topology can make distance weird...

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Michael Penn

Michael Penn

Күн бұрын

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Пікірлер: 37
@MrJoshslatton
@MrJoshslatton 7 күн бұрын
When I was a kid, I had a dream of playing baseball in the Major Leagues. Now I have a dream of being a topologist...and I'm not really sure which is more unrealistic.
@skylardeslypere9909
@skylardeslypere9909 7 күн бұрын
Just gotta have an *open* mind
@RockBrentwood
@RockBrentwood 2 күн бұрын
Well, I picked up a few things fast, here and elsewhere. A separable space is the closure of a countable set. The second countable space is one whose open sets are unions drawn from a countable collection. Euclidean geometries all satisfy that property (one such collection being the interiors of circles of rational radii centered on rational coordinates). 0:00 A metric space, M, has a distance measure A,B ∈ M ↦ AB ∈ ℝ, such that AA = 0, AB > 0 if A ≠ B, AB = BA and AB + BC ≥ AC. Collinearity can be defined by AB + BC = AC (making ABC collinear, with B between A and C). For Euclidean geometries, for such an ABC triple and any point D: AB · CD² - AC · BD² + BC · AD² = AB · AC · BC. (That's also enough to make it possible to ensure that if ABC and BCD are collinear triples, then so are ABD and ACD.) Euclidean geometries are also infinitely extensible: for any points A, B and integer N > 0, there is a point C such that AB + BC = AC = N · AB. They are also infinite divisible: for any points A, C and integer N > 0, there is a point B such that AB + BC = AC = N · AB. They are also Zeno-defying: for any sequence of points A₀, A₁, A₂, A₃, ⋯, such that A₁A₂ ≤ ½A₀A₁, A₂A₃ ≤ ½A₁A₂, ⋯, there is a point A such that A₀A ≤ 2 A₀A₁, A₁A ≤ 2 A₁A₂, A₂A ≤ 2 A₂A₃, ⋯ The above-mentioned properties, together, serve as an *equivalent* axiom set for Euclidean geometry! The largest number of points that can be made equally distant from one another, in it, is the dimension of the geometry plus one. 11:48 For a separable space: "has a metric" = "is second countable". But, what he didn't mention (and what I just picked up) is that it can be stated *unconditionally* as: "has a metric" = "regular" + "Hausdorff" + "countably locally finite". You'll have to look those up, but "regular" + "Hausdorff" = T₃ and "Hausdorff" = T₂, and the proof ("On A Necessary And Sufficient Condition Of Metrizability", by Jun-iti Nagata, Journal Of The Institute Of Polytechnics, Osaka City University, Volume 1, Number 2, Series A) is on-line and is only 8 pages. I didn't know any of this until a few moments ago.
@newwaveinfantry8362
@newwaveinfantry8362 2 күн бұрын
​@@RockBrentwood He stated the theorem incorrectly in the video and so did you. A metric space is separable if and only if it is second countable. That's the correct version. What you and Michael said implies that separable + second countable = metric + separable, but that's false. Second countable implies separable, so that would mean that every second countable space is metrizable, which is false.
@filippburyak8221
@filippburyak8221 6 күн бұрын
I believe your fact at 11:01 is just wrong. Take X={0,1} with trivial topology (only empty set and whole X are open). It's obviously second countable and separable. However, it cannot be metrisable since it is not Hausdorf. That being said, I guess the statement you wanted to use is: A metric space X is separable iff it is second countable.
@newwaveinfantry8362
@newwaveinfantry8362 2 күн бұрын
Yes. He got it wrong. The correct theorem is: separable + metrizable = second-countable + metrizable. Separable + second-countable is just second-countable, as that already implies separable, but not every second-countable space is metrizable and vise versa.
@YT-lu7ct
@YT-lu7ct 7 күн бұрын
B_x is chosen for fixed y. So, B_x is depending on y. Hence, I think the injective map x \mapsto B_x is not defined yet. Maybe we fix any y first. Then, for any x
@MK-13337
@MK-13337 6 күн бұрын
You could amend the argument a bunch of ways but I think it is fine as it stands. For any real numbers x,y where x=/=y if we choose our B_x and B_y correctly they are not equal. Either x
@goodplacetostop2973
@goodplacetostop2973 7 күн бұрын
17:38
@minimo3631
@minimo3631 7 күн бұрын
A good place to stop indeed
@αηομαλψ
@αηομαλψ 7 күн бұрын
Yes! I need topology to understand functional analysis. Thank you 🙏
@timotejbernat462
@timotejbernat462 6 күн бұрын
14:00 so the interval (a,b) is “open”, but by construction the interval [a,b) is also “open”? I’m really confused, how is “open” actually defined? It seems like that term gets abused a lot in topology
@Nameless.Individual
@Nameless.Individual 6 күн бұрын
A topological space is a pair (X, S) consisting of a set X and a set S consisting of subsets of X, in this notion every element of S is said to be an "open set". On the other hand, intervals on the real line have their usual definition of open/closed if they exclude/include their end-points respectively.
