It's amazing what Dr. Schuller did in "Kindergarten".
@mergen98024 жыл бұрын
I tried to write those down in Kindergarten too but the Kindergarten I wanted to sneak in was in covid lockdown
@lezhilo7723 жыл бұрын
My professor also jokes about how Lorentz transformations and four-momentum normalisations are kindergarten physics XD
@daigakunobaku2733 жыл бұрын
@@samuelcosta8pi by that logic integration should be taut to toddlers since it was first invented about 2 000 years ago.
@matthewsekale83323 жыл бұрын
😂 😂
@raphaeledison92363 жыл бұрын
I guess Im asking randomly but does someone know a way to log back into an Instagram account?? I was stupid forgot my login password. I appreciate any help you can give me!
@salvatoregiordano68164 жыл бұрын
Not many people can explain such an abstract topic so clearly. Thank you so much for this amazing lecture!
@rickynoll99285 жыл бұрын
37:20 "You can't make a proof shorter only because the blackboard finishes" hahaha I love this guy
@rtg_onefourtwoeightfiveseven3 жыл бұрын
"I have discovered a truly marvellous proof which the blackboard is too small to contain." -Frederené Schullescartes
@kikiacademy96283 жыл бұрын
but you can if you are writing on a green board
@freyc1 Жыл бұрын
@@rtg_onefourtwoeightfiveseven Fredepierre de Schullermat?
@rtg_onefourtwoeightfiveseven Жыл бұрын
@@freyc1 oops
@LeoHsieh3 жыл бұрын
7:30 Schuller's memory is larger than my 64 RAM computer.
@chetan58484 жыл бұрын
1:21:40, Economics bashing. I love it.
@SantiagoMontouliu5 жыл бұрын
The look on his face at 27:06, when he proves that the empty set is an element of O standard is amazing.
@jerrynews58033 жыл бұрын
Not many can explain such an abstract topic so clearly. Thank you so much for this amazing lecture!
@jayleem9083 жыл бұрын
easily understandable, well-organized, and beautiful lecture with very clear motivation.
@juliensorel14274 жыл бұрын
It is impressive how he committed to memory all those numbers ...
@SCRAAAWWW Жыл бұрын
I believe he strategically placed a small slip of paper with those numbers on the chalk stand beneath the spot on the board where he was writing. Cleverly planned and well executed lecture!
@weinihao3632 Жыл бұрын
@@SCRAAAWWW Indeed. There is a German lecture series where he also presents that table and he mentions that he copied the numbers from Wikipedia :)
@sydamerikaner19487 жыл бұрын
incredible nice writing on the blackboard! i like it
@mindiff8 жыл бұрын
You are a good lecturer. Thanks for uploading.
@damnit2583 жыл бұрын
Yea in kindergarten, the conjunction of U and V, who doesn't remember that from the kindergarten, good times.
@mazenbr86633 жыл бұрын
😂😂😂😂😂😂😂😂😂😂
@paoloziko8 жыл бұрын
Would you provide a link for the problem sheet ? Thank you for these lectures, amazing lecturer
@lorenzodigiacomo25615 жыл бұрын
A little mistake at 49:46: the fact that M\M=∅ is open implies that M is closed (by definition). Similarly M\∅=M being open implies that ∅ is closed. By the way, thanks Prof. Schuller for the awesome lectures
@d3tcovax3 жыл бұрын
Why would that imply M is closed, if i take any interval open, or closed, and remove any number of the elements of that group I am left with another set with no implications on its openness or closure... Please elaborate and show me why you would say otherwise
@YoTengoUnLCD2 жыл бұрын
@@d3tcovax because the complement of M (empty set) is open.
@amirkhan3555 жыл бұрын
Wow, what a gem! I am soooo thankful!
@aakashsharma32163 жыл бұрын
I m so lucky that I got such professor
@airmahathi7 жыл бұрын
Amazing board work. Thank you for sharing such great knowledge.
@joaopaulobrito19933 жыл бұрын
Beautiful class. Congratulations!!
