Errata At 1:02:25, I should have said that it is the *complement* of the Cantor set that has full measure. The Cantor set is what remains after omitting all those middle-thirds sections.
@simohunt2 жыл бұрын
I have no idea of where you intended at pitching the level of these lectures at but as a 46 year old accountant with a decent level of physics knowledge I found them perfect. Thank you.
@NathanWHill2 жыл бұрын
When I was an undergrad I wanted to learn about the hyper reals and no one would talk to me about them. As a consequence I'm very grateful for you putting this out.
@NikolajKuntner3 жыл бұрын
The inevitability of a function concept makes one wonder whether there aren't solid foundations that take such a notion as a primitive. Whether it be category theory, computational calculi or von Neumann set theory-like frameworks... The proposals generally appear to be either less workable or much harder to grasp than the theories that start with sets, types or domains of discourses. Maybe I have that impression because extensionality for sets is a simple starting points and other, younger concepts try to broaden this notion of equality.
@atlas40742 ай бұрын
On continuous induction at 37:20, this can be extended to connected topological spaces, and linear orders are connected only if they are dense orders with the least upper bound property. The more general versions states that if A is non-empty, whenever x is in A there's a neighbourhood of x contained in A (i.e. A is open), and whenever V is an open subset of A then A contains the boundary of V, A is the whole, connected space. The last condition is like saying 'if points arbitrarily close to x are in A, then x is in A.' The second and third properties tell you A is both open and closed, which in connected spaces means A is empty or the whole space.
@columbus8myhw2 жыл бұрын
I read a nonstandard analysis textbook ages ago. I remember being very impressed, but eventually decided that on some level it mostly amounted to swapping some notation out for other notation (replacing lim with st and so on). Reading about Nelson's IST was a trip, though. (Adding nonstandard stuff to the entire set-theoretic universe!) And I remember exclaiming "Induction is broken!" after realizing that, in axiomatic first-order logic, you can only require that induction works for properties that are definable in your language, and you can break it simply by introducing a new thing to your language. (Eg 0 is standard, and if n is standard then n+1 is standard, but we can't prove that all n are standard because induction is only an axiom for things that are defined without the word "standard")
@schmetterling4477 Жыл бұрын
Yes, but in practice absolutely nobody of importance cares. Not the engineers, not the physicists and not even the majority of mathematicians. This is trivial early 20th century stuff and it makes no difference whatsoever for people who are studying actual problems in calculus, functional analysis, differential equations or differential geometry.
@columbus8myhw Жыл бұрын
@@schmetterling4477 New tools open up new modes of thinking, even if they're niche.
@schmetterling4477 Жыл бұрын
@@columbus8myhw The solution to a differential equation simply doesn't depend on nitty-gritty stuff like nonstandard calculus.
@columbus8myhw Жыл бұрын
@@schmetterling4477 Correct. They are provably equivalent. ("Provably" meaning "there is a proof that they are equivalent")
@schmetterling4477 Жыл бұрын
@@columbus8myhw Then whoever came up with it simply reinvented the wheel and you can pick up your mathematics Nobel at the nearest car wash to you.
@columbus8myhw2 жыл бұрын
Continuous induction basically says, "if you're nonempty, open, and closed, and your universe is connected, you're the whole universe!"
@rahibrehman42452 жыл бұрын
These lectures are awesome!
@antoniostojanovski90083 жыл бұрын
Thank you very much for your lessons,they are amazing
@PhillipRhodes4 ай бұрын
Holy shit. I thought I (sorta) understood Epison-Delta, but it wasn't until I saw your diagram at around 05:11 that I think I *really* got it. That one diagram was worth the price of these videos. Oh, wait... still, you get my point. 🙂
@rsm3t3 жыл бұрын
For the problem at 34:00, is it functions with no isolated points? I don't know if this property has a name.
@joeldavidhamkins54843 жыл бұрын
Yes! The third statement requires that every point on the graph is a limit point of the graph of the function.
@RenRealism3 жыл бұрын
Nice lecture!
@leanhdung19943 жыл бұрын
Thank you Sir!
