex Analysis, Big Rudin

i forget how he proves it and don't have a copy of that one. his functional analysis book, which i do have, simply cites R&C for the proof. :) it's not the world's easiest theorem.
math.berkeley.edu/~arveson/Dvi/rieszMarkov.pdf is one route to it. there is some other useful stuff in there.
with particular linear functionals on nice spaces it is sometimes possible to identify the measure explicitly. but these kinds of calculations do not shed too much light on what to do in the general case.

Okay cool, I'll check that link. Thanks!

7:52 AM
it might be the same route rudin takes, for all i know, but at least it isn't written in rudin style.
:)

8:03 AM
got a real analysis proof for y'all
Given nonempty subsets $A, B$ of $\mathbb{R}$, let $C := \{x+y: x \in A, y \in B \}$.
Prove that, if $A, B$ have suprema, then $C$ has a supremum $\sup C = \sup A + \sup B$.
Go! Let $c$ be an arbitrary element of $C$. Then $c = a + b$, where $a,b$ are arbitrary elements of $A,B$. Then:
$a \leq \sup A;$
$b \leq \sup B$
$\therefore a + b \leq \sup A + \sup B$
$\therefore c \leq \sup A + \sup B$
$\therefore \sup C = \sup A + \sup B$
$\blacksquare$
true or false? comments are welcome
btw has anyone else noticed that you'll have a hard time finding two users with an identical avatar
isn't that nifty? I wonder how they're auto generated
also, for the above proof, I haven't assumed the completeness axiom. do I have to?

what you wrote above only shows that $\sup A+\sup B$ is an upper bound for $C$
i.e. $\sup C\le\sup A+\sup B$, you don't have equality yet

oooooh

8:28 AM
thank you, I'll meditate on this some more

@shintuku the approach I can see for the other direction is not as pretty (epsilons and deltas)

you can do it without epsilons and deltas, but it's not like there's a big difference

8:49 AM
Suppose the completeness axiom, and suppose $\sup A + \sup B \neq \sup C$. Then $\sup C$ is less than an element of $C$, which is a contradiction.
soooooooo $\sup A + \sup B = \sup C$
right? I mean, when are we not going to have the completeness axiom anyways
if a set has an upper bound, and this upper bound is in the set, then the completeness axiom tells us the set has a least upper bound, which makes perfect intuitive sense, no?

@Thorgott by epsilon-delta, I mean that we need to consider the difference $\epsilon=\sup A+\sup B-\sup C\gt0$ and find $a\gt\sup A-\epsilon/3$ and $b\gt\sup B-\epsilon/3$ so that $c=a+b\gt\sup C$
or is there a cleaner way?

9:06 AM
I got it!
$\therefore c \leq \sup A + \sup B$
$\sup A \in A, \ \sup B \in B$
$\therefore \sup A + \sup B \in C$
Suppose there is a greater upper bound than $\sup A + \sup B$ that is in $C$. Then there is an element in $C$ greater than $\sup A + \sup B$, which is a contradiction.
$\therefore \sup A + \sup B = \sup C$
no?
relatively clean too
wait nevermind, $\sup A \in A$, $\sup B \in B$ isn't given
those are the maximums

1 hour later…
10:21 AM
i've seriously underestimated the reals

Maybe they should be called complex numbers instead. :-)

uh
is the number $0.0 ... 01$ a real number
my god this is giving me such a tremendous headache, and I initially thought I could do the proof easily

10:36 AM
@shintuku it depends what you mean. If there is a (fixed) finite number of 0 then yes.
0.01 is a real number it is equal to 1/100

right, I meant if there are an infinite number of zeroes in between

@shintuku that's not a real number.

dang it

The digits in the decimal representation of a real number form a sequence.
In an infinite sequence there is no last term.
You cannot have an infinite number of terms, and then something.
It's a bit like saying I take the number after the largest natural number.
It does not make sense as there is no largest natural number.

alright well, I'll close this real analysis book and go construct the reals instead
I can't get myself to believe the completeness axiom for open subsets of $\mathbb{R}$
I feel like an open subset can simultaneously have an upper bound but no least upper bound

10:47 AM
@shintuku Maybe it helps to note that the complement is closed. All the upper bounds are in the complement.
You then have a minimum element there. (It might not really be the minimum of the complement as there could be elements "on the other side".)

