Real Analysis | The limit point of a set A⊆ℝ

[@arvindsrinivasan424]

This reminds me of a topology class I shouldn't have passed

[@sarojpandeya7883]

Thanks Dr Penn for making amazing videos on mathematical analysis

[@tomatrix7525]

These videos are absolutely awesome. You’ve no idea how many people you actually teach Michael, you are basically my teacher

[@goodplacetostop2973]

16:11

[@azhakabad4229]

Sir, when will you start linear algebra ?

[@AQrdmbmaster]

I’m wondering if you would give some motivation regarding the usefulness of these theorems? Even if you merely say “this helps us prove xyz in the future.” Basically, tell us why these concepts matter in the larger picture. Aside: I’m absolutely loving your real analysis series. I look forward to more episodes!

[@yuklungleung620]

They are just so basic just like 1+1=2

[@schweinmachtbree1013]

limit points are used in order to define limits of functions, for example. in the context of "the limit of f at c is L", c is a limit point of the domain of f (by the definition of a limit point, we can "approach" c, getting arbitrarily close to it, but without actually getting *to* it, which is exactly the behaviour we want for defining limits) :)

[@danmwas7967]

@@yuklungleung620 😂

[@manniman82]

@Michael Penn Are the limit points of a sequence the same thing as a limit point of a set? I am using Terrence Tao's Analysis 1 and 2 and he defined the limit point of a sequence in a metric space (X,d) as follows: ''Suppose that x_n (from n=m to n=infinity) is a sequence of points in a metric space (X,d). Let L be in space X. We say L is a limit point of the sequence x_n if and only if for every N>=m and for every (epsilon) > 0 there exists an n>=N such that d(x_n, L) =n} (as I did in step 2) is not correct? Or are the definitions for limit points of sequence and limit points of sets not the same thing?

[@virat.chauhan]

He is definitely the best math teacher at KZbin!

[@parthmani3877]

Please upload a series of measure theory.

[@elasmarsaadallah6126]

6:42 we can also prove that 0 is the only limit point of A ={1/n, n ≥ 1}

[@elasmarsaadallah6126]

indeed for all aₓ = 1/x we can take any £ < d(1/x, 1/x+1)= 1/(x²+x), for exemple d/2

[@clonebin0]

4:26 nice transition

[@izumiasmr]

I wonder, shouldn't we take some care of the situation that some a_n's might be equal, I mean there's nothing wrong with it even according the very definition of a limit of sequence (the induced sequence), however I think we intuitively expect all the terms a_n to be distinct and I think it's good to point out in the very statement of the teorem that it's not an issue (in the end taking a finite number of a_n's won't work as a sequence converging to x, as a_n's should not be equal to x, whereas the fact that we might end up taking particular elements of the set more than once is not itself a problem)

[@scrappybuilds]

Thanks for all your great instructional videos! I was going to hit the like button but this video is currently at pi likes and I didn't have the heart to change that

[@henrywoo1668]

Thank you sir for your great lecture .

[@gregorymacchio4077]

Your content is amazing!

[@hartandrin7565]

Do we need to invoke the axiom of choice in order to choose some a_n in the (possibly) infinite set that you've constructed in your "if" direction? Is there any proof that you know of that doesn't rely on the axiom of choice?

[@schweinmachtbree1013]

just the Axiom of Countable Choice (since we are choosing a_n for each n, so we are making a countable number of choices), which is relatively weak; it doesn't have dramatic consequences like the ordinary Axiom of Choice, e.g. the Banach Tarski paradox. I am 90% certain that Countable Choice is needed, and I'm sure you can confirm that easily with a Google search

[@youssefamen6872]

Pls do imo 2006 problem 5 thx for the hard work.

[@sumittete2804]

Sir...what are the limit points of the set {cosn: n is any natural number}?

[@anirudhranjan7002]

Is the second theorem correct? I found a counter example but not sure if it's correct.... the sequence a_n = Cos(npi/2) /n and the set A be { +-1/n such that n is natural number } U {0} . Clearly sequence lies inside the set A and coverges to 0 but also multiple terms of the sequence are 0. Given that the reverse condition is not satisfied, we still have that 0 is a limit point of the set A. Did i go wrong anywhere? Can someone help me out......

