Do not be confused, people! (0, 3) is not a closed set of the metric space (R, d), but it is a closed set of the metric space ((0, 3), d), which is what the video claimed. Also, that there are Cauchy sequences that do not converge in ((0, 3), d) does not mean (0, 3) is not closed in ((0, 3), d), but that ((0, 3), d) is not complete. This is the gist of what the video is trying to say.
2:30 completeness 3:15 the Cauchy sequence is always a generalization of convergent sequence 4:50 Example of a complete metric space with discrete metric but is incomplete with Euclidean metric
Complete metric spaces? More like "All of these concepts put us through our paces!" Though it can be difficult to grapple with these ideas, I think you're doing a wonderful job of shepherding us through them. Thank you!
For the Functional Analysis series there are a lot of different good textbooks out there. I find it hard to recommend a special one. You can check the one by Rudin and the one by Conway. They don't fit exactly to my course but they can complement it.
6:39 How is b) a complete metric space? Because, not all cauchy sequence converge, as defined above. In this case, if epsilon is >1 then there are so many violent sequences that will be cauchy sequences, because d(x(n),x(m)) is always
@@Chinese_Man23 A Cauchy sequence is a sequence that fulfils d(x(n),x(m)) < eps no matter how small epsilon is chosen. The comment from before only chose epsilson = 1.
I have a question. The idea of having a hole in the space X, also extends to sequences that doesn't converge on R^3 or on R^2 spaces? Should I think it as a real hole eg on the plane (xy) for R^2?
Hello, first 1000 thanks for the videos they are all amazinly explained my question is about the set (0,3) being closed . if we consider the sequence 3-1/n then all its elements belong to X but still its limit (=3) doesn't . ?? so why the set is closed ? please correct me if I am wrong
Remember that the properties of a set being closed or open only hold with respect to a given metric space in this context. What the video is saying is that (0, 3) is a closed subset of the metric space ((0, 3), d).
@@angelmendez-rivera351 i think the subset should be clopen close and open right?? first the subset has same range as set it own all boundary points(so its closed ), and also the open ball at any space in the subset doesnt contain element from the complement of subset .since the complement of subset = ∅.
Why is the set only closed? We proofed in part 4 that set is closed when all of its sequences converge to x belongs this set. Here you showed example of sequence that doesn't satisfy it, I mean (1/n), then why is it still closed?
Hi. Thank you for sharing these videos. I'm sort of confused about whether the sequence summing terms 1/n over the positive natural numbers (is this the harmonic series ?) converges ?! Intuitively it feels like it should right because the terms in the denominator grow to inifinity but I've been reading that it does not converge. Is that true and if yes, is there an easy way to see why ?!
@@jordanfernandes581 It is easy to see why it does not converge using the integral comparison test. This is, lim 1 + 1/2 + ••• 1/(n - 1) + 1/n (n -> +♾) > lim Integral{[1, t); 1/x} (t -> +♾) = lim ln(t) - ln(1) (t -> +♾) = lim ln(t) (t -> +♾) = ♾
In the video we learn that X=(0,3) with the usual metric is closed, and that one argument for this is that the complement the empty set is open. Now in lecture 3 a definition of open sets were given in terms of epsilon-balls. Do the empty set comply with that definition ?
@@brightsideofmaths yeah but a set A is open if for every x in A, there exists an open ball centered at x that is contained in A, so in that example this is fullfilled, right?
Hiiiii :) I Think I miss something :p sequence 1/n is in X but not his limit so X is not closed or we cant apply sequential characterization cuz we dont have cauchy Space ??
You are confusing being closed with being complete. In the metric space ((0, 3), d), the set (0, 3) is closed. If the metric space being considered was (U, d), where U is some other arbitrary subset of R that included (0, 3), then (0, 3) may not be closed with respect to this metric space, but it is closed with respect to the metric space ((0, 3), d), which is precisely what the video claimed. However, the metric space ((0, 3), d) is not complete, even though (0, 3) is closed in it.
Every convergent sequence belonging to the set has there limit with inn the set. The key idea is that we restrict our space to the open interval (0,3) and consider all convergent sequence with limit between 0 and 3. In other words if the limit is not in the set then the sequence does not belong to this set we moves out from our space or the sequence is not defined to our space since the limit does not belong to the open interval therefor is the entire space (0,3) closed.
@@angelmendez-rivera351 i think the subset should be clopen close and open right?? first the subset has same range as set it own all boundary points(so its closed ), and also the open ball at any space in the subset doesnt contain element from the complement of subset .since the complement of subset = ∅.
Why is null set open? Why is (0,3) a closed set in the space ((0,3),d)? We can go any far near zero and find a finite epsilon ball and the same applies for 3 too. So, it is open on both ends. Why is it closed ? And, instead of saying that the space is not complete, why can't we say the space is just open becase 0 is not in it!?
