Supremum Puns

Doing the proof verbally is obviously easy using two sentences from my lectures but I don't know how to formally prove it.

> Early July 1503

The citizens of Antwerp crowded the streets of the Grote Markt, their shouts and cheers warring with the cries of the gulls overhead. The construction of new guild houses lining the packed streets had stopped for the day, and numerous children ambled up the rickety scaffolds to get a better view over the bustling crowd. Several hundred paces away, long blue, red, and gold pennants flowed in the morning’s seabreeze above the masts of a small flotilla of ships now moored at Antwerps busy docks. Suddenly, trumpets blared, announcing some notable’s presence. The street rats looked on with delight as the princely procession made its slow way through the market streets. Prince Philipp of Asturias, flanked by his councillors and guards, rode at the head of the procession, with a lavishly decorated carriage bearing the arms of Habsburg-Valois and the Tudor Rose following close behind.

When the procession arrived in front of the city’s meeting hall - a rather small, ancient building - Prince Philipp dismounted his mare, and waited for a servant to open the door of the carriage. Out first was the young Karl of Ghent, who refused the hand of the servant and rather indignantly climbed out of the carriage backwards. Following him, Princess Mary of England accepted the help of the servant and gracefully stepped out of the carriage. Both followed the Duke of Burgundy into the town hall.

Inside the building were packed dozens of guild leaders, powerful merchants, and local magistrates, who buzzed eagerly with excitement. Most already knew the contents of the vellum scroll Antoine de Lalaing carried to the table before the Duke, but they still awaited its official reveal with bated breath. Turmoil over the collapse of Hanseatic shipping in England and Denmark had threatened to turn the trade city into a riotous frenzy, and this new agreement with England’s King hoped to settle the matter. Prince Philipp raised his hand, bringing the room to silence.

“Merchants of Antwerp, and the Flemish Coast, I bring to you, on this day, a contract.”

> This Convention, Concordat, and Conclusion, as inspected, agreed, and approved by the Parties Signatory: the Serene King of England, Henry; the Archduke of Austria and Burgundy, Philipp; the Captain-Generals and Admirals of Flanders, Holland, Brabant, &c.; the Mayors of London, Antwerp and other Ports and Cities of the Domains of the others.

> ***Article I of the Convention, Concord

Just learned this. I want to make sure I get it:

The maximum, intuitively, is the biggest number in a group. So for (0,1] it is 1, but for (0,1) it does not exist.

The upper bounds are ALL OF the numbers bigger or equal to every number in the group. For (0,1) and (0, 1] both that'd be 1, 2, 1.5, etc.

The supremum is the smallest number that satisfies the condition of the upper bound. For (0,1) and (0,1] both, it is 1.

Is this true? Kind if feel weird about how the maximum and the supremum can actually be the same thing.

Thanks.

Let A = { a + b : -1 < a <= 0, 2 <= b < 3 }. Determine the supremum of A.

I’m trying to show this without the use of the fact that sup(A) = sup((-1, 2]) + sup([2, 3)).

The claim is that sup(A) = 3

Now since a <= 0 I have that a <= 3 - b so a + b <= 3. This means that 3 is an upper bound for a + b.

Now from the definition I need to show that there is an epsilon for which a + b > 3 - epsilon. Is there any obivous choice for epsilon here or what should one think of?

I just wanna know honestly

Let me walk you through my thought process.

The idea came from the notation of ∪X = {x | ∃y . x ∊ y ∊ X}, which doesn't have a subindex.

∪_(i ∊ I) X_i = ∪ {X_i | i ∊ I}

So we could do the same for say a sum. Where ΣX outputs the sum of all of its elements. eg: Σ{1,2} = 3.

Σ_(i ∊ I) x_i = Σ {x_i | i ∊ I}

So for the sum it's not any set which works. It needs a condition. In this case, having a + defined for its elements.

(You may want to use multisets here.)

So now what about lim? It also has a "subindex".

I thought that you could define the lim of a poset as its supremum. If it has only one, it is well defined. Of course that depends on the order you give to it.