@kockarthur79766 жыл бұрын
34:33 This definition of the induced topology, though not necessarily incorrect, technically overcounts identical elements. There could be many U in O such that U ∩ N gives the same result. Also, the notation in this definition gives the immediate false impression that we are using universal comprehension (notice there is no epsilon-relation on the left). These slight issues (or non-issues, depending on how you look at it) can be resolved by defining the subspace topology more formally as {V in P(N) | there exists U in O : V = U ∩ N}. With that, restricted comprehension is manifest. Edit: He actually introduces this definition later at 38:33.
@p0gr4 жыл бұрын
sets dont count their elements! and we dont have an issue with PRC, because it exists by the axiom where a set can be an image of a functional relation.
@martinepstein98263 жыл бұрын
There is no need to worry about universal comprehension when you're working within a given set.
@piercingspear29223 жыл бұрын
Overcounting elements doesn't really matter if you want to build a set. It'll just fix itself up by the axiom of set theory. There is no problem at all with universal comprehension here since U and V already a set and the intersection of two sets is a set. It's just a matter of own taste, whether to use (not necessary) more formal definitions in the whole lectures since he already explained all of these issues in his previous lectures on set theory.
@theInternet6335 жыл бұрын
I was wondering what kind of Kindergarden he went to that teaches set theory :D Sounds like a place i would like to send my child to if i ever have one...
@eldarr0uge4823 жыл бұрын
Hehe, it's part of a movement called "new math" or "modern math", which was a new way to teach math, in a very abstract way to children, starting very very early. It occured mainly in France and the US (that's part of the reason why those 2 are math superpowers, with respectively 12 and 13 Field medals), and to a lesser extent in the UK and Germany (who had lost many brains after WW2), after the Sputnik launch and was meant to boost the science and math curriculum in order to catch up with the soviets who are a great math nation (France already was a math nation but the arrival of new math is due to a member of the Bourbaki group, André Lichnerowicz, becoming head of education ministry.) Problem is, it was so abstract that neither parents nor teachers knew it and could teach it effectively, so, despite highlighting the genius of some students, most children couldn't follow and ended up with a learned helplessness, like a trauma when it comes to math. It thus was abandonned at the end of the 60's.
@piercingspear29223 жыл бұрын
@@eldarr0uge482 Thank you for the knowledge! :)
@rewtnode6 жыл бұрын
After the discussion about the nontransitive dice I wonder: does one need a topology for any type of measurement? What about comparing probabilities or densities on various (event-) spaces? Do they always have to have a known topology that supports this?
@kvazau84446 ай бұрын
Something like a topology called a sigma algebra, yeah. See the measure theory lectures from Schuller's QM playlist
@wildoaklane3 Жыл бұрын
56:00 I believe this is not an equivalence relation on R in its current form? It appears to not be reflexive.
@lorenzopiovan68702 жыл бұрын
Wonderful lectures, congratulations, but it would be really nice and even better if the problem sheets could be available. @Frederic Schuller: do you think you could help your you-tube students by providing the problem sheets in some way?
@millerfour20713 жыл бұрын
7:49, 18:14, 21:12, 24:56, 27:29 (B_r is defined on M itself not the embedded space), 34:21, 43:36, 48:01, 49:28, 53:20, 1:01:57, 1:04:24, 1:09:34, 1:15:41, 1:17:35, 1:20:30 (rock paper scissor for dice), 1:22:18 (wisdom), 1:28:13, 1:33:12, 1:37:32
@PDHo-hl3ld3 жыл бұрын
I love this German precision
@lutherhoward7637 Жыл бұрын
What he means when he says" you learned in kindergarten', is set theory was called the new math in the 70's and students were expected to learn basic set theory in earlier grades. But, that fell apart.
@samueldeandrade85353 ай бұрын
What exactly "kindergarten" means? More precisely, how old are kids in kindergarten?
@alpistein8 жыл бұрын
Are there any links to the problem sheets for this course?
@neelmodi57918 жыл бұрын
I haven't been able to find any, but Wikipedia happens to list many properties of topologies and things like that. I find it useful to prove each of those one by one. Edit: Of course, it is annoying at times not knowing whether the stuff from a particular lecture is enough to prove something.
@UnforsakenXII6 жыл бұрын
Just to confirm: There's no extra book, right? The lectures are self-contained except for the problem sets?
