My first introduction to open sets was in my metric spaces course, this video definitely helped simplify the concept for me. Thank you for the great video.
Could you make a video for the following question: For part (a), show that f(x)=|x| is not differentiable at x=0. For part (b), show that if f: R->R is differentiable at x0, then f is continuous at x0. Tank you very much!
For the first part. Function is differentiable at a point if ANY sequence of points getting closer to the point of interest aproaches the same value of the slope. So, we play a game. I try to find two sequences of infinitly many points, approaching your point, such that the slopes differ (no matter, how close I get). If I find two such sequences, I win and a function doesn't have a derivative at a point. If I fail (slopes of any pair of sequences approach same value) - the function is differentiable. Now. I build two sequences. f(1/n) and f(-1/n). Clearly, the first sequence yields a derivative of 1, the second: of -1. I win and the function has no derivative. At the same time, here "the derivative from left (or from right)" exist. - 1 for the former and +1 for the latter. There are nastie functions like y=sin(1/x) with no left or right derivatives. Even worse. There are functions (like Weierstrass function) that are continuous everywhere, differentiable nowhere. But to understand this function you need "series theory". This is the second year of the university math. Or a series (pun intended) of lectures on youtube. Tl;dr: from definition of a derivative.
I have a question: to say that the set X = [0, 1] it's not open, we have to say that X is a subset of another set, such as R for exemple? Because, if we think that [0, 1] is the entire space ("universe" space) when we make a open ball in point {1} for exemple there's no other space such that a point not belongs to [0, 1], in this case, a open ball will contain points that only belongs to X. Is this correct or there's some error in this argument?
Formaly. Here we limit ourselves to open sets. A number of theorems don't apply to closed sets like [0,1] or semi-open sets like [0, 1). Open sets must have some nice properties. Nice enough, that we study them separately. After all, we don't prove theorems for sets (3,5] (those are not general enough). Tl;dr "What motivates mathematicians to impose such an strange requirement on sets? After all, we just exclude two points (points {0} and {1} from a continuum set of points)?"
If I understand your question correctly the answer is you can't! For example in (a,b) since the interval is open, it is the case that every x in (a,b) has an open neighborhood around it completely contained in (a,b). But x cannot equal a, the boundary, because a is not in the set.
My first introduction to open sets was in my metric spaces course, this video definitely helped simplify the concept for me. Thank you for the great video.
I just wanna appreciate your mic dude.
I'm always trying to make the videos as high quality as possible! So that means 4K face-cam and a high quality mic!!
Great video as always
Thank you doctor you save me at this point before final
this man talks math like an asmrist lol but i love it
Thanks a lot
Most welcome!
Great! Thank you!
Glad to help!
Could you make a video for the following question: For part (a), show that f(x)=|x| is not differentiable at x=0. For part (b), show that if f: R->R is differentiable at x0, then f is continuous at x0. Tank you very much!
Thanks for watching and the request! I can do part a, I would do part b in a separate video, I'll get to it as soon as I can!
For the first part.
Function is differentiable at a point if ANY sequence of points getting closer to the point of interest aproaches the same value of the slope.
So, we play a game. I try to find two sequences of infinitly many points, approaching your point, such that the slopes differ (no matter, how close I get). If I find two such sequences, I win and a function doesn't have a derivative at a point. If I fail (slopes of any pair of sequences approach same value) - the function is differentiable.
Now. I build two sequences. f(1/n) and f(-1/n). Clearly, the first sequence yields a derivative of 1, the second: of -1.
I win and the function has no derivative.
At the same time, here "the derivative from left (or from right)" exist. - 1 for the former and +1 for the latter.
There are nastie functions like y=sin(1/x) with no left or right derivatives.
Even worse. There are functions (like Weierstrass function) that are continuous everywhere, differentiable nowhere. But to understand this function you need "series theory". This is the second year of the university math. Or a series (pun intended) of lectures on youtube.
Tl;dr: from definition of a derivative.
Thanks
Glad to help, thanks for watching!
can you explain in detail about the null set is an open set
u have explanation of closed set?
If { } is open, does that imply the universal set is closed?
Yes it does, though it is worth noting that { } is also closed, and the universal set is also open.
@@WrathofMath why?
I have a question: to say that the set X = [0, 1] it's not open, we have to say that X is a subset of another set, such as R for exemple? Because, if we think that [0, 1] is the entire space ("universe" space) when we make a open ball in point {1} for exemple there's no other space such that a point not belongs to [0, 1], in this case, a open ball will contain points that only belongs to X.
Is this correct or there's some error in this argument?
Zero is not forwarded?
Formaly. Here we limit ourselves to open sets. A number of theorems don't apply to closed sets like [0,1] or semi-open sets like [0, 1). Open sets must have some nice properties. Nice enough, that we study them separately. After all, we don't prove theorems for sets (3,5] (those are not general enough).
Tl;dr "What motivates mathematicians to impose such an strange requirement on sets? After all, we just exclude two points (points {0} and {1} from a continuum set of points)?"
Exclusions opposiaates
Keep going
I do not get why you can find a d for an open set when x=exactly the boundary of the open set... still one d will be out of the set...
If I understand your question correctly the answer is you can't! For example in (a,b) since the interval is open, it is the case that every x in (a,b) has an open neighborhood around it completely contained in (a,b). But x cannot equal a, the boundary, because a is not in the set.
Thanks! I think i got it
Great video but your self-video is too distracting
Let's get real with Wrath of Math! 😀
Open your heart to the open sets!
Before I watch the video i'm hoping that he says that the reals are open by the Archimedean Principle
1:29 Open set: