f inverse is continuous

[@RC32Smiths01]

Very interesting as always man!

[@yooooooooumu2838]

Very helpful! Thx!

[@MrCigarro50]

Le quedo muy agradecido por este y todos los videos que Ud. nos ha brindado.

[@drpeyam]

¡De nada! El gusto es el mío

[@DynamicMateTV]

Rigorous proofs are very satisfying :))

[@andresvasquez5411]

If we have this theorem, that if a function is continuous then the inverse is also continous, can we say that this kind of functions are homeomorphisms?

[@dylank6191]

In metrical spaces, yes, every bijective and continuous function is a homeomorphism. In more general topical spaces, not necessarily.

[@shahinjahanlu2199]

Thx dr.peyam

[@honghong324nt5]

At 7:00 12:00, how would you create a ball in the interval if it was an endpoint?

[@drpeyam]

I think in this case the ball would be a half-interval, basically a ball intersected with an interval

[@honghong324nt5]

@@drpeyam Interesting. Thank you!

[@madhavpr]

Awesome video as always. Just a quick technical question about the choice of epsilon. I understand that epsilon is "small", so we've assumed that it's less than 'r' so that our epsilon ball stays inside the interval I. However, the definition of continuity requires epsilon to be any arbitrary positive number. What if my choice of epsilon was much much much bigger than r ? Example, r = 0.001, but epsilon = 1000. In this case, the epsilon ball centered at g(x_0) would contain the corresponding r-ball. What should we do then? Also, how do we rigorously justify the claim that it's okay to assume epsilon < r ?

[@drpeyam]

min(r,whatever you found)

[@MrRyanroberson1]

i had a thought when i saw this video and it goes like this: consider the function f(v)->w; v and w in R^2 which is continuous everywhere. Let g(v) be defined as Bf(Av), for some 2x2 matrices A, B. If there exist matrices A, B such that g(x, y) -> (p(x), q(y)), what do we know about f?

[@JharkhandMathsZone]

Very interesting please tell me how to achieve America through maths

[@josebeleno1213]

I tried to do the proof. I had the same ideas!!!

[@NonTwinBrothers]

Lol the new thumbnail

[@the_nuwarrior]

very good content

[@dgrandlapinblanc]

Ok. Thank you very much. I figure out the demonstration but for me at 10 minutes there is an element who doesn't work because you said in definitive that g(x0)< or = to g(x) => x0< or = to x and you use the fact that f is increasing for demonstrate that x0 can't be an end point. Ok. But for me it's the fact that f is STRICTLY increasing otherwise it doesn't work. For beginning even using the increasing you must say g(x0) x0< or = to x by definition. Moreover when you suppose g(x0) at the end point of I for demonstrate by contradiction that x0 isn't an end point of J you must use the strictly increasing of f (i.e g(x0) x0

[@Smoothcurveup52]

Happy mathematics day sir

[@thedoublehelix5661]

Here's my attempt: we know f:I->R is 1 to 1 and continuous, so it's either increasing or decreasing (wlog assume increasing). An increasing and continuous function takes open intervals to open intervals. It maps intervals to intervals because it's continuous (intermediate value theorem). Now lets show it preserves openesses of intervals. Assume there was some open interval whose image is not open. Since we know the image must be an interval that is not open, the interval contains one of it's endpoints (wlog assume it's the rightmost endpoint and thus the maximum). Because f is increasing, the preimage of a maximum is another maximum. But the original set is an open interval, so it doesn't contain a maximum element. This is a contradiction which means f preserves openess. Now it's easy to see that the inverse of f is continuous by the topological characterization of continuity.

[@jirihavel2193]

thanks this was very helpful for me

[@MathAdam]

OK, did you change the thumbnail... or do I need to go back to bed?

[@drpeyam]

I did change it 😂

[@chessematics]

When you realize that the thumbnail also got it's inverse function (if only you've seen the very first thumbnail). Edit: and when you realize that Dr. Peyam is always active and gives ♥️ to your comment instantly.

[@drpeyam]

Haha, how meta!!!

[@chessematics]

@@drpeyam you took the ♥️ away instantly as well

[@drpeyam]

Oh no, I accidentally clicked ❤️ twice