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  @slenbi13414 күн бұрын

  Thanks for the series mate, I finally understood that I need to selfstudy more intense)

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  @slenbi13416 күн бұрын

  That was a tough one, thanks) I've got a question about real analysis: which course/book/source would you recommend for it? I just don't have this one at uni but I studied complex analysis so I am little dummy at this one)

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  @CherithLavisetty7 күн бұрын

  How is N an open set? It can be neither open nor close depending on the topology chosen right

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  @matheusjahnke864313 күн бұрын

  18:32 "why is this?"... because picking any point p ∈ X \ {f0}, it is proven that p is in the interior of X \ {f0} (as in, there's a neighborhood of p completely contained in X \ {f0})... which means the interior of X \ {f0} is equal to X \ {f0}... which means X \ {f0} is open;

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  @slenbi134115 күн бұрын

  Hi mate, I didn't get the thing about 'yota-s': is this about sequences of subspace S?

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  @mariusfurter14 күн бұрын

  It should be the inclusion map of S into X, i.e. the function that sends each s in S to itself, viewed as an element of the larger space.

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  @slenbi134114 күн бұрын

  @@mariusfurter oh, now I see the picture, thanks mate, you help me a lot for a research project because I don't have the module in uni)

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  @matheusjahnke864321 күн бұрын

  One thing; I think the topological definition might be redundant; The (i) axiom is redundant if we consider intersections and unions over no sets; Intersection over (no sets) = Universe set = X Union over (no sets) = Empty set (ii) and (iii) imply (i)

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  @richardchapman159222 күн бұрын

  Wondering if these morphing spaces are used by orian to track down the power of money counter accumulation on the black markets.

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  @xanderlewis26 күн бұрын

  These are some of the most thorough and clear videos (on any topic) I've seen so far. I'm amazed this one only has 267 views after three years.

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  @adokwuondoma37Ай бұрын

  I have followed your lectures carefully until now but you didn't explain the meaning of COUNTABLE

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  @mariusfurterАй бұрын

  A set X is said to be countably infinite iff there is a bijection f: N -> X from the natural numbers to X. A set is said to be countable iff it is finite or countably infinite. Intuitively, countable sets are ones in which you can list all the elements one by one in a (possibly infinite) list. Some sets, like the real numbers are too large to be listed in this way.

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  @adokwuondoma37Ай бұрын

  @@mariusfurter so nice and easy. I guess it was so easy that you you overlooked it. Thanks

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  @Maria-yx4seАй бұрын

  I started learning this subject because I thought it was about donuts, turns out its just set theory on steroids, much like abstract algebra. Now I understand the importance of set theory as a foundation for these fields

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  @danielc.martin1574Ай бұрын

  Cool!

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  @Artist-Lover2 ай бұрын

  I think G in the last example is not continuous because the sphere is open and the cube is not. You should change the max to be < 1

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  @Maria-yx4se2 ай бұрын

  thank you for this.

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  @braindead32012 ай бұрын

  For the final example about the connected set that is not path connected, I was having trouble figuring out the detail in the extra notes about the existence of the minimum value a. I managed to figured out a proof which might help others understand. The claim is that there is a minimum a such that g(a)=(0,y). Notice that any value t which satisfies g(t)=(0,y) equivalently satisfies (fg)(t)=0, where f is the projection function along the first coordinate. Hence the set of such t is exactly the set of zeroes for the composite function fg. We know the projection function is continuous and g is continuous by assumption, so fg is also continuous. An important fact to know is that the zeroes of a continuous real function form a closed set. This set is a subset of [0,1], so it is bounded below. This and the fact that it’s closed implies that the set of zeroes contains its infimum. We take this infimum to be the value of a.

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  @braindead32012 ай бұрын

  I don’t think the wedge sum example quite works because we don’t know if the singleton {y0} is closed in Y. If it is closed, we can make the wedge sum, but I don’t think this works for a general topology Y.

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  @mariusfurter2 ай бұрын

  The wedge sum is always defined as X+Y/~ where + denotes the disjoint union space and ~ is the equivalence relation that identifies a point x_0 in X with y_0 in Y. If X and Y are not Hausdorff, the wedge sum may behave unexpectedly. I think the correct statement you want to make is that the wedge sum is not a special case of an adjunction space in general because the conditions on the attaching map could fail if the basepoints are not closed.

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  @youregonnaletityeetyouaway28823 ай бұрын

  love your videos! 1 question, on the basis criterion proof , <= direction, why is it not enough to use the definition of open set that every point in the set has a neighbourhood contained in the set? with B_p as the neighbourhood?

