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While thinking about the set A = { 1/n | n is an element of N }, I got hung up on the proof of the theorem at about 15:00, that 1 is the sup of A, until I made (I think) a leap of insight. When given epsilon > 0, there exists a such that s - epsilon < a. If epsilon is very small, then s must equal a.

One of the best Topology series out there thank you for clearly explaining basis for me took me 3 weeks to know but with you 3 minutes what a lecturer 🔥🔥🔥🔥💯💯💯

is there a way for non -students to access the homework?

Do you also have videos about difference equation ? Mainly finding the General solution, Particular solutation and solving inear differential equations. got a exam in January but can't seem to understand these weird concepts. Any tips would be life saving !

I see neg one third minus one third. For the first example???

Is this for first course ?

it was a bad idea googling DAFPOTRN

6:15 Brute force a hundred vertices in a few days? I think it would be more like a googol years...

At 6:50, because all the mathematicians would switch their votes.

I have a question about the first problem presented in this lecture at 0:30 (but don't feel obligated to answer; I'm not a student in your class, or even at Fairfield!): As a first step, you asked the students to find the dominant terms in the numerator and denominator, and they were to do so without showing any proof. For the numerator, n+2^n they were to recognize that 2^n dominates the n. What does this mean? I think it means that 2^n grows faster than n, i.e., lim [2^n / n ] goes to infinity, or alternatively lim [ n/2^n] = 0. Is this correct? Or maybe it would be preferable to say lim [(n+2^n)/2^n] = 1, but either way.... Then I don't understand your reply to the student who asked (at 4:26 ) if we could avoid using l'Hopital's rule by noting that lim [(n+2^n)/2^n] clearly equals 1.... since we already established that 2^n is the dominate term in n+2^n. Why can we accept (without proof) that 2^n dominates (n+2^n) but not that lim [(n+2^n)/2^n] = 1? They seem the same to me. Thanks for posting your lectures. I really enjoy them!

missing episode 27?

You made me laugh at around 38:00 when you said "Isn't than in the Bible? Either that or Jurassic Park."

What does calculus do with real number? Why do we need to understand real number before understand limit, or derivatives?

At around 12:00, when you made the denominator 9n^2 - 3n^2, couldn't you just throw away the -3 to make the denominator bigger? Then you would have (21n)/ (9n^2) and wind up with 7/(3n).

Aloo sir

Hey i am from India i found this video while search for calculus videos you know your videos are not bad they are helpful

Thank you for the explanation 😊❤

For the epsilon-delta definition of continuity in R, I believe you do not restrict |x-a| to be strictly greater than 0, i.e, x different from a.

My respect to you Prof Staecker!!! Pointset Topology is such an abstract topic and my university only requires linear alg and multivarible calc to take this. So, I have been having a tough time in this course until I found your course. You have made things very well and I hope many students struggling could find your course soon. I will follow through and take notes very closely with your lectures to make sure I finish this semester with confidence in this material! God bless you !!

As a master student & someone switched to math in junior year I never had a chance to take topology during my undergraduate. Topology was a notoriously hard course at my undergrad school. Now I encounter it here&there in commutative algebra class…this series really saved my life. If I had a topology professor as an undergraduate student then I’d definitely like this subject instead of fearing it…prof Staecker you deserve a Nobel prize for teaching topology

Great Professor. Now I can tell people what Real analysis really is as well as point set topology.

There is a mistake in the classes of the torus. There should be a four point equivalence class with all the corners. In the way it was described the classes would not form a partition since the corners are both in the B_1 or 0 and in C_1 or 0.

16:35 (hiking and distance measurement by apps) The way they actually work with timed intervals aside, this seems like one that should be reasonably measurable (with an error range), since every step taken is from a specific position to another specific position, and the nature of your hiking motion is one that even if not linear, does have a limit in the distance traveled between those two points. The exact positions of those points can be refined, but the distance between them when calculated would not go to infinity as the positions and trajectory remain static. I think

She is all that 2:4:2:6

Brea

That's good 👍👍

Yes

32:00 you were fine 😊 🟥🟥| 🟩🟩🟩 🟥🟥| 🟩🟩🟩 🟦🟥| 🟩🟩🟩 🟦🟥| 🟥🟥🟩 🟦🟦| 🟦🟥🟥 🟦🟦| 🟦🟪🟪 🟦🟪| 🟪🟪🟨 🟦🟪| 🟨🟨🟨 🟪🟪| 🟨🟨🟨 🟪🟪| 🟨🟨🟨

where is ep05?

Math can't explain why we are here? Wait, math explains why I am here!

At 4:40: gotcha! log base 2 of 3 is not 8, but log base 2 of 8 is 3.

