Real Analysis | A convergent sequence is bounded.

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We define the notion of a bounded sequence and prove that every convergent sequence is bounded.
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13 окт 2024

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Комментарии : 55

@zaratustra2363 4 года назад

I really love how topics on this channel are in range from competitive mathematics and puzzles to academic mathematics, rigorous stuff, proofs and such. I was searching for something like that on YT ^^

@roberttelarket4934 4 года назад

Thus spoke Michael!

@YaStasDavydov 4 года назад

Hey, I’d love to see a video on Fubini’s theorem. In many videos we just blindly use it to interchange the order of integration, but I think it would be cool to look at the exact statement and maybe some details

@AKhoja 4 года назад

"And that's a good place"

@heewahhin7470 Год назад

Let a_n be a convergent sequence and let lim n →∞ a_n = L. Take ε = 1, then there is N ∈ ℕ such that |a_n - L| < 1 for all n > N. From the reverse triangle inequality ( | |A| - |B| | ≤ |A - B| ), we can show that | |a_n| - |L| | ≤ |a_n - L| < 1 which implies |a_n| < |L| + 1. Now define M = max{|L| + 1, |a_1|, |a_2|, ..., |a_N|}. Therefore we have 0 ≤ |a_n| ≤ M for all n ∈ ℕ.

@panagiotisapostolidis6424 4 года назад

i'd love a video on the jacobi theta functions. You're one of the best teachers on youtube,at least for me, that i've come across with.

@maxpercer7119 2 года назад

The sequence (a_n) converges to a limit L if we can find an index N such that for n > N, or n >= N (depending on your book) , the sequence terms permanently stay within epsilon distance of L. Said slightly different, a sequence (a_n) converges to L if for any epsilon no matter how small, the sequence terms eventually stay within epsilon distance of L.

@brandonread627 4 года назад

Please do more real analysis videos, your demonstrations provide clarity for self-learners like me. From now on when I see a video that you post for real analysis I will like and comment. I'm currently self-studying Tao's Analysis I. In CH 5, section 6, he defines exponentiation with supremum. I think it'd be helpful if you could explain your perspective on this section, because it isn't clear to me how to think about the section such that I'm able to do proofs. Thanks for your content!

@stephenbeck7222 4 года назад

I think you are missing a ‘there exists’ before the n>N part of your negative example at the end. The negation of ‘if P then Q’ is ‘there exists a P such that not Q’ (i.e. a counterexample). Your statement as written sounds like you mean that none of the a_n get close to a proposed L when you really mean that there is always at least one more a_n that is separated from L no matter how far along the sequence you get.

@rafaelpinheiro857 3 года назад

you're correct, which was just the reason Michael only had to prove that |an-L|>=1 for odd n, given L>0 at 10:46. Observe that he didn't have to prove it for even n.

@carstenmeyer7786 4 года назад

@Michael Penn 3:12 There might be an error here - as a counter-example take *a_n = -1 - 1/n* with the limit *L = -1* . We get *| a_n | = 1 + 1/n > | L + 1 | = 0 (Contradiction to last line of **3:12** !)* Similarly, *a_n = 1 + 1/n* with the limit *L = 1* contradicts the second inequality *| a_n | = 1 + 1/n > | L - 1| = 0 (Contradiction to last line of **3:12** !)* *Rem.:* I'd say there are two possible changes to correct the proof: Either change "and" to "or" in that line starting at 2:53 (that is where I felt the proof was going, and using the maximum on the next screen seems to confirm that guess) or use the inverted form of the triangle inequality to get a different upper bound *| x | - | y |

@carstenmeyer7786 4 года назад

@William Boyle Thank you for your feedback! The contradiction is meant regarding 3:12 of the video, where the inequality is indeed stated with "

@carstenmeyer7786 4 года назад

The idea is definitely correct. The problem is not algebraic - it's simply the (incorrect?) use of the logic operator "and" in the last line at 3:12 (should be "or" instead in my opinion). If you take that line literally now, both inequalities stated there must hold _at the same time_ , which I don't think is neither true for all convergent sequences nor was the line meant that way. The counter-examples do each fulfill one of the inequalities, respectively, but not both at once. The idea of the proof is clear and I know I'm probably nitpicking here (especially because the rest after using *max(..)* seems to be correct) - if this channel didn't have such outstanding content, I wouldn't dare :)

@jonaskoelker 2 года назад

I would pick a_n = L = 2 (or = -2), that simplifies the counterexamples. To fix the proof, I would pick integers M1 > -(L-1) and M2 > L+1 by the Archimedean principle. Then -M1 < L-1 < a_n < L+1 < M2. Let M3 = max(M1, M2), so -M3

@wafizariar8555 4 года назад

Could you make a video based on Galois's theory, or his cohomology?