I'll meditate on that a while
I was imagining open subsets by subtracting the open points from the reals, but now that you mention it, thinking about the complement makes more sense
nvm, will continue meditating on that idea

11:09 AM
@robjohn Depends on what you consider clean, of course. I would argue as follows: Assume $c$ is an upper bound for $A+B$. If $B=\emptyset$, we are done, so assume $B\neq\emptyset$ and pick a fixed $b\in B$. Then $a+b\le c$ for all $a\in A$, whence $a\le c-b$ for all $a\in A$, thus $\sup A\le c-b$. Rearranging, $b\le c-\sup A$ and this holds for all $b\in B$, so $\sup B\le c-\sup A$, i.e. $\sup A+\sup B\le c$. Hence, $\sup A+\sup B$ is the smallest lower bound for $A+B$.

11:37 AM
-1

I have some questions regarding step I of Rudin's proof of Riesz Representation Theorem (Real and Complex Analysis). I have included what I've tried. Why is $h_i g \prec V_i$? We know that $g\prec V_1\cup V_2$ and $h_i\prec V_i$. We want to show that $\text{supp}(h_ig) \subset V_i$, $0 \le h_ig... Does an answer to this already exist somewhere? I don't understand why it got downvoted :( I wasn't able to find an answer on the site 1 hour later… 12:56 PM Ah, can someone please help? 1 hour later… 1:59 PM @epsilon-emperor I don't see a reason right off, but some people dislike images of a page. @robjohn Do you recommend typing such long proofs? The pictures are meant to be only for reference, and all the main content of the question is in MathJax. Is that okay? I know, but there are some pro forma downvoters. I don't know what to suggest on that. If you end up getting a lot of downvotes on what look like decent posts, let us know and we can look to see if there is evidence of someone serially downvoting. Got it, thank you! Could anyone drop any hints for this? https://imgur.com/a/1543HKs Seems minor so didn't make a post out of it. I have added my attempt in the link itself, MathJaxed. Ah, it's funny how I get my answers a few minutes after posting :P The tail of the sum goes to zero, and that gives the proof 2:16 PM @quid Sure it is. It is the real number zero. :P 2:35 PM @robjohn @Thorgott Any real number$r$is the supremum of a subset of the reals$(-\infty, r)$. Let$r = a + b$where$a,b$are the suprema of arbitrary subsets$A,B$of the reals. Then,$a+b$is the supremum of$(-\infty, a+b)$, which has a subinterval$C := (\inf A + \inf B, a+b)$. Then$C = \{x+y:x \in A, y \in B\}$, and$\sup A + \sup B = \sup C$. thank dedekind cuts, if this works does it work? I do not follow your two$C$s are far from necesarily the same thing well, isn't anything in$(\inf A + \inf B, a + b)$some$x + y$for$x\in A, y \in B$? why do you think so? 2:51 PM since$a$is the supremum of$A$and$b$is the supremum of$B$, any$x+y: x\in A, y\in B$will be smaller or equal than$a + b$. since$\inf A $is the infimum of$A$and$\inf B$is the infimum of$B$, any$x+y$will be greater or equal than$\inf A + \inf B$this shows that$\{x+y\colon x\in A,y\in B\}\subseteq[\inf A+\inf B,a+b]$(you do need closed intervals here) this inclusion need not be an equality ohhhhh great point, thank you 3:06 PM @epsilon-emperor theres definitely something to that, sometimes I think its a psychological phenomenon where when you type things out to explain it to someone else it becomes easier to reason through the difficulty in the first place ive also had loads of times where i''ve typed something here and then figured it out a minute or so later 3:21 PM is there an analogue of the modular flow on the space of lattices of unit area in$\Bbb R^2$to the space of lattices in$\Bbb R^2_+?$@porridgemathematics Agreed! I think this has something to do with the fact that E_1 and E_2 may not be in the sigma-algebra, but I'm still not sure why U was considered in the solution. I'm referring to the (a) part of Q4. Here is the matrix that acts on the space of lattices in$\Bbb R^2$:$\begin{pmatrix}
e^t & 0 \\
0 & e^{-t}
\end{pmatrix}. $So this is the matrix describing the modular flow on the space of lattices of unit area epsilon: mu is defined as an infimum over open sets. i think his 'U' is implicitly open here. E_1 \cup E_2 doesn't seem to need to be. showing that mu(E_1 cup E_2) is greater than whatever wouldn't show that an infimum over open sets is greater than whatever perhaps. he seems to be importing a lot of notation from a proof that isn't part of the image. i have no idea what M and M_F are. the fact that he's writing mu(E_1 cap E_2) suggests that E_1 and E_2 are measurable, just not in M. is M the family of all measurable sets? or some subset? M_F seems to have something to do with compact sets. 3:37 PM @leslietownes M_F is the class of all subsets E of X which satisfy \mu(E) < \infty, and \mu(E) = \sup\{\mu(K): K\subset