@daigakunobaku2733 жыл бұрын
@@neelmodi5791 @Neel Modi if you are looking for a topology book with problems to solve, I highly recommend the Elementary Topology Problem Textbook by O.Ya. Viro et al.
@Gipsy4u6 жыл бұрын
Thank you, great stuff.
@cl7035 Жыл бұрын
Fabulous courses. thank you!
@xochihuarock2 жыл бұрын
1:11:36 In the convergence definition, you have to say: for all U (not empty) in O: Exists N in the natural numbers that ...
@sirlukelimbo20443 жыл бұрын
I'm addicted to his diction of "there exists".
@LaureanoLuna2 жыл бұрын
Wonderful. Thank you so much, professor. I advice, however, to stress "quódlibet" instead of "quodlíbet" in "ex falso quodlibet".
@hoareg22 жыл бұрын
Excellent lecture. One point though: In your example b) at around 1h28, I think it’s not accurate to say any map phi from M to N where N is equipped with chaotic topology is continuous. The reason is the preimage of N (as an open set in the topology) is a subset of M but not necessarily an open set in M.
@denisapg2 жыл бұрын
Phi^{-1}(N) = M, since, for all p \in M, phi(p) \in N.
@hoareg22 жыл бұрын
@@denisapg Thanks a lot for the reply, my bad. One question (if you have time) Phi^{-1}(N) might not be well-defined as under Phi, two different points in M may hit the same point in N. Am I missing some thing here ?
@denisapg2 жыл бұрын
@@hoareg2, phi^{-1}(N) is, by definition, the set of points of M whose image under phi is in N. But, for all x in M, phi(x) belongs to N. Therefore, phi^{-1}(M) = N. Note that phi does not have to be one to one for this definition to make sense. If phi is one to one, however, then, for every subset B of N, phi^{-1}(B) is equal to the range of B under phi^{-1}.
@hoareg22 жыл бұрын
@@denisapg really appreciate your time.
@bertito957 жыл бұрын
are the problem sheets corresponding to this series of lectures available somewhere?
@jimallysonnevado39738 ай бұрын
In the definition of continuity, should the map be surjective?, If not surjective, is the topology in the target space the topology of the entire space or the induced topology of the codomain of the function? For instance, if f(x)= sin(x) then the codomain is [-1,1] but f in itself is also a function from R->R. Clearly, the set [-1,1] has different properties than R. Does it make a difference?
@elmepelme3 ай бұрын
I think it’s fine either way. For the sin function, if you take any open set in the target R that is not in [-1,1], the preimage will be the empty set. For example the pre image sin^(-1) ( (2,3) ) = empty set which is in the topology of the domain
@tramquangpho3 жыл бұрын
at 30: 21 when he proved the intersection of two set and pick a point in it, I think min(r,v) radius is not enough because assume the point lie in the middle of the intersection there are still ossiblity the radius is too large and lie outside of the intersection.
@piercingspear29223 жыл бұрын
No, his proof is already complete. If the open ball with min. radius should always lie in the intersection. If not, there will be part of the corresponding open ball that lies outside the defined set.
@joefagan93354 жыл бұрын
1:15:45 why does Dr say the sequence 1 +1/n that converges against 1using the normal topology not converge using the discrete topology? Is there not a set in P(R) that contains all off the points on the journey to 1?
@NotLegato4 жыл бұрын
we say that a sequence converges to x if for each NEIGHBORHOOD U of x, there is an index after which all the elements in the sequence are in that neighborhood. {1} is a neighborhood of 1, and absolutely no element of 1 + 1/n is an element of the set {1}, so the sequence does not converge. this is analogous to not being able to find a delta for a given epsilon in a metric space.
@hershyfishman29292 ай бұрын
47:46 Aren't there just 4 possibilities? How are i and ii not included in iii-vi?
@LaureanoLuna2 жыл бұрын
Pllease, help me out. A map frpm R^d to R^f, both sets equipped with the standard topology doesn't seem to recover the usual notion of continuity. Take d = f = 1 and consider a constant map e.g. forall x in R, phi(x) = 1
@hajarikken22125 жыл бұрын
Hello sure, my question is the next: How did you calculate the number of topologies (choices) thant we can have? Thank you.