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  @mariusfurter2 ай бұрын

  Thanks! I understand the confusion. In a general topological space, open sets are primitives (i.e they are part of the data of the space). Hence in that case, we need to prove that the basis criterion holds (a set U is open iff you can find a basis element contained in it). Oftentimes, however, one defines a topology on a space by using a basis (e.g. open intervals on R). In that case the basis criterion holds by definition. But not all topologies are defined in this way. The implication is not trivial because it holds for any basis for a given topology, regardless how that topology was defined.

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  @braindead32013 ай бұрын

  Awesome video. I just have one nitpick. Whenever you talk about open sets in the product topology, like in 15:50, you write the open sets as unions of open sets in the component spaces. I don’t think it’s necessary to do all that given that each union would be an open set in each component space already. All you really need is that U=U1xU2.

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  @mariusfurter3 ай бұрын

  It's possible I missed some opportunities to simplify. But beware: Open sets in the product topology are not all of the form U1 x U2 for opens U1, U2 in the component spaces. Such products U1 x U2 only form a basis for the product topology, so opens in the product topology are in general unions of such sets. For instance, in R^2 the open unit ball B_1(0) is not of this form.

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  @braindead32013 ай бұрын

  @@mariusfurterThat’s a good point. I had not thought about that.

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  @nektariosorfanoudakis22703 ай бұрын

  For a less intuitive but simpler version we can take the identity function from a space to itself, but with two different topologies; the topology in the domain being strictly finer than the one in the codomain. For example X is a space with at least two points, the discrete topology in the domain, the trivial topology in the codomain. This example is the same, we can change the topology in the interval so that it becomes homeomorphic to the circle, but it won't be the subspace topology in R^2!

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  @sidharthdembi44003 ай бұрын

  Great set of videos. Thanks.

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  @braindead32013 ай бұрын

  At 32:40 you should not assume that the points are not colinear.

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  @dutonic3 ай бұрын

  the textbook I'm using says the open cover only needs to contain the space X. Not that it needs to equal it.

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  @mariusfurter3 ай бұрын

  Yes, there are two possible definitions. For all practical purposes, they are equivalent. If you have an open cover containing X, you can get one equal to X by intersecting it with X. Conversely, every open cover that equals X also contains X. In proofs one usually goes back and forth between both options, depending on what's most convenient.

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  @sidharthdembi44003 ай бұрын

  Wonderful explanations Thx

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  @Shubham-ic5tx3 ай бұрын

  Thank you professor ❤

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  @SiyabongaPhiofillasMthimkulu3 ай бұрын

  Ada Boy!

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  @eamon_concannon4 ай бұрын

  30:06 I think that f has also to be shown to be continuous to conclude that f is an open map.

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  @anishagrawal3322 ай бұрын

  f is a bijective local homeomorphism, which means it is locally continuous, by a previous proposition, every locally continuous map is globally continuous, so we have that f is continuous.

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  @shengzheyang744 ай бұрын

  I'm confused about the beginning example in lecture 7 that 2 is also a limit point of set {2,2,2,2...}. According to the definition of limit point in lecture 3, a point p is a limit point of a set if any open set containing p contains points of the set other than the point p itself. There are two open sets that contain 2, which are {1,2,3} and {1,2}, and neither them meets the requirement of the definition, then how could 2 be a limit point of the constant set {2,2,2,2,...}?

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  @mariusfurter4 ай бұрын

  You might be confusing "limit point" with "limit". Not all limits are limit points. It is true that 2 is not a limit point of the set {2,2,...} = {2}. It is however a limit of the constant sequence (2,2,2,...) since every neighborhood of 2 must include 2, and hence all of the sequence elements. In the example it is also the case that 1 is a limit of the sequence. In fact 1 is also a limit point of the set {2}.

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  @shengzheyang743 ай бұрын

  After some searching work, I now know that the definition of 'limit point of a set' is not the same as 'limit of sequences'. I thought them were totally same not being two distinct definitions. Than you very much!@@mariusfurter

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  @eamon_concannon4 ай бұрын

  27:15 Part of this is tricky to prove. Suppose by way of contradiction that a nbdh B of y is completely contained in X\A, so B contains no points of A. Since B is open, X\B is closed. Now A ⊆ X\B and since Ā is the smallest closed set containing A, we have that Ā ⊆ X\B. This means B ⊆ X\ Ā = Ext A, hence y ∈ Ext A, which contradicts y ∈ ∂A.

