interesting, i never realized that having "for any" before "there exists", as in "for any ε>0 there exusts δ>0, s.t....." is strictly stronger than the reverse order.
@@Meghana_Nallamilli"for any" and "for all" are equivalent. For all x there is y, p(x, y) means that for any x that you choose, exist an object y that satisfies p(x, y). For example, for x=X1, there must be y=Y1 that satisfies p(X1, Y1). Similarly for x=X2 m, there is an y=Y2, p(X2, Y2) Now, there is a y that for all x p(x, y) means that there ab object X, that satisfies p(x, y), no matter which y you choose. It's the same object x that works for all y. In contrast with the other expression that all y has an x, but not necessarily the x's are the same for all y.
Most of the problems in my "Intro to Higher Math" playlist come from Daniel Velleman's book "How to Prove It" Third Edition. This problem, I believe, is not in the book, but it best resembles a problem you would see in chapter 3.
Counter example: Define P(x,y) as y = x+1. So this is basically true if x is one less than y. Since every number has a number one smaller, the "if" part is true. But it is not true that there is a number that is one less than all other numbers.
![∀x(P → Q(x)) iff P → ∀xQ(x) Proof [ILIEKMATHPHYSICS]](http://i.ytimg.com/vi/D2VQUO7zo6c/mqdefault.jpg)
![∀x(P → Q(x)) iff P → ∀xQ(x) Proof [ILIEKMATHPHYSICS]](/img/tr.png)
![Prove if lim(xn+1/xn) ﹤ 1, then lim(xn) = 0 [ILIEKMATHPHYSICS]](http://i.ytimg.com/vi/Ih5iQ73KFsE/mqdefault.jpg)
![Why you didn't learn tetration in school[Tetration]](/img/1.gif)
Could you please also do a video about jacobin Determinants and their origin, thanks.🎉
Uniform convergence implies pointwise convergence
interesting, i never realized that having "for any" before "there exists", as in "for any ε>0 there exusts δ>0, s.t....." is strictly stronger than the reverse order.
It’s actually the opposite right?
In the video here, we’re talking about “for all” and not “for any”. I don’t get the notion of “there exists delta >0 for any epsilon > 0”
@@Meghana_Nallamilli"for any" and "for all" are equivalent.
For all x there is y, p(x, y) means that for any x that you choose, exist an object y that satisfies p(x, y). For example, for x=X1, there must be y=Y1 that satisfies p(X1, Y1). Similarly for x=X2 m, there is an y=Y2, p(X2, Y2)
Now, there is a y that for all x p(x, y) means that there ab object X, that satisfies p(x, y), no matter which y you choose. It's the same object x that works for all y. In contrast with the other expression that all y has an x, but not necessarily the x's are the same for all y.
i enjoy this videos, by what book these problems comes from?
Most of the problems in my "Intro to Higher Math" playlist come from Daniel Velleman's book "How to Prove It" Third Edition. This problem, I believe, is not in the book, but it best resembles a problem you would see in chapter 3.
@@iliekmathphysics thank u
I think, probably, the reverse version ∀y ∃x P(x,y) ⇒ ∃x ∀y P(x,y) is also true.
It's not true
Counter example:
Define P(x,y) as y = x+1. So this is basically true if x is one less than y.
Since every number has a number one smaller, the "if" part is true. But it is not true that there is a number that is one less than all other numbers.
@@toddblackmon Thanks! I got it now.