E, K compact\}. @leslietownes M is the class of all E \subset X such that E\cap K \in M_F for every compact K @leslietownes Hmm, my sense is that nothing will go wrong if we just put U = E_1\cup E_2. What do you think? i think he's just trying to relate the things to the definition of mu. E_1 and E_2 are not assumed to be open sets. i assume that U is (has he been using letters like U, V for open sets?) so he's bringing open sets into it so he can appeal directly to the definition of mu. which is in terms of open sets. Ah, actually, he has defined \mu for all sets as: \mu(E) = \inf\{ \mu(V): E\subset V, V open\} So we don't really need E to be open i'm not saying that E needs to be open. i'm saying that the definition of mu(E) is in terms of mu of open sets. he is likely bringing open sets into it so he can point right at the definition. if i'm reading this correctly, mu(E_1 cup E_2) is defined to be inf {mu (open sets containing E_1 cup E_2)}. I dont think U is open why would he call V an open set explicitly, and then just forget to call U one if he consistently uses U and V for open sets and not other kinds of sets (see the proof that he is citing back to), it could be just a hiccup in the prose. it tracks the definition of mu, anyway. the inequality that precedes the statement "where the first inequality follows by..." is true, by definition of mu, if U is assumed to be an open set. if U is just some set i guess i agree i don't understand why he's bringing it in "where the first inequality follows since" i should have said. 3:46 PM if$A \subset B$are any sets, open or not,$\mu(A) \leq \mu(B)$as per his definition, because for any open set containing$B$,$U_B$,$U_B$contains$A$, so we certainly have$\mu(A) \leq \mu(U_B)$, so$\mu(A) \leq \inf \{ \mu(U_B): B \subset U_B ; \}$where$U_B$are open sets, thus we have$\mu(A) \leq \mu(B)$then if$U$is any set at all containing$E_1 \cup E_2$, certainly$\mu(U \cap V_1) \geq \mu(E_1)$and the analogous inequality for$E_2$will hold hes just being verbose, once you have this just set$U = E_1 \cup E_2$to get the other side of the inequality and (a) is proved @porridgemathematics Yeah true @leslietownes That's right @leslietownes He consistently uses V for open sets, and never U. if a set is called U and not open, it violates the geneva convention 3 @porridgemathematics Okay great, cool! i'm with thorgott on this one i'll call the ICJ Wait, there is a problem @leslietownes @porridgemathematics Since E_1 and E_2 are disjoint, if we put U = E_1 \cup E_2 in the beginning, aren't all the inequalities actually equalities? or am I jumping ahead of myself and that's what we're trying to prove? 4:00 PM I believe that is what were trying to prove, but you are probably more familiar with what you're reading than I am yes, that's what you're trying to prove there's nothing sophisticated going on here but he is i believe attempting to tie a proof about what mu(E_1 cup E_2) is to the definition of mu(E_1 cup E_2) there are so few inferences involved in getting the inequality that it seems like almost nothing else is happening Hmm, so can you explain why \mu(E_1 \cup E_2) \ge \mu(E_1) + \mu(E_2)? 'its almost too easy!' It's not LMAO please explain? just do rudin's proof but put "an open" in front of "subset" in "let U be a subset" 4:07 PM By the way, that's not Rudin's proof, that's a solution manual Also if we do the proof with U, how does the first step come if we do not assume that \mu is additive over disjoint unions? \mu(U) = \mu(U \cap V_1) + \mu(U\cap V_2) - this requires justification too oh yeah, lol, you are right, and i dont know how they split that sum how is mu defined on open sets The infimum thingy whats puzzling is, why do we even have$U = U \cap V_1 \cup U \cap V_2$? all we know is$U$contains$E_1 \cup E_2$surely there can be stuff in$U$outside$V_1 \cup V_2$? Yeah, this is crazy 4:12 PM it should be \geq Wait, I have an idea Is there a problem if we assume U to be open? No, right? ok i found an infringing ebook of this book mu is defined on open sets by sup {Lambda f: lambda prec V}, where lambda is a positive linear functional @leslietownes Yes but this is equivalent to the infimum definition, as stated in the next few lines So no difference really well it seems like additivity of mu for disjoint open sets then follows from additivity of Lambda on functions @leslietownes There's a sup there Won't work I guess? In Step III, Rudin proves that M_F contains every open set with finite measure. If we suppose U is open, and V_1,V_2 are already open, we have U\cap V_1 and U\cap V_2 are also open. Then in Step IV, Rudin shows additivity for disjoint sets of M_F. @leslietownes Can you check Steps 3 and 4 from your copy of Rudin, and see if my argument works? 