For instance, R with its usual order has ∞ as sup. That is, lim R = ∞.

This should be equivalent to lim_(x -> ∞) x. For x ∊ R.

And lim N = ∞ too, and this should be equivalent to lim_(n -> ∞) n, for n ∊ N.

If we want the lim as x -> a ∊ R, we simply change the order.

We may have -∞ ≤ ... ≤ x - k ≤ ... ≤ x - 1 ≤ x, and also ∞ ≤ ... ≤ x + k ≤ ... ≤ x + 1 ≤ x. So you get convergence from both sides.

We may call this poset R_a .

The trick is that one can apply functions to posets, and induce an order on the image of the function. (x ≤ y => f(x) ≤ f(y).)

This way one could have lim_(x -> ∞) f(x) = lim f(R).

It makes the notation so much more compact and elegant in my opinion.

We would like:

lim_(x -> ∞) f(x) = lim {f(x) | x -> ∞},

whatever that last "x -> ∞" means.

But here lies the problem. Take f:N->R s.t. f(0) = 0 and f(n) = 1/n for n > 0.

Here the limit should be 0. However this notion of supremum doesn't do that. It gives you the whole set, since everyone is ≤ than everyone else.

I believe, maybe not unexpectedly, that some topology is needed. Because we care about a supremum but only locally.

Do you know how to give the proper definition in these terms?

I have this Problem:

>Let (a_n) be a bounded sequence.
>
>(a) Prove that the sequence defined by y_n = sup{a_k : k ≥ n} converges.

My question is: how does this sequence y_n look like, or what are its elements?

The supremum of a_n is well defined, right? There is ONLY ONE supremum of a_n. So should the set y_n only contain the sup of a_n?

Or maybe there are different supremums, depending on the subsequence a_k? And so each element y_k is just the corresponding supremum of a_k? That would only work with a_n being monotone and decreasing, right?

Or maybe something else?

Thanks for your help.

(And if you have a hint to solve the rest of the exercise, feel free to share it)

Prove/Disprove.

Any sequence a of n, n=1 to infinity has either an infimum or supremum or both.

The supremum has to be greater than or equal to all the elements in the subset, right? Then technically, even a ω-chain with all elements the same has a supremum. So what does it mean to be a ω-complete partial order?

The supremum axiom states that Real numbers must always have a supremum, and from there we can get the infimum axiom that states that we always get an infimum in R (that's at least what I learned). But my question is, you can't get an infimum in -1≤r² can you? So does this mean the infimum axiom is false?

Think that although this arc is showing an advanced form of Conqueror's, this isnt the most advanced form. It is simply a capstone to Armament haki, which was what the arc was about.

A supremum in mathematical terms is defined as the least upper bound of a set. The key difference between simply calling it a maximum is that the supremum has no obligation to actually be within the set. For example, the square root of 2 can never be expressed as a fraction, but the set of all fractions whose square is less than 2 have it as their supremum.

So CoC coating is the completion of CoA.

So if Luffy has an arc with Shanks, he will probably learn a few more tricks that he can do with CoC. Afterall, as a Yonkou, Shanks needs some way of rolling up to a threat in his territory and altering the battlefield like a god.

So this isnt the end of CoC's development.

Suppose A, B ⊆ R is nonempty

If for all x ∈ A, y >= x for some y ∈ B. I know I can say that y is an upper bound for A, what do I also need in order to conclude that y>=sup(A)?

Apocalypsis in Cassari: Supremum Lupos

At some point in the past, I made a fan RPG based on Cultist Simulator (and Fallout, but that's not important right now). It was based on Shadowrun, Dungeon World, and some Word of Darkness. My main gaming group enjoyed the game I made. I decided to run it again, adding some bits, and making the story a bit more interesting (plus I was able to add more things thanks to Exile and knowing history). This time, I decided to share with people what we are doing in case people are interested. If you want to check out the fan RPG, it's here. In any case - the time is: late night in November, 1932. The place: Alsace, France.

