Problems in Real Analysis | Ep. 2

[PhDVlog]

For the third problem, I confused myself with the epsilon towards the end of the proof. Although the result still holds, my reason for the epsilon being there was incorrect.

[gabrielherman]

You’ve become on of my favorite YouTubers in a matter of days. Keep the content coming

[Leo_.]

Man i’m a mechanical engineering and this channel is like a drug I swear. Public everyday mate, ur videos are so relaxing

[Jaeghead]

24:00 I think what happens here is that we split up the integral again: On [1, 1/ε]\E the integrand is 1/x² which is less than 1, and integrating 1 just gives us m([1, 1/ε]\E]) (without the interval (1/ε, ∞) as that has Lebesgue measure ∞). The integral over (1/ε, ∞) of 1/x² we can calculate directly, giving us the "+ ε". So it does not come from the size of the set but from the value of the integral of 1/x².

[shuhaotian]

13:15 I think it feels a bit off since you technically need to specify two different Ns: an N_0 that you pick first such that n>N_0 implies |b_n|<ε/2, and then another N_1 such that n>N_1 implies a_n<ε. N_0 and N_1 don't necessarily have to be equal from the outset, but they were both ambiguously denoted by the same symbol N in the video. So to make it clearer, you can specify that N_0 was chosen first such that |b_n|<ε/2, then N_1 was chosen later such that S_1<ε/2 and |b_n|<ε/2 at the same time, i.e. N_1 > max(2(1+b_1+...+b_(N_0-1))/ε, N_0)

[bonky]

I am an undergraduate student taking a symbol logic class. We just learned about De Morgan's and seeing it used here was pretty cool I think.

[jorex]

Love this type of videos! I would like to see other series of this type for different branches of maths as well! (maybe for linear or abstract algebra, topology, differential geometry or functional analysis.) Keep up the good work!

[VinOnline]

This is completely incomprehensible to me LOL

[larryyonce]

It helps to have a passion for inequalities!

[xaxuser]

these actually are really easy problems, the first one is really an application exercise on our course of topology in my second preparation years for greath engineering schools in france

[mycatsnameisgizmo]

I’m a high school student who’s not currently even on the Calc Train and it’s legitimately like watching someone speak a different language. Interesting nonetheless but I have not even an iota of an understanding lmao😂

[jidrit]

i am not a student but i am doing from book understanding analysis by abbot. i am on continuity chapter. so still a newbie. hope i will reach your level one day

[simonreiff]

For #2 I would finish it off by saying that since |b_n| tends to zero as n grows large, and since (S_1]/n clearly tends to zero for all n greater than or equal to N for a sufficiently large N, we can choose N such that (S_1)/n < (eps)/2 and |b_n| < (eps)/2 for all n greater than or equal to N. Then, since (S_2)/n is less than or equal to ((n-N)/n)*[(eps)/2]=(eps)/2 by the triangle inequality, we have that a_n is strictly less than ((eps)/2)+((eps)/2)=epsilon, so |a_n| < epsilon for all n greater than N, as required.

[stratosvas]

Where can I find those problems with solutions?

[dilation]

I used to real analysis proofs especially in limits and continuity, sometimes the epilson delta proofs were kinda annoying to do and I abandoned them ( i used to study real analysis due to interest) there is something very special about real analysis which is basically purely on the concept of sequences and series. Though I must say something like differential topology is harder real analysis, kinda has a special place in the SET of hard topics in matsh.

[aravartomian]

For problem 1. I have a direct proof. We know that the irrationals are uncountable and hence can not be written as the countable union of countable sets. Therefore, one of the closed sets has to be uncountable. Taking that set, F, we know that the rationals are dense in R and therefore F can’t have a nonempty interior. But this implies that F is the disjoint union of singleton sets and also uncountable which we cannot write it as the countable union of closed sets. Thus, our original countable union of closed sets contains a closed set which cannot be written as the countable union of closed sets.

[Olexquicity]

Great videos! I really enjoy all of them. Just one question, what kind of pen do you use?

[ahmetboran]

GOD BLESS YOU....ALHAMDULILLAH MASHA ALLAH..