@00TheVman4 жыл бұрын
I don't have any background computing the number of topologies, but from what I've heard it is really hard to do. I believe the number of topologies on a finite set is computed through brute force on a computer. I'm not sure but I think that there isn't a general formula for the number of topologies on a set of size n. If one imposes a condition on the topology, that it be T_0 (meaning no two distinct points are members of the same open sets) then counting the number of T_0 topologies on a finite set becomes equivalent to counting the number of partial orders on the set (which is understood better but I believe still hard).
@someguy33353 жыл бұрын
You can Look at all the subsets of the powerset and consider the topology that is generated by a given subset. Then u need to sort of classify subsets that generate the same topology and the number of those classes is the number of topologys. This is the naive approach. One might be able to derive a formula for finite sets this way
@danielrosales93164 жыл бұрын
Hey, great video. Just one question: What is the implication of the arrow symbol in 27:00?
@unopinionated18233 жыл бұрын
The condition to check was that for all points in the empty subset, there exists a ball of radius r such that the ball contains all such points. However, since the empty set contains no such points, whatever condition was the subsequent of the “for all points in the empty set” predicate would trivially hold Ex-Falso Quodlibet is a Latin term that describes the fact that a false predicate can imply anything.
@malawigw3 жыл бұрын
check lecture 1 on logic.
@mohamedabdelfateh8072 Жыл бұрын
Thank you Dr . 🇩🇿🇩🇿🇩🇿
@running_cactus3 жыл бұрын
Hello @Frederic Schuller, I notice that you defer some problems/ideas to problem sheets. I did find a few interesting and would like to check my approach is acceptable. Would you be able to provide a link to those?
@brendawilliams80622 жыл бұрын
Thankyou
@manimusicka25 жыл бұрын
Thank you.
@nahuu44812 жыл бұрын
Auf Deutsch sind Sie der beste, auf Englisch auch.
@sanketthakkar4496 Жыл бұрын
Lectures are great.. how can i find problem sheet regarding this course for a practice purpose ?
@edgars4nt0s2 жыл бұрын
I like so much this lesson
@nawazishali5025 жыл бұрын
sir, there is any formula to find total numbers of typologies from a set??
@00TheVman4 жыл бұрын
I don't have any background computing the number of topologies, but from what I've heard it is really hard to do. I believe the number of topologies on a finite set is computed through brute force on a computer. I'm not sure but I think that there isn't a general formula for the number of topologies on a set of size n. If one imposes a condition on the topology, that it be T_0 (meaning no two distinct points are members of the same open sets) then counting the number of T_0 topologies on a finite set becomes equivalent to counting the number of partial orders on the set (which is understood better but I believe still hard).
@peggy7673 жыл бұрын
How do you calculate the number of topologies you have on a set?
@d3tcovax3 жыл бұрын
Plain old combinatorics. If you have a set of n elements, the powerset must contain: { {}, {1}, {2}, {1,2}, ... {1,2,...,n} {2,3,...,n} {1,3,...,n} ... } So it is simply the number of possible combinations of elements one can construct without regard to the number of elements in the set or the order of elements in the set (given that no elements repeat). I believe the exact equation is: /Sigma_{i=0}^{n} i! Someone feel free to suggest another equation if I am wrong I am usually bad with pattern representation in maths 🙃
@peggy7672 жыл бұрын
@@d3tcovax I don’t think that’s true because all combinations don’t make topologies right ?
@deltango123453 жыл бұрын
V good stuff.
@pedidep3 жыл бұрын
I think in axiom II of topology you should add finitely many intersections. If an infinite intersection were allowed one could end up with a ball with radius zero in the standard topology.
@CrusaderTube3 жыл бұрын
He is only requiring intersection of two open sets to be open. It implies finite intersection by induction, but not countable intersection.