@@brightsideofmaths can i ask you how is it possible for x=(0,3) with descret metric space to be complete ?? the sequence still converge to constant 0. but there is no 0 in our set.
![Functional Analysis 10 | Cauchy-Schwarz Inequality [dark version]](/img/n.gif)
Do not be confused, people! (0, 3) is not a closed set of the metric space (R, d), but it is a closed set of the metric space ((0, 3), d), which is what the video claimed. Also, that there are Cauchy sequences that do not converge in ((0, 3), d) does not mean (0, 3) is not closed in ((0, 3), d), but that ((0, 3), d) is not complete. This is the gist of what the video is trying to say.
it because the set is(0,3) thats why the subset is closed right?? because it has the same range of its set .
What is R here, Is (0,3) itself or Real numbers?
@@VinVin21969 Yes, if I understand you correctly. Sorry for the very late response.
@@Shahid_With_Art R is the set of real numbers.
@@angelmendez-rivera351 Thank You..
These videos are amazing for building intuition.
2:30 completeness
3:15 the Cauchy sequence is always a generalization of convergent sequence
4:50 Example of a complete metric space with discrete metric but is incomplete with Euclidean metric
You don't know how thankful I am.
Complete metric spaces? More like "All of these concepts put us through our paces!" Though it can be difficult to grapple with these ideas, I think you're doing a wonderful job of shepherding us through them. Thank you!
Take any closed subset A of a complete metric space. Then, the set A with the same inherited metric is a complete metric space in its own right.
Alles super erklärt. Ich freue mich jetzt schon auf die nächsten Videos!
So good. Thank you!
Dr. Großmann is there any textbook you would like to recommend, for which contains exercises which corresponds to your lectures.
For the Functional Analysis series there are a lot of different good textbooks out there. I find it hard to recommend a special one. You can check the one by Rudin and the one by Conway. They don't fit exactly to my course but they can complement it.
6:39 How is b) a complete metric space? Because, not all cauchy sequence converge, as defined above.
In this case, if epsilon is >1 then there are so many violent sequences that will be cauchy sequences, because d(x(n),x(m)) is always
Hello! Maybe you misinterpreted what a Cauchy sequence actually is: You need the property for *all* epsilon.
@@brightsideofmathswhat do you mean?
@@Chinese_Man23 A Cauchy sequence is a sequence that fulfils d(x(n),x(m)) < eps no matter how small epsilon is chosen. The comment from before only chose epsilson = 1.
@@brightsideofmaths thanks for explaining it! I had to use the Snapchat AI because I was so confused
would you be so kind and make a videoseries about stochastiks?
would really appreciate it
I have a question. The idea of having a hole in the space X, also extends to sequences that doesn't converge on R^3 or on R^2 spaces? Should I think it as a real hole eg on the plane (xy) for R^2?
R^2 and R^3 with the standard metric are complete :)
Im not writing Analysis this semester, but looking forward to learn with this next semester.
How have your studies been going?
@@PunmasterSTP terribly frustrated with differential equasions
Hello, first 1000 thanks for the videos they are all amazinly explained
my question is about the set (0,3) being closed . if we consider the sequence 3-1/n then all its elements belong to X but still its limit (=3) doesn't . ?? so why the set is closed ? please correct me if I am wrong
For showing closedness, you need to choose a convergent sequence with limit inside the set. Your sequence simply does not fulfill this requirement.
Your sequence doesnt have a limit (X = (0,3))
The sequence you chose does not have a limit in X, but this does not mean no sequence with a limit in X exists.
Remember that the properties of a set being closed or open only hold with respect to a given metric space in this context. What the video is saying is that (0, 3) is a closed subset of the metric space ((0, 3), d).
@@angelmendez-rivera351 i think the subset should be clopen close and open right?? first the subset has same range as set it own all boundary points(so its closed ), and also the open ball at any space in the subset doesnt contain element from the complement of subset .since the complement of subset = ∅.
Why is the set only closed? We proofed in part 4 that set is closed when all of its sequences converge to x belongs this set. Here you showed example of sequence that doesn't satisfy it, I mean (1/n), then why is it still closed?
That is a good question. Maybe you can state the condition from part 4 again. If you state it correctly, you will understand why here it's different.
The set will be closed if we cannot leave the set from the inside using the sequences. Is this the correct way of thinking?
Hi. Thank you for sharing these videos. I'm sort of confused about whether the sequence summing terms 1/n over the positive natural numbers (is this the harmonic series ?) converges ?! Intuitively it feels like it should right because the terms in the denominator grow to inifinity but I've been reading that it does not converge. Is that true and if yes, is there an easy way to see why ?!
