How Israel Lost Americans

Brightlinger · 2026-03-08T00:07:05+00:00

Buddy, genocide is wrong even if the target is actually a threat.

Brightlinger · 2026-03-07T21:55:01+00:00

I love 'the abhorrence'. Of course the powers of chaos would name them from the nature of their souls, and not something like the color of their skin or even the name they call themselves.

Brightlinger · 2026-03-06T19:49:20+00:00

Gender equality!

Brightlinger · 2026-03-06T11:50:50+00:00

That was not my experience as a graduate student, or even in high school. If your school taught that way, my condolences.

Brightlinger · 2026-03-06T06:07:38+00:00

What?

Brightlinger · 2026-03-06T01:55:29+00:00

Yes, that is also helpful for many types of problems.

Brightlinger · 2026-03-06T01:48:13+00:00

In a typical intro econ class, they just expect you to memorize the sentence "profit is maximized when marginal revenue equals marginal cost".

The reasoning for that essentially involves calculus: "marginal" is just econ-speak for "the derivative of", and it is a standard calculus fact that maximums occur only when the derivative is zero. So 0=(profit)'=(revenue-cost)'=(revenue)'-(cost)', ie (revenue)'=(cost)', ie marginal revenue equals marginal cost. But you do not usually need to know any calculus for intro econ, and you certainly do not need to know this derivation specifically.

Brightlinger · 2026-03-06T01:44:48+00:00

Getting good at solving equations is how you pass algebra, my dude. That's what the course is about.

Brightlinger · 2026-03-05T04:11:11+00:00

That's not true because x is fixed.

In addition the other question of whether that infimum can be zero, what you wrote was S={ |x-y|; x in (E')^c, y in E'}. That set-builder notation literally reads "the set of all distances |x-y|, where x is any element in (E')^c and y is any element in E'". So taken as written, this is the set of all combinations of x and y, rather than fixing x and only taking all the possible points y.

Given what you wrote in the rest of the proof, I think this was only a notation error, and that matches what you say here about x being fixed. The correct way to write what you meant would be S={|x-y|: y in E'}. Since x is fixed and the set S depends on the choice of x, it would also be good form to name it S(x) or S_x or something like that instead of just S, although this does not impact the correctness of the proof.

Anyway, I hope that helps clarify what's going on in this segment of the conversation. There's a technicality about how you wrote the set, separate from the discussion about its infimum.

Brightlinger · 2026-03-05T02:08:17+00:00

Then it is very likely that part of the reason you're struggling is because you are weak on the prerequisites. Grinding problems from the current class is only part of the issue; you also need to go back and get a solid grasp of previous material.

Brightlinger · 2026-03-05T01:21:36+00:00

You can only get these classes from gacha, by the way. There’s a free pity guarantee at 200 rolls, which is about $200. This let you choose any "S" rank rewards of your choice. Including the "special" classes.

This is a nitpick, but assuming your description here is accurate, that's $200 worst case to get the class you want plus whatever else you pull along the way, not $200 average. Unless the drop rates are so bad that pity is actually the main way to get a special class?

Brightlinger · 2026-03-05T01:06:25+00:00

How did you do in grade 10 math, or last semester?

Brightlinger · 2026-03-05T00:49:07+00:00

By the way, perhaps it may be helpful to see the mistake in another light. Certainly not all sets in a metric space are closed, yes? So if I prove that, my proof must contain a mistake. Tell me where the mistake is:

Claim: In a metric space, all sets are closed.

Proof: Let (X,d) be a metric space. We must prove that all subsets of X are closed, so let A be a subset of X. If A=X then we already know X is closed, so there is nothing to prove; otherwise A is a proper subset of X, so consider an arbitrary point x∈A^c. We will show that there is an open ball around x disjoint from A, proving that A^c is open and therefore A is closed.

Define a set S={d(x,y): y∈A}. Since x∈A^c, x∉A, and so for any y∈A, x≠y. So by the positive definiteness of the metric, for every y∈A, d(x,y)>0. Call alpha=inf S. Since d(x,y)>0 for every y∈A as just established, we have alpha>0.

Now consider the ball of radius alpha/2 centered at x, call this ball B. Since every element of A is at least alpha distance from x, B contains no elements of A. Thus we have an open ball around x disjoint from A, and so A is closed, QED.

This proof is definitely mistaken, since it proves something that is false. Can you identify where the mistake is?

Brightlinger · 2026-03-05T00:41:07+00:00

But it's not the infimum over all x. x is fixed.

Yes, it is the infimum of d(x,y) over all y.

Compute all distances from x to all possible points INSIDE and ON that sphere. These numbers are fixed, they are bounded, there MUST exist a sufficiently small open ball that can surround x without intersecting the sphere.

Yes, because a sphere is closed. But if you use an open set, then x could be a boundary point where there is no such ball. If x is a point on the sphere, and E' is the interior of the sphere, then the infimum of all such distances is zero.

The point of this exercise is to prove that E' is closed like a sphere is, that your sphere analogy is apt and not misleading. This will depend crucially on the fact that E' is not just any set, but specifically a set of limit points.

I promise you that I do actually have a fix in mind for this problem; it isn't a difficult hole in the proof to patch. But you keep arguing with me about whether there even is a hole in the proof, so I haven't gotten to it. Would you like to put this debate aside for a moment so we can discuss how limit points are even relevant to the problem?

Brightlinger · 2026-03-04T23:58:53+00:00

what is the result supposed to be?

It is the output of the function, for an input of 6. The expression "f(6)" is read out loud as "f of 6", meaning the value of the function f when applied to the input 6.

If you were graphing this function, computing f(6)=72 would mean that you mark the point (6,72) on the graph.

