Real Analysis | The Heine-Borel Theorem
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- Опубліковано 19 вер 2024
- We prove that the topological definition of compactness is equivalent to a set of real numbers being closed and bounded.
Here is the last bit of the proof:
• Real Analysis | If [a,...
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I just woke up to notice this video cuts off in the last step. Sorry everyone! My files have been strangely exporting recently... I will make a video with the last step and post it unlisted later today -- linked at the end of this video.
Here is the new video: ua-cam.com/video/p9nKuqz6D9w/v-deo.html
17:54 “So this is...” not a good place to stop
GOT EM
lol
One of two things happened here: 1) Cauchy sequences do not necessarily converge in the UA-cam space. 2) Gödel's famous theorem was being demonstrated in some way.
One minor note is that we must show that S (from 12:08) is nonempty before determining the existence of its supremum. This is done by taking some U_i containing a, and finding some x_i such that [a,x_i] is contained within U_i, then this single U_i provides a finite (containing only one element) subcover of [a,x_i], so x_i is in S. For those of you preparing for exams, this is the kind of detail that I have gotten dinged on in my oral exams.
The rest is left as an exercise to the viewer.
The rest of the solution is trivial and left as an elementary exercise
@@raphaelkelly861 a tensor is something that transforms like a tensor.
Can someone help explain why c < b is a contradiction of s @16:15 thank you
Case c' != b. Since c' = min{ s + eps, b } --> c' = s + eps --> s = c' - eps --> s < c' which is in S --> s is not an upper bound of S --> s != sup S, which contradict that s = sup S.
Lemma: the proof look like an induction. But don't you have to prove first that S is not empty. In that way, could we say that [a,a] is obvious part of S.
Nice treatment of tougher proof, but there is an editing problem near end .
Video just stopped 😩😩
Eagerly waiting for the ending 😩😩
At 10:26, we contradicted to the hypothesis by showing y must be in K, but I could not see how y must be in K. In fact, we actually contradicted to the hypothesis that some terms of {y sub n} is not in K, and it makes more sense to me. Is there a mistake?
We started off with a K that satisfies the property that every open cover of it has a finite subcover. We also assumed that the K has a limit point 'y'. If K does not have a limit point then K is simply closed. From the proof, we know that if such K exists, i.e. satisfying the two conditions then y not being an element of K is not possible. Therefore, if this K exists then we must have all of its limit points to be in K, which makes K a closed set.
Well, this answers my question from yesterday. Apparently I just need patience.
You are saving me...my master exam is next month and it's all real analysis! Thank you for your classes!
*Imagine watching the whole video waiting to see how it ends 😢*
I couldn't understand anything. And doesn't makes sense , how is it useful?? Can someone explain in what are the uses of these things? Any use in physics?
That's basically just a part of the common language of advanced analysis. (Physics loves it)
The result isn’t applied directly but is important to use in physics sure
Complex numbers is a nice field with compactness property
A notion of finiteness is nice
Many theorems use compactness (see compact support)
where can I find that proof for Rⁿ ?
"Ok, great". I think we'll call you the Hitchcock of the maths :)
At 9:58,
Why is it true that yN doesn't belong to any of U sub xi?
By definition, the limit (introduced around 7:18) must be within an infinitely small distance from the terms in the sequence. However, U sub xi caps this distance at 1/2 min {(x1-y), ...., (xn-y)}.
I do not understand why c'= min(s+epsi,b). The sets of finite cover Ui are not constraint to be in S, nor in [a,b], so the upper bound of the cover could be anything bigger than c.
Your demo is a special case wher [a,c] = Union of Uin.
The aim is to prove finite cover of [a,b], by proving a finite cover of [a, s+e] where either s+e=b and so the required finite cover exists (that it may cover more than [a,b] is irrelevant)
My favorite theorem
I have only one concern: when we define epsilon using the minimum function, aren't we assuming that a minimum exists? Why should this be true in general?
The set is finite by the finite subcover condition, so finite sets always admit a minimum element by the well ordering principle.
Good
UA-cam: *This is a good place to stop*
ahhh more analysis
youtube thinks Im smart enough to understand any of this
this looks so useless