@debendragurung3033 3 года назад

OMG. That negation statement is quite a Mindfull

@dadrunkgamer_007 3 года назад

Why is |an| < |L-1|. Surely after N sufficiently large, it is above |l-1|

@arma5166 3 года назад

Well if L

@bachoundaseddik250 9 месяцев назад

but he did write and not "or"@@arma5166

@JB-ym4up 4 года назад

Take abs(L)+1 rather than abs(L+1) or abs(L-1).

@joaohax52 3 года назад

I did so: |an - L| < 1 |an| = |an - L + L|

@deepakbhatt4143 4 года назад

Hey Michael , can you make a simple video proving limit points of sin n. I will be really grateful.

@vanneswijaya9787 4 года назад

Do you do Euclidean geometry ? I would love to see some euclidean geometry theorem proving or problems. Good video btw.

@barendbadenhorst7245 3 года назад

Thank you so much! This is a great video, really really helped me a lot

@thehym1420 Год назад

Damn this guy makes it look easy , thanks sir for this video

@sidharthdembi4400 2 года назад

At 3.10 why is mod(an) < mod(L-1)

@backyard282 4 года назад

Why did you take epsilon = 1? Could you have chosen any other epsilon, say 2.4?

@mrsv5287 4 года назад

Late answer, but because he supposed that the sequence converge. This mean that the inequality will work with any value of epsilon. The ultimate goal is to show that the sequence is bounded so you can choose any value of epsilon like 1 or 2.4. Really, he should have kept epsilon, but I think he wanted to simplify the board. And fixing epsilon = 1 doesnt make the proof wrong in the end!

@folorunsoolawale1828 2 года назад

Great lesson What textbook are you using🥺✏️

@ogreeni 8 месяцев назад

Abbott

@roberttelarket4934 4 года назад

If it has a limit, then that's a good place where it "stops"!

@skylardeslypere9909 4 года назад

What about the sequence a_n = (-1)ⁿ * (1/n) It converges tot 0 but it isn't bounded? Edit: nvm I thought of functions, ofcourse it's bounded by -1 below and 1 abovr

@cletushumphrey9163 4 года назад

it's bounded by 1 since every number in the sequence is less than or equal to 1

@piyushkumarutkarsh9124 Год назад

can't thank you enough

@RandomBurfness 4 года назад

Isn't it much simpler to prove the contrapositive statement here, i.e. that an unbounded sequence is divergent?

@ashishKjr 4 года назад

I don't think so. If you have a unbounded sequence (a_n) then you need to show there is no real number x such that (a_n) converges to x. It doesn't make it easier.

@ireallydontknow3299 4 года назад

@@ashishKjr It's actually rather simple. Unbounded: for every positive real M, and for every natural N, there is a n>=N such that |a_n| > M. (assuming otherwise would make the sequence bounded by max{a_1, a_2, ..., a_N, M}) Assume: (a_n) is unbounded and converges to L. Then for every e>0 there is a natural N such that if n>=N, then |a_n - L| < e. But then: |a_n| - |L| |a_n| - |L| < e => |a_n| < |L| + e for all n>=N. But we know that there is K>=N such that |a_K| > |L| + e. Contradiction!

@ashishKjr 4 года назад

@@ireallydontknow3299 don't you think you did the same amount of work as in video?

@sarojpandeya9762 4 года назад

Thanks for this!!!! Amazing

@adarshyadav253 4 года назад

I love your number theory lectures it is better than any other channel I don't know why your subscribers are low people are not smart enough to understand these things

@L0wLevel01 4 года назад

how about by contrapositive ?

@augustusfalloon9884 2 года назад

nice channel

@kuba-xz3rl 4 года назад

I have a nice question for you. I would love to see your solution on this channel. Let n be an arbitrary positive integer and (p_i)_{i=1}^{\infty} be a sequence of prime numbers, such that p_{i+2} is the greatest prime divisor of p_{i}+p_{i+1}+2n for i>0. Prove that (p_i) is bounded.

@carstenmeyer7786 4 года назад

Did you forget to give two initial conditions (e.g. *p_1* and *p_2* ), or are they arbitrary primes as well?

@kuba-xz3rl 4 года назад

@@carstenmeyer7786 They are arbitrary