Session 1:

The party start out as a large wolfpack. There are almost a dozen of them and they are the last wolfpack in France. Food is getting scarce, true, but on the other hand, their territory is huge, which means that they are able to roam as far as necessary. It is their downfall as the pack is seized with some kind of invisible force, binding their legs, tails, and muzzles. About a dozen figures emerge from the forests, led by three. One smells like raw blood and birthing fluid, another smells like a frozen corpse, and the third smelled like ash, obsidian, and a smell that none of the wolves had smelled before.

The pack was dragged by regular humans to a clearing where several cars were idling and split into three groups. The smallest group of three went to the amniotic fluid guy, eight of them went to the frozen corpse guy, and one went to the strange-smelling guy. The party was the group of three. The car drove off and took about two-and-a-half hours ((that they were able to infer by the position of the Sun)) when they were carried by the birthing blood guy and an assistant into cages specially prepared for them.

Thus began the strange captivity of the wolves. On the very first night the man came in with a knife, seizing one of the wolves with the strange magics yet again. He dragged the wolf, not gently, but not painfully either into his workshop, where he stuck the knife into the wolf's flesh and began peeling the skin off. Surprisingly to the wolf, it wasn't painful, just very itchy and cold. Then everything shifted, as the wolf suddenly lost it's sense of smell and its vision became sharper and full of colors. It did it's best to look around with eyes and saw that it looked similar

I am asked to find a counter example to the following statement:

The supremum of a bounded above set is the greatest of its limit points.

I'm working in R1. I have thought for a while, but really cannot think of anything.

So my teacher explained us this but none of my class understand this and i don't know how to calculate this. Can someone explain me or advice a video or a book for me?

Sorry for bad english.

Can someone explain the concepts of infimum and supremum like to a 3 year old?

My understanding so far is that we can view them as min and max equivalently. A bit more formally, infimum is the largest lower bound while supremum is the least upper bound.

The confusion arises from texts and lecture notes in statistical learning theory where learning is defined as the empirical minimisation risk.

So good so far

Here's where things get a bit fuzzy for me, the definition of ERM is to choose f with the smallest risk such that:

f* = arg inf of R(f) for f: X-->Reals

From the sake of the argument, lets say we represent the set of all functional outputs of R with the following set F of numbers {-18, -20, -5, 0, 3, 15}.

What would be f*?

What would be inf of F?

Thanks!

Let An be a sequence of real numbers such that An <= An+1 (a sub n less than or equal to a sub n +1) for all n, and suppose alpha is the supremum of the sequence. Show that for all epsilon > 0, just finitely many terms are <= (less than or equal to) alpha - epsilon

A few questions, am I correct in assuming the supremum is the limit of the sequence? To me the problem sounds like a rearrangement of the convergent sequence definition, as in I need to find an n to make this true. I'm not sure where to start, but I was thinking of making epsilon the distance between successive terms, but I'm not 100% sure, the "finitely" many terms part is throwing me off. Any guidance would be appreciated, and this is for a real analysis course.

Thank you!

Consider a set S. Sup(S) = Max(S) or it equals the limit of a sequence composed of members of (S) that converges to sup(S). Moreover, if sup(S) exists and does not equal max(S), we can assume that such a sequence exists.

Is this a fact that you can take for granted and use to prove other theorems, or is this something that needs to be proven? If so, is the proof a challenging one or is it fairly simple?

I'd like to use this fact to prove other theorems, but something tells me that that would be too easy.

A = {n/(n+1) | n is part of N*}

The function is strictly increasing which means inf(A) = f(1) = 1/2.

However I also need to find the supremum for my homework, and I can't see any other way than by calculating the limit of n as it goes to infinity, which we weren't taught yet so I'm assuming there's a way to do it using only algebra etc. so how can you do it?

I just don't understand how to get the sup of lower sums. I understand that it is supposed to be the greatest possible lower sum but how is that calculated?

Without using that we can find c,C > 0 s.t. one norm is bounded in the other. I guess we must use open sets?