I love seeing these voting system-related math explorations so close to elections ☝One minor note 5:00 ... "C = 3 · 0 + 2 · 1 = 1" ? ( I kept waiting for the correction, or a student to catch it. It doesn't affect the IIA issue you were demonstrating. )

Is a revolving door open or closed?

I think a distinction needs to be made between a basis for a topology and a basis for the topology T (already established on X). The conditions given were for B to generate a topology, not for it to generate the topology T.

This video made me cry. It saved my Econ PhD.

Such a put-together video!! i have been trying to understand this concept since few days and this video helped me grasp the concept quiet well.

Thank you so much for having these lectures and the course website publicly available!! This was great

I see from math overflow someone says that in the projective plane, the corner cells of tic-tac-toe are diagonally self-adjacent, so you only need two moves. I can kind of see where they're coming from. Say you had the top left corner cell and you break the diagonal move into two parts: an up (which takes you to the bottom right) and then a right which takes you back to the top-left! I suppose this is an "up-left" diagonal from the top left because the second part of the move, to the right, would have been to the left before flipping over. So going diagonally from the top left takes you to the top left, so you can use that one twice! The question is called "Tic-Tac-Toe on the Real Projective Plane is a trivial first-player win in three moves", and the answer is by Parcly Taxel. I won't post a link because then TH-cam eats the comment!

I never actually took Real Analysis but I did take a class on topology, so I approached this whole subject of open and closed sets from the other direction, with the general definition of a topological space, where you just arbitrarily take certain subsets of it as open, to define the topology, as long as they satisfy certain axioms. And then it was kind of a mental stretch to see how this very general definition corresponded to the ideas of open and closed that I learned back in junior-high algebra class. But it's interesting seeing the stuff I learned as the *axioms* of open sets derived as theorems from the standard topology on the real line. I guess from a topological perspective, these theorems constitute proofs that the real line with the standard topology is a topological space.

This pattern of "introduce the theorem at the end of the lecture, talk about it less formally, then prove it at the beginning of the next lecture" is an interesting one that I don't think I ever tried when I was teaching stuff. Gives the class some time for the statement to sink in.

when talking about which of those diamond sets are open in the given topologies, (around 26:00) i was confused as to why the diamond with only its lower left edge included was not open in the lower limit X lower limit topology. obviously the interior points of the diamond can be completely contained within a basis set, but since the basis sets include their lower left corners, couldn't a point on the diamond's lower left edge stay completely inside a basis set if you put it at that set's lower left corner?

I saw the cone but not S^2. But for me that's progress.

There's an extra wrinkle to watch out for in the diagonal proof: it's best to construct your "choose another digit" formula so that it avoids choosing 9s, just to make sure you can't get caught by the .9999... = 1 ambiguity in decimal numerals, which implies that some different decimal expansions can represent the same number. That's easy to do, though. You can just choose 0 if the digit is either 8 or 9, for example. And assume that in your list, you never use the representation that ends in an infinite string of nines. I think that as long as you do that, you can show that the diagonal argument is really the same thing as the nested-intervals proof. Choosing an nth digit that is different from the nth number is just a specific procedure for choosing nested closed intervals that avoid the sequence of numbers. Another quibble: I think alef-one is defined to be the "next" cardinality after alef-zero, so the cardinality of the real numbers is only alef-one IF the continuum hypothesis is true.

26:30 I think the "when added to itself" is specifically where the natural numbers come in. He's talking about taking some fixed interval, then adding it to itself, that is, *multiplying* it some natural number of times. And that can be made to exceed any other magnitude, that is, any real number. So Archimedes is really saying that for any real numbers a and b, an > b for some n in N. That implies the modern version of the Archimedean property if we take a=1.

Sup of the evening, beautiful sup

I keep wondering about the first proof in this one, particularly the reverse case. We can imagine sort of reasonable definitions of |x| and > and < on the line with two origins: there's a second origin 0* which is less than any positive number and has absolute value 0*. Then the theorem is incorrect because the right side is true for 0*. So there must be some property of the real numbers we're implicitly assuming and using here that excludes the line with two origins, and I can think of a lot of them, but most of them are kind of close to assuming the thing we're trying to prove. Maybe that isn't so bad because it's not supposed to be a complicated proof. I guess the proof assumes that if x is not 0, |x| must be greater than zero, which is not true for the definitions I came up with for the line with two origins (|0*| must not be greater than 0, or else the proof will still work because we could just set epsilon to 0*).

Euler wasn't a real person. Can't believe this is still being spread

Gem of a series, awesome teacher

thank you very much for these lectures! I found them perfect and very useful... Thanks!!!

Thanks professor❤😊