@pedidep3 жыл бұрын
@@CrusaderTube your point fits to my intuition nicely as soon as we are in R. But I have a problem. Consider that I have no idea of what open set is. Then the definition suggest that every collection O (curly) is an open set. This is where I was confused. I still think one should add finitely many intersection to avoid the problem that I mentioned. I also take a look at Alan Hatcher notes on point set topology (available on his webpage). He wrote "The intersection of any finite collection of sets in O is in O"
@CrusaderTube3 жыл бұрын
@@pedidep the second axiom says that if U and V are in O for two sets then U intersection V must also be in O. It is possible to prove by induction that this means intersection of any finite collection of sets in O is in O. The axiom presented by Hatcher you mentioned and the axiom (ii) are then equivalent, since the implication clearly holds the other way too (as {U,V} is a finite collection).
@pedidep3 жыл бұрын
@@CrusaderTube I can assume O with infinite subsets then their intersection is again in O regarding Prof. Schuler's definition. All I can understand is that it contradicts Hacher's definition. Can you explain to me how do you avoid infinite intersection in present axiom?
@CrusaderTube3 жыл бұрын
@@pedidep The definition says “intersection of any TWO subsets (of M) that are in O must be in O”, NOT “intersection of any collection of subsets in O must be in O”. I can’t explain it more clearly than this. Read my previous responses and think for yourself. There is no problem with the definition in the lecture.
@malawigw3 жыл бұрын
I need to re-live my life as a german where axiomatic set theory is taught in kindergarden
@Aphrodite1121210 ай бұрын
I wish there was a translation for this video
@alijoueizadeh84776 жыл бұрын
So the intersections of sets whithin the topology and the whole set M need not to be in the topology. 14:00
@jacksonh20832 жыл бұрын
3:30 (Not an important time stamp or anything, just have to step away so thought I’d time stamp it for later.)
@amrmajul4597 жыл бұрын
@ kzbin.info/www/bejne/Z6jcgKKCis-Im6s "In Kindergarten you wrote this U intersect V" Kindergarten me ate glue :| not did set mathematics.
@notachannel78815 жыл бұрын
Did he define what an open set is? From his use it seems to mean any element in the equipped topology of the set in consideration. Is there a more precise definition than this.
@jandejongh5 жыл бұрын
No, as that is exactly the definition of an "open set": a set in the topology. The choice of open sets (obeying the three rules) for a space is a topology.
@d3tcovax3 жыл бұрын
I have always found it funny that in topological set theory technically anything can be a topology so long as the three axioms hold, but that's kinda the point, it's based on a notion of free logic meaning you want the theory to be as free from assumption as possible except where necessitated by the function of the theory. Unfortunate for us that means an open set gets defined when the topology gets chosen for a given space of sets! Basically it comes down to a similar idea: define an element, or set, as lovely as possible... A set is a collection of elements which satisfy that all elements are members of the set... It's crazy but it's the free-est logic we can use so that sets can be used in a wide range of applications while still achieving what we built them for in the first place 🙂
@cartmansuperstar2 жыл бұрын
@Notachannel: this is the way, i understand it: "Open sets" and their choice in a topological space define (!), what it means for elements to be "close" together in a space, that actually has no notion of "distance". In a metric space, where you do have a notion of distance, the open sets (at least, when the topology is induced by the metric) are those, that constitute a neighbourhood for all its elements, so that for any element of that open set the others literally surround it (you have an epsilon-ball around each element fitting completely into the open set). Not a single element lies on its boundary. By choosing a topology in a purely topological space, and thus choosing a specific open set, you are also saying: these elements (shall!) constitute a neighbourhood for each other (although you couldn´t have said that without defining it, because there is no notion of distance). If you have an even smaller open subset of that open set, you are saying: these elements of the subset are even closer together - like the elements of the smaller subset are like the inner layer of the neighbourhood, while those, that stay outside are the outer layer. Hope, that´s helpful, but as i said: my personal interpretation and not completely sure, it is correct. I had a hard time myself, trying to make sense of it.
@cocoaaa2680 Жыл бұрын
1.24.00 - continuity of a map btw teo top. spaces
@cocoaaa2680 Жыл бұрын
1.34.00 isomorphism
@mohammedkhalili11547 жыл бұрын
Can we have the problems' sheets please??
@finaltheorygames17815 жыл бұрын
At 49:30 he says M=M/0, subtract the empty set, but I thought the empty set was in ever set so you can't subtract the empty set from a set?
