Real Analysis | Open subsets of ℝ.

[@cletushumphrey9163]

I believe the mistake at 6:10 is putting a minus sign in front of the parentheses instead of a plus

[@izumiasmr]

I wonder however do we need 1/2 coefficient when picking ε, just getting the minimum seems to be enough 🤔

[@JMac___]

@@izumiasmrthe minimum is sufficient, but in the case of taking just the minimum you would have to do case work to show the end result. Ie, case min:=a-b and case min:=c-a. By taking the 1/2, we have that 1/2min{a-b, c-a} is less than both a-b and c-a automatically, which is nicer to work with in the end inequality.

[@sanjursan]

Michael Penn and the three "C's" of Real Analysis. Clear, Concise, and Comprehensive.

[@lucascaique2943]

This is such a great playlist that I'm actually looking forward for any new video.

[@loopingdope]

Like a netflix series, but better

[@Manuel-pd9kf]

This deadass is a great playlist, keep it up!

[@murielfang755]

life saver playlist. Clear explanation and nice speed.

[@thesecondderivative8967]

12:32 I believe we can make epsilon as small as we want since the intersection of open intervals is open.

[@PMBUNESA-wj3li]

Why are you take the epsilon as the half of the minimum? Not just the minimum?

[@goodplacetostop2973]

18:32

[@e.m.winter500]

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.

[@abhijitharakali]

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.

[@willyh.r.1216]

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.

[@spencerpencer]

R is most certainly not compact my friend

[@willyh.r.1216]

Thank you for correcting me. It's been a long time I did this real analysis.

[@bobajaj4224]

@@spencerpencer that's true, that's why we used the Alexandroff's extension

[@bobajaj4224]

and the limit of rational means that Q is dense in R

[@willyh.r.1216]

@@bobajaj4224 Alexandroff compactidication of R. I have recollection of that.

[@xoppa09]

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.

[@jonaskoelker]

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.

[@izumiasmr]

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

[@hopegarden7636]

Great video as usual also them gains are showing itself

[@tomatrix7525]

These are really good, thanks!

[@BlueRobair]

Thank you !

[@GKinWor]

so helpful

[@__hannibaalbarca__]

I love General Topology; and A Counterexample in GT

[@ishaangoud3180]

Is this topic connected to Metric Spaces?

[@freddyfozzyfilms2688]

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

[@사기꾼진우야내가죽여]

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.

[@freddyfozzyfilms2688]

@@사기꾼진우야내가죽여 The fact that each set is open means that the epsilon has been chosen for us right?

[@matsnordstrom8584]

Is this course or playlist following Rudin's "principles of mathematical analysis "? Great work. Will follow!

[@spicyy812]

late reply, but its following Abbotts understanding analysis.

[@izumiasmr]

​@@spicyy812thanks 🙏

[@moorsyjam]

For the proof of the finite insection of open intervals, doesn't that only hold if the intersection is non-empty?

[@griffine6111]

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.)

[@moorsyjam]

@@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.

[@lucascaique2943]

It's implied that the intersection is non-empty, since we know the empty set is open.

[@thesecondderivative8967]

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.

[@rafael7696]

It's a very simple concept

[@rafael7696]

@@mr.knight8967 very easy too

[@sthetatos]

Take a=Pi and epsilon = sqrt of 2. How to calculate (a-epsilon, a+epsilon)? Is this neighborhood well defined? Thanks.

[@vardaandua3585]

Sir please tell how to find the integral of (x/(xsinx +cosx ))^2 without using integration by parts

[@romeoaubrey4119]

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]

Yes, the empty set is closed

[@tahafakhech7712]

This will be the greastest playlist of all time

[@arvindsrinivasan424]

🔥🔥🔥

[@athelstanrex]

im first

[@mrrashedali]

And I'm second