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  @matheusjahnke864320 күн бұрын

  ∂A=X\(Å ∪ Ext A)=(X \ Å) ∩ (X \ Ext A) = (X \ Å) ∩ Ā = Ā \ Å Which means if y is in the boundary of A.... it is equivalent to being on the closure of A, which means (c) applies, while not being in the interior which means the equivalent statements in (a) are false; (a) and (c) may be used to prove (b)

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  @AnoraWu4 ай бұрын

  You just can't imagine how those videos saved my life. Thanks so much for making those awesome videos.

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  @angelmontespalma83864 ай бұрын

  Gracias men, me hiciste entender mucho sobre este tema, que lo tenía como mi nemesis

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  @danielsinderson10474 ай бұрын

  Excellent video. This is not a simple topic, but your explanations are very clear. Thank you!

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  @user-uk9us8wl5l4 ай бұрын

  please can you give me research topological space

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  @juniorcyans29884 ай бұрын

  Very informative! It helped me review many relevant important concepts. Thank you very much!

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  @mariusfurter4 ай бұрын

  You are very welcome!

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  @juniorcyans29884 ай бұрын

  Thank you so much for making these wonderful videos! You've been saving countless lives.

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  @MortyInARobe4 ай бұрын

  these videos are so amazing i really hope more people get to see them, they have really nice explanations and this has been an amazing way to learn topology!

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  @Gatopreto133144 ай бұрын

  Nice

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  @Giulia-mi5ds5 ай бұрын

  Thank you so much for your videos about topology! I found them very helpful

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  @shengzheyang745 ай бұрын

  What does so called restricted map f : U -> Y (where U is a subset of set X) mean? Why isn't it a map f: U -> V where U is a subset of X and U is a subset of Y? how could f map points of U to the whole set Y?

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  @mariusfurter4 ай бұрын

  The notation f: X -> Y means we assign every x in X a point f(x) in Y. Hence the image of f does not need to cover the entire set Y (i.e be surjective). The restricted function f: U -> Y is defined by sending every u in U to f(u) in Y. Again, it does not cover all of Y. As an example, think of the squaring function (-)^2 which sends x to x^2. We can think of (-)^2 : R -> R as a function from the reals R to R even though not every real number is the square of some number. We can also restrict the domain of this function to for instance view it as a function [0,1] -> R. Restricting the codomain of a function is also possible, but one needs to be careful that one doesn't remove points that the function assumes. In fact, by restricting the codomain of any function, you can make it surjective.

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  @shengzheyang744 ай бұрын

  Thank you very much! @@mariusfurter

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  @darrenpeck1565 ай бұрын

  Does this mean there is also a one to one correspondence amongst closed sets and sets that are both closed and open?

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  @mariusfurter4 ай бұрын

  Yes! Homoemorphic spaces share all topological properties. This includes the set of closed subsets and the set of subsets that are both open and closed.

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  @darrenpeck1565 ай бұрын

  Can you show how continuity would breakdown if one open set had a closed pre-image or here if one closed set had an open preimage? There must be a problem of convergence on the limit of the function. These lectures are awesome. Thank you.

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  @mariusfurter4 ай бұрын

  Remember that "closed" is does not mean not open. Rather it means that the complement of the set in question is open. Hence it is possible for the preimage of an open set to be closed, even if the function is continuous. In this case the preimage would be both open and closed. As an example of such a continuous map, consider the constant map R -> R which maps each x in R to 0. Then the preimage of each open set is either empty or all of R. Both of these sets are both open and closed in R. Regarding the preimages of closed sets the same thing holds. If a function is continuous, then the preimages of all closed sets are closed, but they might also be open at the same time. In the specific case of R, the only sets that are both open and closed are R and the empty set. This follows from the fact that R is connected. Hence in R (or any connected space) you can indeed infer that a set cannot be open and closed at the same time if it is nonempty and not all of R.

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  @juniorcyans29885 ай бұрын

  Thank you very much for making this series! Very informative and helpful!

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  @saraperestrelo21705 ай бұрын

  I am following your videos to understand Topology for the first time. Very good explanations. My question: In a discrete space, why is every subset open? It's a question also from the first lecture, I didn't understand. Why can't we find a closed subset? Ex: a set of points.

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  @mariusfurter5 ай бұрын

  In a discrete space every subset is open by definition, in particular singleton subsets are open. Conversely, if all singleton subsets are open, then every subset is open since one can express an arbitrary subset as the union over singletons (its elements). Hence a space is discrete if and only if all singleton subsets are open. By definition, a subset is closed if its complement is open. Hence in a discrete space, every subset is closed as well (since every subset, in particular its complement, is open)

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  @user-qy8ib4ef1g6 ай бұрын

  Thanks

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  @dutonic6 ай бұрын

  Your lectures are super clean and have been such a huge help for my semester first studying topology and my introduction to upper level pure math. You're awesome dude. Thanks a bunch

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  @mariusfurter6 ай бұрын

  Thanks! I'm glad they could help you.