@ForlornSnake4 жыл бұрын
The empty set 0 (as you wrote it) is not an element of every set, rather, it is a subset of every set. What he is doing is subtracting the elements in the empty set, which contains no elements. Therefore M = M\0. Hope this clears up your confusion.
@hershyfishman29292 ай бұрын
The elements of ∅ are indeed in every set, but that's what's special about nothingness that you can subtract it as many times as you like and it will still be there
@franciscoaguero90284 жыл бұрын
Is the definition of continuity a theorem or an axiom? Or just a proposition?
@martinepstein98263 жыл бұрын
It's a definition. It's not a theorem, axiom, or mathematical assertion of any kind. It just says we use a certain word a certain way.
@rajarshichatterjee86366 жыл бұрын
how good is that blackboard , where from i can get one like that ???? well thanks for such helpful lessons !!
@sidddddddddddddd5 жыл бұрын
That's a green chalkboard, bro.
@joefagan93354 жыл бұрын
You have to go to Germany to get those greenboards, washers and sqeezy things.
@atouihalim22886 жыл бұрын
Zoom in blackboard for a good vision.
@LaureanoLuna7 жыл бұрын
Like the irony at 1:17:37
@sabarishnarayan91005 жыл бұрын
At 4:22 What does he mean by "You can't intersect more sets than fit into a set?"
@TimTeatro5 жыл бұрын
Watch the earlier lectures. There are objects which are too big to be sets, and we can't take their union and expect to get a set. I don't think he mentions the name of these bigger animals, but they are _classes_.
@taraspokalchuk72568 жыл бұрын
1:27:56. 1. What if there is no preimage of some element in N? N is not being hit entirely. Then the map is not continious? 2. What if all N is being hit, but we choose a topology, which contains an element, or a set, which contains an element than doesn't lie in N? Is the map not continous?
@blackflan7 жыл бұрын
1. The map is still continous. "Having no preimage" is equivalent to "The preimage is the empty set" or "No elements from M maps to some "a" belonging to N, hence a set of them is empty" and we know that the empty set is on any topology in M 2. A topology on a set N can't contain sets with elements outside N because the topology, by definition, MUST be a subset of P(N), the set of all subsets of N. If some "a" doesn't belong to N, it doesn't belong to any subset of N and hence doesn't belong to P(N), so it can't be in the topology.
@woistdasniveau82902 жыл бұрын
Can we download the exercise sheets somewhere?
@michaellewis78614 жыл бұрын
1:26:37 what are the requirements of phi is it bijective?
@00TheVman4 жыл бұрын
phi just needs to be a function from M to N, nothing more. As an example the map phi(x) = x^2 from the reals to the reals (with the standard topology both times) is continuous, and the preimage of the open interval (-1, 4) is the open interval (-2, 2). This is because some real number x has x^2 in the interval (-1, 4) exactly when x is in the interval (-2, 2). One can check that the preimage under phi of any open set in R is again open.
@dancingstarsfirstlove22272 жыл бұрын
How can one define a topology induced by a metric space
@cartmansuperstar2 жыл бұрын
He did that in one of the previous lectures. A set S is then defined open, if for each element x in S, you can find a (positive) r, such that B_r(x) is a subset of S.
@johnarnold3126 жыл бұрын
Help if you can? What is the difference between a subset symbol with one line under it and a subset symbol with two lines under it? (Google doesn't know)
@ftangdude6 жыл бұрын
I think they're equivalent! Both are used to denote subsets where it may or may not be a proper subset.
@kockarthur79766 жыл бұрын
No, he is using two lines to denote proper subset and one line to denote subset (not necessarily proper).
@kockarthur79766 жыл бұрын
One lines mean subset whereas two lines means proper subset.
@p0gr4 жыл бұрын
@@kockarthur7976 two lines most certainly dont mean proper subset! those two lines are an equal sign and mean that the subset might be equal to the full set. to make it proper, those two lines need to be crossed, like an unequal sign. the subset sign with one line is a variant where the bottom of the sign above serves as the top line of the equal sign. also the sign without any lines usually means the same. so 0,1,2 lines, all the same. to make the subset proper, one or two lines but crossed.