@JPK314
@JPK314 7 күн бұрын
15:44 this doesnt prove that there exists a basis set which both contains x and is a subset of [x,x+1) though, right? The basis set which you're guaranteed that is a subset of [x,x+1) doesn't need to contain x.
@charlottedarroch
@charlottedarroch 7 күн бұрын
The usual definition of a basis of a topology (more frequently called a base) is that every open set is a union of elements of the base. As every open set is a union of base elements, and as x is contained in the open set [x,x+1), x is contained in a union of base elements, each of which is a subset of [x,x+1). So there must be a base element containing x and included in [x,x+1).
@JPK314
@JPK314 7 күн бұрын
@charlottedarroch ah, it's the definition of a basis / base which I was missing
@potato1793
@potato1793 7 күн бұрын
Is there any way to show that R with the lower limit topology isn't second countable without the axiom of choice?
@MK-13337
@MK-13337 6 күн бұрын
We never needed the axiom of choice when proving that R_l is not second countable. An injective function existing from R to B means B is uncountable because R is. R is uncountable proven by (for example) the Cantor diagonalization argument.
@Nameless.Individual
@Nameless.Individual 7 күн бұрын
One careful comment to add is that the theorem should state "does not possess a metric compatible with the topology", since a metric structure can generally be defined on any set. For example, just take d(x, y) = 0 iff x = y and 1 otherwise.
@atreidesson
@atreidesson 7 күн бұрын
You didn't really need to use "iff"
@erikfauser2418
@erikfauser2418 7 күн бұрын
I think it’s pretty obvious what is meant by “has a metric” if it’s clearly about the topological space and not just the set
@Nameless.Individual
@Nameless.Individual 7 күн бұрын
@@erikfauser2418 I have personally lost weeks of research time chasing false propositions because of incomplete statements or implicit assumptions, so I have grown a penchant for pedantry. This rings especially true as this channel is pedagogical in nature and hence ought to promote best practices, though that is my own opinion.
@erikfauser2418
@erikfauser2418 7 күн бұрын
@@Nameless.Individual the more accurate wording would be “metrizable”
@Павал-л8ч
@Павал-л8ч 6 күн бұрын
It is enough to say "a topology space X has a metric" because if the metric is not compatible with the topology, it is not a metric on this topology space. The notion of topology space "fixes" the topology.
@Alan-zf2tt
@Alan-zf2tt 7 күн бұрын
Math is beautiful ⭐
@pierreabbat6157
@pierreabbat6157 7 күн бұрын
What does "separable" mean? I thought it was a kind of differential equation.
@atreidesson
@atreidesson 7 күн бұрын
I stared at the second line a minute to realize that the inner brackets are around "iff", not around " and" :)
@Nikolas_Davis
@Nikolas_Davis 7 күн бұрын
One of the bummers of being a physicist and doing general relativity is that you constantly work with a "metric" which isn't a metric (according to the standard definition). Curved spacetime has a local (infinitesimal) structure which is usually called the "Minkowski metric", and can be thought of as defining spacetime distance; but it can be positive, zero or negative; distinct points can have zero ("null") distance between them; and the triangular inequality is inverted: straight lines are the *longest* paths between two points in spacetime.
@bizopca
@bizopca 7 күн бұрын
And this is a bummer, why?
@Nikolas_Davis
@Nikolas_Davis 7 күн бұрын
@@bizopca Oh, just joking... part of a tongue-in-cheek rivalry between mathematicians and physicists, about how formal exactly one should be with terminology ;-)
@Happy_Abe
@Happy_Abe 7 күн бұрын
Funnily enough I’m in a math PhD program and in my differential geometry class we just defined a Riemannian metric and I was so confused trying to understand how it’s a metric. Turns out it isn’t a metric in the normal metric sense, yet it’s still being referred to as “metric” throughout the course 😂
@Nameless.Individual
@Nameless.Individual 7 күн бұрын
Nothing quite as difficult as shedding away decades of pre-existing traditional terminology due to how the topics historically developed. Unfortunately, the pseudo-Riemann only look like ducks but don't quack like one.
@Happy_Abe
@Happy_Abe 7 күн бұрын
@@Nameless.Individual yeah, I love when terminology accurately reflects what the thing is and is consistent across different subjects in the field. Everyone should agree what a duck is and how it quacks ideally 😂
@topquark22
@topquark22 7 күн бұрын
I have a proof about the Mandelbrot set, about how to iterate using different exponents. You know that for x -> x^2 + c, then x -> infinity if |x| > 2. Well, I worked it out for any exponent. It would be cool if you could explain my proof on your channel. How can I contact you?
@hqTheToaster
@hqTheToaster 7 күн бұрын
Here to chill out and show good vibes. I am a simple person. I see 'DNE', I appreciate the poetry.
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