Least Upper Bound Property

[@beatoriche7301]

An interesting piece of trivia: the intermediate value theorem, a famous consequence of the least upper bound property, actually also implies the least upper bound property - that is, they are logically equivalent. In other words, it would be equally valid (though I've never read an analysis textbook that does that) to use the IVT as an axiom for the construction of the reals. So saying that "there are no holes in the real numbers," as the axiom of completeness is often stated on an intuitive level, can be interpreted rather vividly using the IVT - in the rationals, a continuous function like, say, _x^2 - 2_ is negative on one side (say, x = 1) and positive on another (say, x = 2) but never actually hits 0 because it basically passes through a hole on the real number line. The completeness of the reals ensures this isn't possible.

[@mrh4t]

Good morning Dr! I broke up recently, and I'm filling myself with math and university stuffs, your last mention touched me! Thank you so much

[@rutvij9]

This course is so much fun. I eagerly wait for the youtube notifications of Dr Peyam and then click the video right away. Thank you for the amazing series

[@erkanbey4504]

you re such a teacher love you so muchhh

[@drpeyam]

Thank you!!!

[@iabervon]

I think, in order to complete that proof, you need to know that there's no smallest rational whose square is greater than 2. It's obvious if you've proved the the rationals are dense in the reals, but not if you've only formalized the rationals so far. In this case, I think the easy direct proof is that, for any rational upper bound, Newton's method will give you a lower rational upper bound.

[@Nikhil_Kumar_Math]

2:53 LUB U 2

[@BlokenArrow]

Last time I was this early, pi was 22/7

[@mariomuysensual]

There is always a better student than me in my courses, I guess I'm not the sup :(

[@drpeyam]

LOL, maybe the sup is infinity!

[@sunshine6119]

Thank you for making this video .❤❤

[@GlorifiedTruth]

The example of the set that does NOT have the LOB really helped this make sense for me. Thanks!

[@mateorestrepo9750]

The def of sup must be wrong, because if S=(a,b) and we let the sup(S)=H=(a+b)/2 the Mindpoint of S then its true that for all M1 < H there exist a S1 in S such that S1>M1 especificly H since by def H>M1

[@mateorestrepo9750]

Thought IT would be eseasly fixed If we just add that for all x in S the sup(S)≥x

[@drpeyam]

No, the definition of sup includes the assumption that M is an upper bound of S, so your (a+b)/2 example wouldn’t work

[@katereggageorgewilliam5908]

Great lesson Dr. Thanks.

[@thomasrascon1086]

Oh my God, He's Kyle from NELK

[@mariomuysensual]

Thanks Dr!

[@dgrandlapinblanc]

Ok. Thank you very much.

[@elosant2061]

4:34 I have read in other texts that a set that isn't bounded above doesn't have a supremum (nor of infinity), unless you consider the extended real numbers?

[@drpeyam]

A set that is not bounded above has sup(S) = infinity

[@hOREP245]

It's a convention thing. It's similar to how when a series diverges to +infinity, we also say the series equals infinity. Obviously, it's not a real number, it's just convention.

[@kevinfung6697]

Hi, Dr Peyam. I got a question which confused me for quite a long time. I saw in a proof in real analysis that the author assume that the open interval is bounded(which is bounded below and above according to what I learned.). So my problem is, isn’t open interval (a,b) always bounded? Why we have to “assume” that it is bounded?

[@drpeyam]

The problem is that (a,infinity) is also an open interval, but it is not bounded. My guess is that the author assumes bounded to make sure to mean (a,b) where a and b are finite

[@kevinfung6697]

Dr Peyam Thank you so much!!! Forgot that (a,infinity) is also an open interval too. XD

[@starter497]

Concerning the set you created in the rationals, why would u not be able to pick some rational number that is very close to square root 2? say something like [sqrt(2)- epsilon] where epsilon > 0 is irrational such that [sqrt(2) - epsilon] is rational. Can we not make some sort of construction for our supremum?

[@drpeyam]

But then sqrt(2)-(epsilon/2) (or something like that) is a rational number bigger than sqrt(2)-epsilon, so sqrt(2)-epsilon cannot be an upper bound

[@hyperboloidofonesheet1036]

@@drpeyam You called the real numbers "complete" for this reason; does this make the integers "complete"? For example, take the set { x ∈ 𝐙 | x² < 2 }; in this case you can say that 1 is an upper bound for this set, since there aren't any integers greater than 1 whose square is less than 2.

[@drpeyam]

Yep, the integers are indeed complete! But they don’t form a field, that’s why they’re not useful for analysis

[@dhunt6618]

Please relate this to Lorne Greene's theorem relating Cylonic integrals to double integrals including the Laplacian to surface integrals :)

[@drpeyam]

What? 🤣

[@nournote]

@@drpeyam Are you aware of Wildberger's criticism of the construction of real numbers?

[@dhunt6618]

@@drpeyam Sorry, Lorne Green was the star of Battle Star Gallactica... Combing Green's theorem I Couldn't resist the bad pun :(

[@tomkerruish2982]

@@dhunt6618 Yes, that was quite a Bonanza of humor.

[@eliyasne9695]

Is it necessary to be uncountably infinite for a set to be compleat?

[@drpeyam]

No, {1} is complete

[@frogstud]

please talk about R-Modules

[@drpeyam]

Ughhhh no

[@Happy_Abe]

By 7:29 why do we say bounded above by 3 and not 2 since sqrt(2)

[@drpeyam]

We could have said 2. Both are upper bounds. Even pi or 5 are upper bounds, but there’s just one least upper bound

[@Happy_Abe]

@@drpeyam thanks!

[@paulfoss5385]

How about for the Cantor set?

[@drpeyam]

The sup is 1, since it is a subset of [0,1] and 1 is in it

[@paulfoss5385]

Dr Peyam And the complementary set of the Cantor set on zero to one?

[@drpeyam]

That’s a great question! I still think that sup is 1, because if M1 < 1, then you can find a point not on the Cantor set that’s between M1 and 1, so by definition of sup, the sup is 1

[@paulfoss5385]

@@drpeyam Okay, that makes sense. Just going with a couple weirder sets to check my understanding of the concept. Thanks.

[@yungegor]

Just lube it up...