Least Upper Bound Property

Least Upper Bound Property

In this video, I state the least upper bound property and explain what makes the real numbers so much better than the rational numbers. It's called Real Analysis after all!

[beatoriche]

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.

[mrht]

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

[rutvij]

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

[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.

[erkanbey]

you re such a teacher love you so muchhh <3

[BlokenArrow]

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

[mateorestrepo]

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

[elosant]

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?

[mariomuysensual]

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

[GlorifiedTruth]

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

[starter]

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?

[kevinfung]

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?

[dhunt]

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

[eliyasne]

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

[Happy_Abe]

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