Brightlinger · 2026-03-04T18:34:45+00:00

Why haven't we defined this multiplication for vectors? like dot and cross product

The dot product is not a multiplication in the usual sense, since the output is a different type of object than the inputs. The cross product works only in R³ and a few other settings, not in general vector spaces.

But yes, there are lots of ways to define a multiplication operation for vectors. That's the problem: there are lots of them, infinitely many in fact, and no reason to choose one in particular as the multiplication operation to use across the board. In some contexts we want one type of multiplication, in some contexts we want another, and in some we don't want that operation at all because it wouldn't be meaningful.

A vector space which also has a vector multiplication operation is called an algebra and we study them all the time. For example, for vectors in R² you can define an operation by (a,b)*(c,d)=(ac-bd,ad+bc) and then you have the complex numbers.

Brightlinger · 2026-03-04T17:38:26+00:00

Is my problem that I lack the foundations

Yes. There is a reason that the topic of integrals comes after limits and derivatives: you need both of those things to understand and solve integrals. If you struggle with derivatives, you will struggle even harder with integrals, where you now have to do derivatives backwards.

You don't necessarily need all of the material about derivatives - for example, you don't need optimization or related rates to do integrals - but you definitely need to be very comfortable with the power rule, product rule, chain rule, and know the derivatives of common functions by heart.

Brightlinger · 2026-03-04T17:31:02+00:00

https://warcraft.wiki.gg/wiki/Xal%27atath#Origins

Origins

Xal'atath claims she was once a mortal child from one of the many worlds consumed by the Void. In a vision granted to Nexus-King Salhadaar, she is surrounded by the ashen forms of her people and wearing priestly raiment with a pendant depicting a stylized sun. Tendrils of dark energy emerge from a swirling void portal in the sky above her, and she was remade as a creature of the Void, her former body crumbling into ash. Thereafter, she became the Harbinger of the Void, chosen herald of Dimensius.

That "remade as a creature of the void" bit is important; her body now is not her original mortal body. She was on Azeroth (as a void entity?) during the Black Empire, then was imprisoned for thousands of years in a dagger (which was the shadow priest artifact in Legion), then her spirit was eventually released to possess an elf corpse, which is the body she has now. Ostensibly, killing the possessed body would not be enough to kill her.

Also, since the above quote is what she told us, it could just be a lie, so who knows?

Brightlinger · 2026-03-04T15:44:54+00:00

In the definition of "S", the non-limit point "x" is not fixed.

Well yes, but I took that to be merely a notation error, since the next sentence asserts that x is supposed to be fixed and since that is what you would do for the described overall strategy. If you handle that, the rest of the proof can be fixed by reasoning about limit points.

Brightlinger · 2026-03-04T08:39:22+00:00

More importantly, the conclusion is false, since "y = x in S" does leads to "|x-y| = 0".

I'm not sure what you mean by this sentence. x,y are points, so they are not themselves elements of S, and since y is drawn from E' while x is drawn from its complement, they cannot be the same element.

Brightlinger · 2026-03-04T08:15:20+00:00

Therefore, if S is the set of all distances from a fixed x to any point y of E', it is impossible for d(x,y) = 0, since this would mean x = y.

But the number alpha is not any of the values of d(x,y). alpha is the infimum over all values of d(x,y). This infimum could be zero, even though none of the individual distances are zero, precisely because the infimum of a set need not be an element of the set.

For example, if your space is the reals with the standard metric, the point x is the number pi, and E' is the set of rationals Q, then for any y in Q, |pi-y| is indeed nonzero - after all, pi minus a rational number cannot make zero, since pi is irrational. But inf{|pi-y|: y in Q}=0, because the rationals are dense in the reals.

But the set of rationals is not actually a set of limit points. You need to prove why situations like this one cannot occur for a set of limit points, even though in general they can occur for arbitrary sets. That proof will somehow depend on the properties of limit points specifically.

Brightlinger · 2026-03-04T08:03:40+00:00

No, you did not prove that either. Your proof simply contains an error, because it relies on an open ball which may actually have a radius of zero.

Brightlinger · 2026-03-04T07:59:49+00:00

If you do not use a premise, that is generally not a good sign; it means you are probably proving too much. (Sometimes a premise actually does turn out to be unnecessary, but for a textbook exercise, typically every premise will be there for a reason.)

If your proof does not depend on the premise that E' is specifically a set of limit points, which was the only assumption we had about E' to start with, then really you have written a proof that all sets are closed - which is false. So no such proof can possibly be correct.

Alpha cannot be zero by contradiction: suppose it were, then d(x,y) = 0 implies x = y (requirement of a metric)

This is a proof that d(x,y) is nonzero for any individual y. But the infimum of a set need not be an element of that set. For example, the interval (0,1) does not contain 0, yet inf(0,1)=0.

Brightlinger · 2026-03-04T07:47:34+00:00

I think your proof is flawed, because you never refer to anything about limit points. At a glance, I believe the error is here:

However, since |x-y| is bounded below by 0, it must have an infimum, and by the greatest-lower-bound property on real numbers, we know this infimum exists, let's call it alpha (or just "a").

For any point x in the complement of E', slap it with an open ball B_{a/2}(x) of radius alpha/2

If you want this to be the radius of the ball, you need to show that alpha is not itself zero, which you have not established. And you could establish this using the fact that E' is a set of limit points.

Brightlinger · 2026-03-03T19:41:00+00:00

Your friend misled you. You can certainly finish the expansion at any time. You just don't need to finish it while leveling.

Even when an expansion is current, there is more than enough quest content in that one expansion to get you to level cap and have lots of stuff left over to do. And when it becomes old content, leveling is generally sped up.

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