[@punditgi]

Really helpful to get a handle on point set topology of the set of real numbers. Thanks for getting this ball rolling! One issue that I hope you can clarify is this: Various authors I have read have lumped together limit point, accumulation point, and cluster point. However, I came across at least one author who tried to define all 3 somewhat differently but I can no longer find that reference. Are you able to draw any distinctions among those 3 terms?

[@schweinmachtbree1013]

Wikipedia should help you (or the references therein if you don't like wikipedia itself)

[@johnalley8397]

Another great video. Wonder if you might be on the lookout for opportunities to deliver an informal explanation of the intuition of a thing. The proofs get a bit dry and abstract and sometimes i just need a sentence or two on the jist of a thing to orient my thinking a bit. When I heard the learn’d astronomer, When the proofs, the figures, were ranged in columns before me, When I was shown the charts and diagrams, to add, divide, and measure them, When I sitting heard the astronomer where he lectured with much applause in the lecture-room, How soon unaccountable I became tired and sick, Till rising and gliding out I wander’d off by myself, In the mystical moist night-air, and from time to time, Look’d up in perfect silence at the stars. - WALT WHITMAN

[@yuklungleung620]

Real analysis is rigorous, no informal explanation

[@johnalley8397]

@@yuklungleung620 Brilliant observation. 100% true if you want to reach only the folks generally reached. If you want to expand the appeal of these videos to a wider audience, lead with a sentence or two about the intuition. 3b1b is the only necessary supporting argument here.

[@RandomBoy_0o0]

Sir I think that at 3:43 you meant "2+e/2 is definitely not equal to 2..."

[@hyperboloidofonesheet1036]

Really grinding up the chalk today.

[@noumaneelgaou1624]

Tank you techer for this vidios can you plise explained graph theory

[@cicik57]

emm i may be wrong but you do not need to have epsilon >0 since the Ve ist symmetrical around zero and you need just epsilon not equal to zero

[@rafael7696]

In Spain is called acumulation point

[@cah220]

An interesting question (gleefully lifted from an exercise in baby Rudin) that I'd like to see a detailed solution to (been 20 years since I took undergraduate Analysis). "Construct a bounded set of real numbers with exactly 3 limit points." I get lost in a few of the details.

[@billjoe5991]

The easiest way of solving that that I can think of off the top of my head is by finding 3 sequences that all aren’t “close” to one another and that all converge to different points. You can then just intersect those 3 sequences or smack them together as subsequences into one main sequence, and you’ll have a set with 3 different limit points. An example would be the set of points of the sequences: 1) A= {5-1/n : n in Z} 2) B={10 -1/n : n in Z} 3) C= {20 + 1/n : n in Z} Then the set D = A U B U C has 3 limit points which are {5, 10, 20}

[@debjitmullick7004]

Sir ... Is it necessary, that limit point should be ... In side the set ....???

[@XavierMacX]

No it does not have to be. He addressed this last video :) cheers.

[@lynettemojica6503]

5:30 won't we be out of the interval if epsilon is great than 2. Should we put a cap on epsilon so it stays in the interval?

[@birdboat5647]

this is so hot

[@hybmnzz2658]

AGREED

[@cletushumphrey9163]

yes

[@soranuareane]

Maybe first? Doubt it.

[@failsmichael2542]

That’s a good place to start.

[@soranuareane]

@@failsmichael2542 You win today, Michael, but alas, one day, I will one-up that.

[@muuubiee]

Isn't that definition wrong? This part confused me so much, because the definitions given made no sense, and was obviously flawed. It must contain x AND at least another point other than x... Not just any point which is not x (which this definition suggests), because then every single possible point is a limit point.

[@numericalcode]

Chalk drop

[@james3440]

Ummmmm, What?

[@johnstroughair2816]

About time to move your merchandise off Teespring given their association with extreme rightwing issues?

[@MichaelPennMath]

I haven't heard of this... let me look into it.

[@johnstroughair2816]

Michael Penn Currently trending on twitter that Teespring is selling Free Kyle (Rittenhouse) T-shirts. By the way I should add I am really enjoying your videos, since my son pointed them out to me.

[@kanewilliams3613]

he needs to apply the brakes...!!