@williamsimpson46704 жыл бұрын
Following on from p0gr's reply: at 33:28 Schuller erases the bottom line on the subset symbol ({\subseteq}) to denote "proper subset" ({\subset}), explicitly saying so.
@codework-vb6er2 жыл бұрын
@9:23 Frederic Schuller alters the space-time continuum of his own youtube video. In fact, I argue that he formed a discontinuity in the continuity of this space time. LOL
@EdgarSanchez-cg9lf5 жыл бұрын
It is funny when he says "it's of kindergarten "
@nashrabanocutepop43346 жыл бұрын
Nice vedio thank you so much sir. Please tell me Msc mathematics all books name?
@joefagan93354 жыл бұрын
Most taught (as opposed to research + thesis) MSc courses will provide a selection of courses from which you have usually some mandatory and several optional subjects.
@Deepakyadav-vp8xx Жыл бұрын
Sound is not clear
@sidddddddddddddd5 жыл бұрын
@26:46 According to which principle? Did he say something in German?
@43hi5 жыл бұрын
Siddhartha Gupte Principle of (Explosion)ex falso quodlibet
@p0gr4 жыл бұрын
latin, not german.
@williamsimpson46704 жыл бұрын
Lecture 1 Propositional Logic 28:30 "ex falso quodlibet" written on the board. (You sometimes see "ex falso sequitur quodlibet" in texts - "from false follows anything you like".)
@taraspokalchuk72568 жыл бұрын
At 34:40 why did we use a subset symbol for N and M but an є relation for U and O?
@EddingDefault8 жыл бұрын
Because O is a set of subsets of M (equivalently a subset of P(M)).
@jovaha8 жыл бұрын
When you define S^1 whit the equivalence relation. how do you get the topology on S^? because you cant induce it from R^1 since now S^1 consist of equivalence classes not elements of R^1 and you cant use your definition of the standard topology since it requires the norm which requires square root and multiplication. Also you said that any norm generates the same standard topology. Is there a more general norm which can be used to define the standard topology for for example equivalence classes? or more simply what operations are necessary for a norm since square roots aren't closed under all sets whit multiplication? Anybody's answer is appreciated :)
@00TheVman4 жыл бұрын
This is a bit late but there is a way to inherit a topology through an equivalence relation. As the professor mentioned this is called the quotient topology. If one has an equivalence relation, like x ~ y whenever y = x + 2*pi*k for some integer k (the equivalence relation I believe the professor meant to write) then one can induce a topology on the set of equivalence classes R/~. One way to understand this topology is by considering the "quotient map" q:R to R/~ which sends a real number x to its equivalence class say [x]. Then one induces the topology by saying that a subset V of R/~ is open if and only if the preimage q^{-1}(V) is open in R (this is also called the final topology on R/~ induced by the map q). Essentially given a subset V of R/~, meaning a collection of equivalence classes, one is looking at the points in R that are a member of some equivalence class in V (one could say the union of the equivalence classes in V). Then the induced topology is just saying that a collection V of equivalence classes is open exactly when this union is open in R. Hope that helps and let me know if anything is unclear :)
@shardaclasses95186 жыл бұрын
Prove that for all n>=1, πn (An)=Z. Sir pls solve this .
@OhadAsor8 жыл бұрын
Utility (quality) function is usually required to be convex. That's wrt economy. For measuring better scientific theories, here's a tip of an iceberg www.cyberneum.de/fileadmin/user_upload/files/publications/TR_145_[0].pdf
@juhokupiainen55157 жыл бұрын
I find your comment intriguing. Do you know of some keywords that would allow me to search for more information on what you are saying about the requirement of the utility function to be convex?
@markvyber24584 жыл бұрын
Is there the Product Topology somewhere??
@alexbaykov92217 жыл бұрын
But why can't we equip the space of all theories with chaotic topology? This will make every sequence of theories convergent. But yeah, they will converge to a complete nonsense as well. Why do we define almost constant as constant everywhere, except finite amount of points? Shouldn't it be countable amount?
@henrywang69316 жыл бұрын
Doesn't finite imply/mean countable?