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  @SpotterVideo7 ай бұрын

  Conservation of Spatial Curvature: Both Matter and Energy described as "Quanta" of Spatial Curvature. (A string is revealed to be a twisted cord when viewed up close.) Is there an alternative interpretation of "Asymptotic Freedom"? What if Quarks are actually made up of twisted tubes which become physically entangled with two other twisted tubes to produce a proton? Instead of the Strong Force being mediated by the constant exchange of gluons, it would be mediated by the physical entanglement of these twisted tubes. When only two twisted tubules are entangled, a meson is produced which is unstable and rapidly unwinds (decays) into something else. A proton would be analogous to three twisted rubber bands becoming entangled and the "Quarks" would be the places where the tubes are tangled together. The behavior would be the same as rubber balls (representing the Quarks) connected with twisted rubber bands being separated from each other or placed closer together producing the exact same phenomenon as "Asymptotic Freedom" in protons and neutrons. The force would become greater as the balls are separated, but the force would become less if the balls were placed closer together. Therefore, the gluon is a synthetic particle (zero mass, zero charge) invented to explain the Strong Force. An artificial Christmas tree can hold the ornaments in place, but it is not a real tree. String Theory was not a waste of time, because Geometry is the key to Math and Physics. However, can we describe Standard Model interactions using only one extra spatial dimension? What did some of the old clockmakers use to store the energy to power the clock? Was it a string or was it a spring? What if we describe subatomic particles as spatial curvature, instead of trying to describe General Relativity as being mediated by particles? Fixing the Standard Model with more particles is like trying to mend a torn fishing net with small rubber balls, instead of a piece of twisted twine. Quantum Entangled Twisted Tubules: “We are all agreed that your theory is crazy. The question which divides us is whether it is crazy enough to have a chance of being correct.” Neils Bohr (lecture on a theory of elementary particles given by Wolfgang Pauli in New York, c. 1957-8, in Scientific American vol. 199, no. 3, 1958) The following is meant to be a generalized framework for an extension of Kaluza-Klein Theory. Does it agree with some aspects of the “Twistor Theory” of Roger Penrose, and the work of Eric Weinstein on “Geometric Unity”, and the work of Dr. Lisa Randall on the possibility of one extra spatial dimension? During the early history of mankind, the twisting of fibers was used to produce thread, and this thread was used to produce fabrics. The twist of the thread is locked up within these fabrics. Is matter made up of twisted 3D-4D structures which store spatial curvature that we describe as “particles"? Are the twist cycles the "quanta" of Quantum Mechanics? When we draw a sine wave on a blackboard, we are representing spatial curvature. Does a photon transfer spatial curvature from one location to another? Wrap a piece of wire around a pencil and it can produce a 3D coil of wire, much like a spring. When viewed from the side it can look like a two-dimensional sine wave. You could coil the wire with either a right-hand twist, or with a left-hand twist. Could Planck's Constant be proportional to the twist cycles. A photon with a higher frequency has more energy. ( E=hf, More spatial curvature as the frequency increases = more Energy ). What if Quark/Gluons are actually made up of these twisted tubes which become entangled with other tubes to produce quarks where the tubes are entangled? (In the same way twisted electrical extension cords can become entangled.) Therefore, the gluons are a part of the quarks. Quarks cannot exist without gluons, and vice-versa. Mesons are made up of two entangled tubes (Quarks/Gluons), while protons and neutrons would be made up of three entangled tubes. (Quarks/Gluons) The "Color Charge" would be related to the XYZ coordinates (orientation) of entanglement. "Asymptotic Freedom", and "flux tubes" are logically based on this concept. The Dirac “belt trick” also reveals the concept of twist in the ½ spin of subatomic particles. If each twist cycle is proportional to h, we have identified the source of Quantum Mechanics as a consequence twist cycle geometry. Modern physicists say the Strong Force is mediated by a constant exchange of Gluons. The diagrams produced by some modern physicists actually represent the Strong Force like a spring connecting the two quarks. Asymptotic Freedom acts like real springs. Their drawing is actually more correct than their theory and matches perfectly to what I am saying in this model. You cannot separate the Gluons from the Quarks because they are a part of the same thing. The Quarks are the places where the Gluons are entangled with each other. Neutrinos would be made up of a twisted torus (like a twisted donut) within this model. The twist in the torus can either be Right-Hand or Left-Hand. Some twisted donuts can be larger than others, which can produce three different types of neutrinos. If a twisted tube winds up on one end and unwinds on the other end as it moves through space, this would help explain the “spin” of normal particles, and perhaps also the “Higgs Field”. However, if the end of the twisted tube joins to the other end of the twisted tube forming a