4:18 PM well doesn't this prec business mean that the f for which f prec V and the f for which f prec W (V, W disjoint open sets) basically operate independently of one another so the sup is indeed the sum @leslietownes Could you explain more? f \prec V means that 0\le f\le 1, the support of f is contained in V, and f\in C_c(X) given f for which f prec V cup W, define f_1 = the restriction of f to V, and 0 elsewhere, and f_2 = the restriction of f to W, and 0 elsewhere. then f = f_1 + f_2 and f_1 prec V and f_2 prec W, and vice versa some details there but i think that's right so mu(V cup W), a sup over all lambda(f) satisfying a certain condition, is a sup over all lambda(f_1) + lambda(f_2), with f_1 and f_2 satisfying a certain condition, which should indeed be mu(V) + mu(W) @leslietownes mathb.in/55480 Here, I've typed it out. How do you go ahead with your argument from here? 4:34 PM 40 If$A,B$non empty, upper bounded sets and$A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that$\sup(A+B)=\sup A+\sup B$? there's a minor detail that our f_1 and f_2, as defined, are indeed in C_c(X), but that comes from the assumed openness of V, W and the support of f being contained in V cup W what a coincidence also note that my argument was in terms of open sets containing E_1 and E_2, not E_1 and E_2 themselves my whole vibe of this problem has been to reduce facts about mu on arbitrary sets to facts about mu on open sets. i've been staring at the same corner of the handheld whiteboard I use to write proofs trying to solve that same damn proof for over hours maybe not the only way to do it, but that's what's been in my head as i look at it @leslietownes Oh but you can use E_1, E_2 directly right Is this good? I think this proof is complete 4:37 PM if you want to use E_1 and E_2 'directly' it's up to you. i was working in terms of open sets because mu(anything) is defined as an infimum of values of mu on open sets where we have another characterization of mu, coming from the hypotheses of the riesz theorem Oh right E_1 and E_2 are not open So we can't use them directly @leslietownes Your proof is good for open sets, but how do you solve Rudin's problem from here? mathb.in/55481 @leslietownes Yes, but the result we just proved doesn't seem useful here (or I'm missing something) again, the thing in there needs to be reworked in terms of open sets, not E_1 and E_2 Just assume that the "for open sets" section only cares about the case when E_1 and E_2 are open Then we turn to the general case below How do we proceed? if$U$is an open set containing$E_1 \cup E_2$in the solution outline you're given,$\mu(U) \geq \mu((U \cap V_1) \cup (U \cap V_2))$follows from the deifnition of mu as a sup of values of lambda.$\mu((U \cap V_1) \cup (U \cap V_2)) = \mu(U \cap V_1) + \mu(U \cap V_2)$from that argument about sup(A+B) and the characterization of mu on open sets. and those are$\geq \mu(E_1) + \mu(E_2)$by definition of$\mu(X)$as an inf i'm not assuming E_1 and E_2 open here. there was too much ambiguity in the image about what's being assumed open and what isn't. E_1 and E_2 are just the sets in rudin's statement of the problem Yeah E_1 and E_2 aren't open. Thanks for the help, let me try to write down your argument 4:52 PM @leslietownes would you know how come$\sup(A+\sup B) = \sup(A) + \sup(B)$in the thread you linked? @leslietownes Just to confirm, the first part comes from (U \cap V_1) \cup (U\cap V_2) \subset U right? yeah. shin, i don't know why that solution takes that approach; proving that isn't that different from proving the result directly. sup(A + t) \leq \sup(A) + t is clear because sup(A) + t is an upper bound for A + t [short proof if needed]. the other direction is similarly clear because sup(A + t) - t is an upper bound for A [again, short proof if needed] if$\sup(A + \sup B) = \sup(A) + \sup(B)$, it sort of looks like if you can skip the limit-taking-like procedure 5:10 PM @leslietownes Finally, it is done http://mathb.in/55483 @porridgemathematics Tagging you since you were part of the discussion 5:23 PM @leslietownes actually, yeah. it completely skips over the$\epsilon$part:$(x, y) + z = (x+z, y+z)$. So,$\sup [(x,y) + z] = y+z$, but$y + z = \sup[(x,y)] + \sup[(w, z)]$nifty 5:41 PM 0 (Minimum modulus theorem) Suppose$f(z)$is analytic in a domain$D$and that$f(z)\neq 0$in$D$. Then$|f(z)|$cannot attain a minimum in$D$unless$f$is constant. If$f$is also continuous on$\overline{D}$,$\overline{D}$compact, the$|f(z)|\$ attains a minimum on the boundary. Question i...

I have a question related to maximum/minimum modulus theorem
I think I almost solve it but stuck

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