@henrywang69316 жыл бұрын
No, I was wrong. There can be countably but infinitely many element in a set. But I think his point is that if there are finitely many exceptions, then we will eventually skip through all exceptions for some large N, which would converge the sequence. This we cannot do for countably many, as we can have infinitely many elements in the exception set.
@rewtnode6 жыл бұрын
Great lecture, some funny jokes, a few too many hence, and sometimes funny pronunciation: e.g. ep-Cylon - is that one of those from Battleship Galactica?
@davidwright57195 ай бұрын
This lecture desperately needs to explain the motivations for these definitions. WHY define topologies via open sets? WHY are the 3 requirements what they are? WHY not define closed as not open?
@aakashsharma32163 жыл бұрын
Sir please make a video for CSIR NET mathematics for all
@taraspokalchuk72568 жыл бұрын
Why do we define a topologie as we do? Why do we define continuity the way we do? It's not intuitive, i don' understand help
@lucasdarianschwendlervieir37148 жыл бұрын
Because it is the most general definitions which turn out to give a good intuitive picture of a space and continuous maps, in many cases. This is something you only understand after seeing examples. This is not to say all examples will be intuitive. But some very important examples will be, like the standard topology on R^d. Also continuous mappings (according to the topology definition) from R^m to R^n in the standard topology corresponds with the usual definition of continuity that you will have seen in multivariable calculus. If you wish, you may post your questions in the Physics Forums, since it is easier to type mathematical notations in there. I will try to find your post and if I don't, someone else will probably have a good answer for you.
@taraspokalchuk72568 жыл бұрын
Thank you very much, but one more wuestion, what forums do you recomend?
@lucasdarianschwendlervieir37148 жыл бұрын
www.physicsforums.com/
@abhishekaggarwal27126 жыл бұрын
Anyone interested in forming a study group (in London) ?
@jinggan7125 жыл бұрын
me!
@segilolaphilip63955 жыл бұрын
I'm interested but unfortunately I'm in Nigeria
@ocularisabyssus96284 жыл бұрын
hell no
@jonny1996ahh7 жыл бұрын
What? There is no wy you have 29 topologies with 3 elements in a set. You have 9 of them!
P(M) has 8 elements, which means there are 256 candidates for O, but many get ruled out for not satisfying the axioms.
@freyc1 Жыл бұрын
9 which are not topologically equivalent, but 29 in total.
@mathiasbarreto96333 жыл бұрын
Ppprrrppp! 1:10:55
@whdaffer13 күн бұрын
I find his side comments about "gender and diversity" troubling. In no way does this attract from his quite obvious of the subject. I simply subtract those comments and move on.
@whdaffer13 күн бұрын
Attract==distract
@akashbasak26313 жыл бұрын
Discalculia
@EmperorZelos6 жыл бұрын
why are you using superscript instead of subscript to denote varaious x i'th!? it looks like you are trying to take them to a power!
@disgruntledwookie3696 жыл бұрын
You've obviously never heard of covariant indices...
@EmperorZelos6 жыл бұрын
I have, it looks shit.
@AB-uz8sq6 жыл бұрын
Fortunately, you aren't in charge of standard notation.
@EmperorZelos6 жыл бұрын
You like introducing confusion? Ok!
@raydencreed15246 жыл бұрын
OmnissiahZelos Don’t get mad at other people because you’re too stupid to remember to remind yourself of the distinction.
@theflamingsword6 жыл бұрын
I really love this guy's style (no homeo-ba dum tsh), sad to see that gender studies are very probably even affecting the math and physics departments with their bs.
@KipIngram2 жыл бұрын
41:00 - I'm the right age to have had set theory in lower school. I saw it several times, actually. I never saw the point, and I honestly still don't. Not because it's "not relevant," but because it is *never put to use* in a normal "average person training program*. It's just time spent that never gets called upon again. I think it should just be left to be introduced when a student embarks on a program of learning that actually makes relevant use of it. Also, naive set theory should NEVER be taught - there's just no point in teaching something that isn't fully correct and that modern math is NOT based on. Maybe the reasoning was "it will be easier for youngsters to understand," but that's pointless - just postpone it until it's for sure the student will need it, and then teach it RIGHT.
@chemistrywithmariab60858 жыл бұрын
thanks sir but topology is really hectic ...............i really dont like