twisted torus (neutrino), would this help explain “Parity Symmetry” violation in Beta Decay? Could the conversion of twist cycles to writhe cycles through the process of supercoiling help explain “neutrino oscillations”? Spatial curvature (mass) would be conserved, but the structure could change. ===================== Gravity is a result of a very small curvature imbalance within atoms. (This is why the force of gravity is so small.) Instead of attempting to explain matter as "particles", this concept attempts to explain matter more in the manner of our current understanding of the space-time curvature of gravity. If an electron has qualities of both a particle and a wave, it cannot be either one. It must be something else. Therefore, a "particle" is actually a structure which stores spatial curvature. Can an electron-positron pair (which are made up of opposite directions of twist) annihilate each other by unwinding into each other producing Gamma Ray photons? Does an electron travel through space like a threaded nut traveling down a threaded rod, with each twist cycle proportional to Planck’s Constant? Does it wind up on one end, while unwinding on the other end? Is this related to the Higgs field? Does this help explain the strange ½ spin of many subatomic particles? Does the 720 degree rotation of a 1/2 spin particle require at least one extra dimension? Alpha decay occurs when the two protons and two neutrons (which are bound together by entangled tubes), become un-entangled from the rest of the nucleons . Beta decay occurs when the tube of a down quark/gluon in a neutron becomes overtwisted and breaks producing a twisted torus (neutrino) and an up quark, and the ejected electron. The production of the torus may help explain the “Symmetry Violation” in Beta Decay, because one end of the broken tube section is connected to the other end of the tube produced, like a snake eating its tail. The phenomenon of Supercoiling involving twist and writhe cycles may reveal how overtwisted quarks can produce these new particles. The conversion of twists into writhes, and vice-versa, is an interesting process, which is also found in DNA molecules. Could the production of multiple writhe cycles help explain the three generations of quarks and neutrinos? If the twist cycles increase, the writhe cycles would also have a tendency to increase. Gamma photons are produced when a tube unwinds producing electromagnetic waves. ( Mass=1/Length ) The “Electric Charge” of electrons or positrons would be the result of one twist cycle being displayed at the 3D-4D surface interface of the particle. The physical entanglement of twisted tubes in quarks within protons and neutrons and mesons displays an overall external surface charge of an integer number. Because the neutrinos do not have open tube ends, (They are a twisted torus.) they have no overall electric charge. Within this model a black hole could represent a quantum of gravity, because it is one cycle of spatial gravitational curvature. Therefore, instead of a graviton being a subatomic particle it could be considered to be a black hole. The overall gravitational attraction would be caused by a very tiny curvature imbalance within atoms. In this model Alpha equals the compactification ratio within the twistor cone, which is approximately 1/137. 1= Hypertubule diameter at 4D interface 137= Cone’s larger end diameter at 3D interface where the photons are absorbed or emitted. The 4D twisted Hypertubule gets longer or shorter as twisting or untwisting occurs. (720 degrees per twist cycle.) How many neutrinos are left over from the Big Bang? They have a small mass, but they could be very large in number. Could this help explain Dark Matter? Why did Paul Dirac use the twist in a belt to help explain particle spin? Is Dirac’s belt trick related to this model? Is the “Quantum” unit based on twist cycles? I started out imagining a subatomic Einstein-Rosen Bridge whose internal surface is twisted with either a Right-Hand twist, or a Left-Hand twist producing a twisted 3D/4D membrane. This topological Soliton model grew out of that simple idea. I was also trying to imagine a way to stuff the curvature of a 3 D sine wave into subatomic particles. .---------------

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  @ABCABC-fk7ut8 ай бұрын

  The python example demonstrates foldl instead of foldr.

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  @yomo_138 ай бұрын

  thnx 💗

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  @kamalhaider3978 ай бұрын

  Hi Marius, are you going to make further videos from chapter 5 onwards?

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  @dexteraronstam94149 ай бұрын

  Do you have any plans to continue this series with the mentioned next video looking at when quotients of Hausdorff spaces are Hausdorff?

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  @mariusfurter9 ай бұрын

  I'm teaching topology again this semester so this might be a good opportunity to continue. But I'm also quite time constrained at the moment though.

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  @akanksha83119 ай бұрын

  Can you make playlist on algebraic topology ???!! pleeeease

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  @mariusfurter9 ай бұрын

  It's possible that I might make some videos on homotopy and the fundamental group in the next months but I can't make any promises.

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  @sinclairabraxas355510 ай бұрын

  awesome series thanks