I believe the mistake at 6:10 is putting a minus sign in front of the parentheses instead of a plus
@izumiasmr10 ай бұрын
I wonder however do we need 1/2 coefficient when picking ε, just getting the minimum seems to be enough 🤔
@lucascaique29434 жыл бұрын
This is such a great playlist that I'm actually looking forward for any new video.
@loopingdope4 жыл бұрын
Like a netflix series, but better
@sanjursan2 жыл бұрын
Michael Penn and the three "C's" of Real Analysis. Clear, Concise, and Comprehensive.
@Manuel-pd9kf4 жыл бұрын
This deadass is a great playlist, keep it up!
@jonaskoelker2 жыл бұрын
Note that for the finite intersection of open sets, it's enough to show that the intersection of two open sets is open. [This principle holds for any class of objects closed under some combination of two of them: from two you get any finite quantity.] If it holds for two, it follows that (((U_1 * U_2) * U_3) * U_4) is open, and so on by induction [where * is the combination of two objects, here the intersection of two sets]. This would probably make for a slightly easier proof, at least with respect to notation.
@izumiasmr10 ай бұрын
Thanks nice remark! I guess Michael's approach might be in a way instructive sort of low level approach, and then it comes as a bummer what you said, sort of automating via induction
@e.m.winter500 Жыл бұрын
Very much appreciated all the editing done to save a few seconds here and there. Must have been a lot of work but it makes the video much more smooth. Beautifully done thank you.
@willyh.r.12164 жыл бұрын
What a beautiful refresher for me Michael. Thank you. This reminds me my first college math back in 1987, french math curricula. Dedekind's approach is also very interesting for the construction of R. R is bounded and complete (with Bolzano-Weirstrass theorem). All Cauchy sequences are convergent. And we can also prove that any real number is a limit of a rational sequence. Meaning, the set of rational numbers denses in R. Those key results came back spontaneously to my mind while watching your video. Thank you.
@spencerpencer4 жыл бұрын
R is most certainly not compact my friend
@willyh.r.12164 жыл бұрын
Thank you for correcting me. It's been a long time I did this real analysis.
@bobajaj42244 жыл бұрын
@@spencerpencer that's true, that's why we used the Alexandroff's extension
@bobajaj42244 жыл бұрын
and the limit of rational means that Q is dense in R
@willyh.r.12164 жыл бұрын
@@bobajaj4224 Alexandroff compactidication of R. I have recollection of that.
@murielfang7553 жыл бұрын
life saver playlist. Clear explanation and nice speed.
@abhijitharakali4 жыл бұрын
Thanks Prof. Penn. I'm glad you are making these videos. They are valuable for us and I hope you'll continue to post such videos.
@thesecondderivative8967 Жыл бұрын
12:32 I believe we can make epsilon as small as we want since the intersection of open intervals is open.
@goodplacetostop29734 жыл бұрын
18:32
@xoppa09 Жыл бұрын
This guy is brilliant and his proofs are unassailable ( i cant find any mistake). He is the like jesus of math , saving undergrad math majors from failing. :P I don't even have to speed up the video because he moves so quickly through the proofs. Never a dull moment.
@PMBUNESA-wj3li10 ай бұрын
Why are you take the epsilon as the half of the minimum? Not just the minimum?
@hopegarden76364 жыл бұрын
Great video as usual also them gains are showing itself
@BlueRobair2 жыл бұрын
Thank you !
@__hannibaalbarca__4 жыл бұрын
I love General Topology; and A Counterexample in GT
@tomatrix75253 жыл бұрын
These are really good, thanks!
@GKinWor2 жыл бұрын
so helpful
@rafael76964 жыл бұрын
It's a very simple concept
@rafael76964 жыл бұрын
@@mr.knight8967 very easy too
@vardaandua35854 жыл бұрын
Sir please tell how to find the integral of (x/(xsinx +cosx ))^2 without using integration by parts
@ishaangoud31802 жыл бұрын
Is this topic connected to Metric Spaces?
@sthetatos4 жыл бұрын
Take a=Pi and epsilon = sqrt of 2. How to calculate (a-epsilon, a+epsilon)? Is this neighborhood well defined? Thanks.
@matsnordstrom85844 жыл бұрын
Is this course or playlist following Rudin's "principles of mathematical analysis "? Great work. Will follow!
@spicyy812 Жыл бұрын
late reply, but its following Abbotts understanding analysis.
@izumiasmr10 ай бұрын
@@spicyy812thanks 🙏
@romeoaubrey4119 Жыл бұрын
Im confused. I thought the empty set is equivalent to the complement of R and so since R is open then it's complement which is the empty set is closed. Am I wrong?
@Lucashallal Жыл бұрын
Yes, the empty set is closed
@freddyfozzyfilms26883 жыл бұрын
when take an epsilon from each set in the finite intersection, does this step require the axiom of choice? Since there could be an uncountable number of epsilons
@user-uw1ut4ss2q3 жыл бұрын
Although there exist uncointably many choices of epsilon for each set, there are only finitely many sets from which we choose epsilon, so we don't need the axiom of choice . I think the choice of epsilons from each of finitely many sets can be done by mathematical induction.
@freddyfozzyfilms26883 жыл бұрын
@@user-uw1ut4ss2q The fact that each set is open means that the epsilon has been chosen for us right?
@moorsyjam4 жыл бұрын
For the proof of the finite insection of open intervals, doesn't that only hold if the intersection is non-empty?
@griffine61114 жыл бұрын
If the intersection of a finite number of open sets is the empty set, that still works since the empty set is open! If you want to talk about the "empty intersection" which is when you are intersecting no sets, this is the whole set, which is also open. (This is like multiplying no copies of a number together and getting 1. Written usually as x^0 =1.)
@moorsyjam4 жыл бұрын
@@griffine6111 I get that the theorem still holds, since he showed the empty set is open. He's just taking an element of the intersection for the proof, which is kinda difficult if it's the empty set.
@lucascaique29434 жыл бұрын
It's implied that the intersection is non-empty, since we know the empty set is open.
@thesecondderivative8967 Жыл бұрын
I believe the proof implies that the intersection is non-empty. If the intersection were empty, then we use the fact that the empty set is open.