No, the harmonic series is not convergent. That's not the topic for this video but maybe I will make another video about this.
@@brightsideofmaths Alright. Thanks for replying. I will look forward to that video 🙂
@@jordanfernandes581 It is easy to see why it does not converge using the integral comparison test. This is, lim 1 + 1/2 + ••• 1/(n - 1) + 1/n (n -> +♾) > lim Integral{[1, t); 1/x} (t -> +♾) = lim ln(t) - ln(1) (t -> +♾) = lim ln(t) (t -> +♾) = ♾
In the video we learn that X=(0,3) with the usual metric is closed, and that one argument for this is that the complement the empty set is open. Now in lecture 3 a definition of open sets were given in terms of epsilon-balls. Do the empty set comply with that definition ?
Yes, for all points in the empty set you can find an epsilon :)
@@brightsideofmaths But the epsilon ball will not be a subset of the empty set, because it is empty?
@@sheldonchou8830 Maybe the confusion comes from the logical use of the conditional (🡲). If the premise is false, then the conditional is always true.
@@sheldonchou8830 You can find the corresponding video here: thebrightsideofmathematics.com/start_learning_logic/overview/
@@brightsideofmaths it seems that URI is broken
I am wondering why some of your video has no english subtitle? do you make it or is it automatic?
Why is (0,3) a closed set in ((0,3), d)? Isn't it that for every x in (0,3) any ball centered at that point is a subset of (0,3)?
"Closed" is not the opposite of "open".
@@brightsideofmaths yeah but a set A is open if for every x in A, there exists an open ball centered at x that is contained in A, so in that example this is fullfilled, right?
@@marcbanfi4985 Yes, which means the set is open. But before you asked about why it is closed.
@@brightsideofmaths can a set be open and closed at the same time?
@@marcbanfi4985 Yes, of course. You know one example now :)
Hiiiii :) I Think I miss something :p sequence 1/n is in X but not his limit so X is not closed or we cant apply sequential characterization cuz we dont have cauchy Space ??
You are confusing being closed with being complete. In the metric space ((0, 3), d), the set (0, 3) is closed. If the metric space being considered was (U, d), where U is some other arbitrary subset of R that included (0, 3), then (0, 3) may not be closed with respect to this metric space, but it is closed with respect to the metric space ((0, 3), d), which is precisely what the video claimed.
However, the metric space ((0, 3), d) is not complete, even though (0, 3) is closed in it.
@@angelmendez-rivera351 Thank you for your answer its clean :-)
Why (0,3) is closed?
Every convergent sequence belonging to the set has there limit with inn the set. The key idea is that we restrict our space to the open interval (0,3) and consider all convergent sequence with limit between 0 and 3.
In other words if the limit is not in the set then the sequence does not belong to this set we moves out from our space or the sequence is not defined to our space since the limit does not belong to the open interval therefor is the entire space (0,3) closed.
(0, 3) is closed with respect to the metric space ((0, 3), d), not with respect to the metric space (R, d), which is the one we usually work with.
@@angelmendez-rivera351 i think the subset should be clopen close and open right?? first the subset has same range as set it own all boundary points(so its closed ), and also the open ball at any space in the subset doesnt contain element from the complement of subset .since the complement of subset = ∅.
@@VinVin21969 Yes, that is exactly right.
Normally, the name Cauchy always gives me a headache. But this time at The Bright Side, it isn’t the case.
If limit is 0 then we can say that the sequence is not convergent?
No, if there is a well-defined limit, we have a convergent sequence by definition.
Why is null set open? Why is (0,3) a closed set in the space ((0,3),d)? We can go any far near zero and find a finite epsilon ball and the same applies for 3 too. So, it is open on both ends. Why is it closed ?
And, instead of saying that the space is not complete, why can't we say the space is just open becase 0 is not in it!?
It is because the subset is the entire set and a sequence in it can't escape the set, so it is closed by definition.
All closed sets are complete but all complete sets may not be closed. Correct?
From an intuitive point of view, why is important that all Cauchy sequences converge?
Intuitively, Cauchy sequences are sequences that *should* converge. Therefore, spaces with this property are better to handle :)
@@brightsideofmaths can i ask you how is it possible for x=(0,3) with descret metric space to be complete ??
the sequence still converge to constant 0. but there is no 0 in our set.
@@VinVin21969 Don't forget that your sequence has to consist of members from the set X.
@@brightsideofmaths could you make it more clear please?? i already tried to find the answer but still have no clue
@@VinVin21969 Try to formulate your question more concretely. Then I can answer in a clear way :) What is your counterexample for X = (0,3)?
without subtitile ，it s so hard to understand for un-native speakers