@StephenPaulKing9 ай бұрын
What downsides might obtain by taking delta to be nilpotent?
@ndmath3 жыл бұрын
Thanks!
@dimitardimitrakov28413 ай бұрын
I am listening on my laptop full volume and somehow struggle to hear. May be it's a good idea if you can edit the video and raise the volume up a bit
@Achrononmaster2 ай бұрын
@41:15 The Asgaardians did one better than Yuen Ren Chao, they figured out the any amount of beer *_or_* mead is allowable theorem.
@Achrononmaster2 ай бұрын
And Abraham Robinson died of beer thirst because he only allowed himself a hyperreal ball amount of beer.
@ryanj7483 жыл бұрын
About the definition at 33:45--does the definition say that there exists a dense subset D of the domain of f such that the restriction of f to D is continuous?
@joeldavidhamkins54843 жыл бұрын
This is not quite strong enough. The third property asserts that every point on the graph of the function is a limit of other points on the graph. So if you change the value of a continuous function at one point, it would have your property, but not the property I stated.
@jrshipley3 жыл бұрын
I don't have the quote handy, but I'm sure there's a passage in Berkeley where he, talking about algebra (use of imaginary numbers, I think), endorses a sort of formalist, instrumentalist theory of symbolic significance (the sign gets meaning from the role it plays in a system of calculation, rather than from the idea it stands for). It's long been a project in the back of my mind to try to get to the bottom of why that was fine for algebra but not calculus. Anyway, I teach 5/5 at a community college, with little time for such projects, so if any of you young phil math hotshots out there can get to the bottom of that and write a nice article for me to read.... ?
@joeldavidhamkins54843 жыл бұрын
Perhaps one possible difference might be that Berkeley thought the infinitesimal theory might be inconsistent? After all, he was mocking them for allowing methods that could also be used to derive nonsense.
@columbus8myhw2 жыл бұрын
I suspect the deal was, it was not clear what the rules are. With i, we can manipulate it purely formally. (Various bits from nonstandard analysis, if memory serves, involve formally manipulating _quantifiers,_ not just pieces of equations, which is an added level of complexity.)
@scythermantis2 жыл бұрын
@@joeldavidhamkins5484 Signs DO get their meaning from not just the 'role they play' but more the relationship between them and *OTHER* signs, according to Derrida; in fact, there *IS* nothing 'behind' that the signs signify; there is no 'signified', only 'signifiers'. See kzbin.info/www/bejne/fKWvfoGZpcyMe9U Curious as to how Bennington and you would discuss this.
@Dystisis Жыл бұрын
@@scythermantis What role they play *includes* their so-called "relation" to other signs, which is a more vague notion that the notion of what role they play (in calculation and in general use).
@Achrononmaster2 ай бұрын
@1:21:00 who is your colleague in New York? Carl ????
@mtgodziszewski3 жыл бұрын
It seems that the statement at 33:58 defines the class of Darboux functions, i.e. the functions with the intermediate value property, right?
@joeldavidhamkins54843 жыл бұрын
Not quite. Although every Darboux function has the property, there are also other functions with the property. It is slightly more general.
@mtgodziszewski3 жыл бұрын
@@joeldavidhamkins5484 Thank you, right, I see: we do not require that f(y) is a fixed value, but rather that it is close to f(x). An example is any bijection between the unit interval and the set (2,3) \cup (3,4). The image of the interval is not an interval itself. In the condition you ask for, the image of every interval is only dense-in-itself (i.e. with no isolated points), not necessarily an interval. But I do not know the name of this class of functions
@joeldavidhamkins54843 жыл бұрын
@@mtgodziszewski These are the functions whose graph has no isolated points.
@evangorstein43453 жыл бұрын
Is the statement at 33:58 another definition of continuity?
@joeldavidhamkins54843 жыл бұрын
Not quite. Although the continuity of f implies the property, it is not equivalent to continuity. There are discontinuous functions with the property. For example, the Conway base-13 function has the property.
