Metric Spaces

[@isxp]

Thank you for a clear explanation and visually appealing graphics. The font, color choice, and animations are all nice to look at. You're helping an ME senior secure his Math minor with these videos

[@DrWillWood]

Thanks so much! Good luck with the numerical methods class (One of my favourite topics and very difficult but so useful!).

[@蔡小宣-l8e]

Brief and clear! Thank you. 谢谢。

[@5Stars49]

Lovely explanation 😮

[@t.ganesh1692]

Excellent video! Thank you Will!

[@ibmalik84]

your channel is really helpful!!

[@mcgrewgs]

At around 2:10, that arrow could be bidirectional, right, since the two statements must be equivalent? I.e. d(x,y)=0 x=y If it weren't bidirectional, I could define, for example, d(x,y)=1 for all x,y and that would fulfill all the requirements.

[@einstien311]

Correct, the impilcation there should be bidirectional. If the distance between two elements is zero, then the two elements must be the same. Conversely, if two elements are the same element, the distance between the element and itself must be zero. If you are trying to prove some proposed function is a distance function, then you have to show forward implication and the backward implication. Interestingly, there is a distance function called the "discrete metric" (I think) that is defined as d(x,y)=0 if x=y, and d(x,y)=1 if x≠y. Essentially, the discrete metric just is a toggle that indicates whether the inputs are the same or different.

[@edgarbonet1]

Indeed. As written, d(x, y) = 1+|x−y| would be a distance over ℝ, which is wrong.

[@stipepavic843]

subbed instantly, great videos!!!

[@JordanWeitz]

Hrm. at 9:07 Will seems to claim a < b ==> |a| < |b|. It is true in this case, but he doesn't explain why.

[@FreeAsInFreeBeer]

Yes, this argument is not correct. (counter example: -3 < 2, but |-3| > |2|) An easy way to prove it is like this. These inequalities hold by definition: -|a| ≤ a ≤ |a| -|b| ≤ b ≤ |b| Add the inequalities together: -(|a| + |b|) ≤ a + b ≤ |a| + |b| We solve by cases: If a+b not negative, then: a + b = |a + b|, and so a + b = |a + b| ≤ |a| + |b|. If a+b negative, then a + b = -|a + b|, and so -(|a| + |b|) ≤ a + b = -|a + b|, multiplying with -1 gives: |a + b| ≤ |a| + |b|. So in both cases it is true that: |a + b| ≤ |a| + |b|. Now, let a = x-y, b = y-z. |x-y + y-z| ≤ |x-y| + |y-z| |x-z| ≤ |x-y| + |y-z| Done!

[@zapzya]

This is true if 0

[@MT.632]

Excellent video!!🌸

[@choukriadan4706]

Thank you for the explanation Dr.will . But this units of so much though

[@edgbaston149]

Thank you so much You're amazing 👏💗

[@Atistatic]

Dr. Wood i would like to ask if anyone who want understand metric spaces, knowning well multi variable calculus is a prerequisite??

[@DrWillWood]

I don't think so! Knowing the basics of set theory is enough to get started

[@BHuman2024]

Would you please suggest any book for learning metric space more.

[@DrWillWood]

Hi! I used Introduction to Topology by Bert Mendelson (Dover Books) for preparing this video which has a chapter on metric spaces and is a nice book in general. I've also heard positive things about Metric Spaces: iteration and application by Victor Bryant but I haven't actually looked at that book myself!

[@zetacrucis681]

8:53 This step is invalid. You can't just take absolute values of both sides of an inequality, e.g.: -2 < 1 does not imply |-2| < |1|, i.e., 2 < 1.

[@josephcampbell4877]

He explains in the video that it's valid since the right hand side of the equation is non-negative

[@bartomiejpotaman6973]

@@josephcampbell4877 the problem is that we don’t know the sign of the left side, zeta crucis is right about that one

[@user-vg7zv5us5r]

More rigorous vectors sans direction parameter.

[@alminaahmed_2552]

It's ❤

[@magroubezpieczeniasp.zo.o.2137]

d(x, y) = d(y, x) ... not in family court

[@tombouie]

Thks, ?????but what is the clear difference between a metric vs normed space???? en.wikipedia.org/wiki/Metric_space vs en.wikipedia.org/wiki/Normed_vector_space

[@DrWillWood]

Hi! A normed linear space (NLS) IS a metric space but the definition but made a little bit more specific to vector space. Its the property (3.) d(x,y) = d(y,x) that is changed. i.e. the distance is same in either direction. for NLS, if we reverse the direction of a vector X we multiply by -1. By property 3 of a NLS we have ||-1X|| = |-1| ||X|| = ||X|| and so its the same with NLS that direction doesn't matter. But we also want that intuitive property of vectors built in that 2 times a vector should by twice as long and so on so that's we why have ||aX|| = |a| ||X|| for constant a in general. The way I think about it is more simple: that a NLS is a metric space that happens to also be a vector space!

[@tombouie]

@@DrWillWood Ooooooh, thks-yous ever so much.

[@Joffrerap]

distance has 2 inputs and norm only has 1 so that's one big difference already.