Real Analysis | Subsequences

We introduce the notion of a subsequence and prove a few simple results including the Bolzano-Weirstrass Theorem.

If you are going to use an ad-blocker, considering using brave and tipping me BAT!

Books I like:

Abstract Algebra:

Differential Forms:

Number Theory:

Analysis:

Calculus:

My Filming Equipment:

[kademeyer]

Dude I just wanted to say that I fell in love with this channel within a single video. I was looking for help on a homework for my intro to analysis class and stumbled across your video on cauchy sequences. I was almost in disbelief at how efficiently you explain these concepts. Keep rocking on man!

[tyjensen]

Btw the fact that every bounded sequence has a convergent subsequence is called the Bolzano-Weierstrass Theorem

[JoshStadler]

Thank you for doing these videos. These are the best real analysis lectures I can find!

[thatdude_]

I just wanna mention, a sequence is not a set! In a set the order or repeated appearance of the member is irrlevant, i.e {1, 3, 2, 3, 3, 4}={1, 2, 3, 4}, whereas in a sequence these factors indeed play a role. Thats why the notation (a_n) is more common than {a_n}, since formally, a K-valued sequence is a map IN --> K, hence an element of the IN-fold cartesian product of K, thus more akin to a vector with infinitely many entries, than a set.

[doodelay]

I'm on my way to real analysis right after Calc 3, so I cannot thank u enough for having advanced topics like this ready for me and everyone else who'll need them.

[jonaskoelker]

Every increasing sequence of natural numbers is unbounded:

Let n_i be an increasing sequence of natural numbers (N = {1, 2, ...}), and assume for contradiction that there is a natural number M such that n_i < M for all i.

Since the n_i are increasing we have
n_i < n_i + 1 <= n_{i+1} ;; since n_i + 1 is the smallest number greater than n_i
and n_{i+k+1} >= n_{i+k} + 1 >= (k + n_i) + 1 = n_i + (k+1)
so in general n_{k+d} >= d + n_k by induction.

But then n_{M+1} >= M + n_1 > M, contradicting n_{M+1} < M.

[victorserras]

These lectures are incredible, the explanations are so clear. Thanks for these videos.

[muwongeevanspaul]

Michael Pen, I looked at this video my first time and I quit it coz possibly my mind had not yet settled but just again have looked at it, I have clearly understood what is a subsequence and how to come up with one. This is so easy. Thanks alot Pen, it is a puzzle that has been bothering me alot. I have understood the idea. Thanks so much. May the Good LORD maker and creator of heavens and the Earth bless u and really bless u much.

[doodelay]

Also professor if u don't mind the question, are u able to do research papers when running a channel like this? or has the channel become a full priority now? thx! I'm just trying to gauge how time consuming it is to be a researcher

[rizalpurnawan]

I was roaming around to this channel to find some inspiration to prove that, in the case of metric space, the complement of a closed subset S of a metric space (M, d) is open, using Cauchy sequence. And 04:45 on this video gave me the idea. Great video! Thanks sir!

[thesecondderivative]

5:37 I'm unsure if this is correct but here's my proof. Assuming n_k is bounded, Since it is strictly increasing, n_k has a limit. Therefore;
L - e < n_k<L+e for an arbitrary epsilon and
L -e < n_(k +1) < L +e(1)
L-e +1<n_k +1< L +e +1
1/(L +e +1) < 1/(n_k+1)< 1/(L -e+1) (2)
Multiplying (1) and (2);
(L -e)/(L +e + 1) < n_(k +1)/(n_k +1) < (L + e)/(L - e +1).
Since n_k is in the naturals and is an increasing sequence,
n_(k +1) >= n_k +1
Therefore n_(k +1)/(n_k +1) must be greater than or equal to 1.
The right hand side of the limit is greater than 1 only when e > 0.5. That means you can find an epsilon small enough but greater than zero for which our expression is less than 1. This leads to a contradiction.

[razvbir]

I was taugh a more complicated proof of the last result. 😅 This video was very useful. 😊

[ashlaw]

in the proof of B-W theorem, you must take n1<n2<n3<…….

[nestorv]

If M/2^(k-1) is the length of the I_k th set, then shouldn't the length of the I_K th (k>K) set be bigger? This means that the step at the end ( 21:16 ) when he says that the length of the I_K th set is less than epsilon is not correct.
length of I_k < length of I_K
So even if (length of I_k ) is less than ε, this does not imply that the length of I_K is also less than epsilon.

[NoOne-wbxr]

Good Place to Start 0:14
Good Place to Stop 22:14
Needless to say, the whole video is a Good Thing to Watch

[rahul_k_a_g]

Michael can you please help me on a question where we have a constant sequence and are asked about number of subsequences of it. Do we count the subsequences as distinct or is it only one?

[lucassaito]

The notation is quite confusing, but fortunately now I got it, thanks!

[devnull]

I don't really know any set theory. But, question: Was that the Axiom of Choice in the second proof?

[uffe]

I think a strictly increasing sequence can be bounded?

[ochinglam]

How do you make sure that the n_1, n_2, ... you picked is strictly increasing?