@luyombojonathan66887 ай бұрын
Thanks alot
@mtgodziszewski3 жыл бұрын
At 50:33 you make an argument that "we can give an account of - at least in simple cases - of abstract existence, for example to say what does it mean to say that the empty set exists". Can I ask what do you take to be this account? Is it some philosophical, semantic (extra-mathematical?) account? An immediate naive reply by just giving the definition "it is a set that has no elements" is obviously not satisfying, since it relies on the assumption that we actually can give an account of the 'existence of sets', which, if explained simply in terms of the axioms of some set theory (defining a notion of set), still leave unanswered the question of the existence of the objects of the theory - I mean, one can of course say that a set is any object defined uniquely by what are the elements of this object, which again opens the question of giving an account of 'being an element'. Therefore, do you take this account to be something along the lines of hilbertian or if-thenist theories that give an account of existence in terms of consistency along the lines of e.g. conceptualist theory, according to which existence of mathematical objects 'somehow' relies on human/rational mental operations, or something completely different?
@joeldavidhamkins54843 жыл бұрын
What I had in mind is something like this: Whatever there is, we can consider the property that never holds---every predication instance fails. This property seems perfectly sensible, whatever one's ontological commitments might be, and this property provides a way of understanding what the empty set is. This account does not seem burdened with a requirement to develop an elaborate set theory of any kind. It seems possible to talk coherently about the empty set or the never-holding property or the never-true predicate, regardless of what objects exist and what nature they have. In contrast, to describe what it means for an object to exist physically seems burdened with all manner of complicating and quite mysterious issues concerned with the nature of substance and material. In light of the fundamental confusing aspects of quantum mechanics etc., there are very elementary aspects of physical existence that we cannot easily explain.
@mtgodziszewski3 жыл бұрын
@@joeldavidhamkins5484 Thank you. I completely agree that 'physical existence' is rather mysterious. But do I understand correctly that ontology of sets you would offer is that they are references of properties? Or what you say rather provide an account of 'simple' abstract concepts? Whatever the answer, would you say that there is also an easy way to give a (similar?) philosophical account of infinite sets?
@joeldavidhamkins54843 жыл бұрын
@@mtgodziszewski I was only claiming the simple case. But of course, my belief is that one can carry on for quite a long way, including infinite sets, but of course, the account becomes more and more difficult, and therefore doesn't serve the point in the argument I was making here, namely, that some abstract objects (such as the empty set) seem to have simpler and more thorough accounts than physical existence.
@lunavicta41722 жыл бұрын
33:35 For any x, there is a point y arbitrarily close to it such that f(x) == f(y). For example, if we had the following piecewise function: x != 0: f(x) = sin(1/x) x == 0: f(x) = 0 I don't think it's *continuous* at x = 0 because for epsilon 1/2, no matter how small your delta, there are points within it with f(x) = 1, which is further than 1/2 from 0. However, there is a *particular* point y however close you like to x = 0 such that f(y) == 0, satisfying the condition that f(y) is within epsilon of f(0). Is that moving in the right direction at least?
@schmetterling4477 Жыл бұрын
Sounds about early 19th century correct.
@pinklady71843 жыл бұрын
You spell "rigor" in your video title. American English spelling. Are you British?
@joeldavidhamkins54843 жыл бұрын
I am American. But having lived now in the UK for several years, my usage is mashed potatoes! My students say x, y, zed, and so do I, sometimes, but not always, and saying "parentheses" confuses them, so it is "brackets". And don't get me started on the issue of using pants as trousers!
@chonpincher2 жыл бұрын
@@joeldavidhamkins5484 The typographical terms for ( ), [ ], and { } respectively are parentheses, brackets, and braces. Calling parentheses "brackets" loses a useful distinction. Another point: A "space-filling curve" can't be a continuous map from the unit interval onto the unit square, which is impossible. In particular, such a map would violate the conservation of dimension by continuous injections between euclidean spaces. What may be said about the given sequence of curves in this respect is that points in the unit square can be arbitrarily well approximated by points on the track of a curve sufficiently far down the sequence. Also, the sequence of curves exhibits a Cauchy-type sequential convergence (with respect to a natural metric) - without admitting any limit curve. Cauchy-convergent sequences are convergent in finite-dimensional euclidean spaces, but not in the space of continuous maps from the unit interval to the unit square.