Hey Epic. Your video on the group of matrices with all entries equal was really nice. It's a great example to really focus on understand the definition of a group. Good stuff.
As for the combination of both spaces having the discrete metric, the thing is that all functions whose domain uses that metric are uninterestingly locally constant, since they only locally have one input.
let f map to a discrete space and assume f is continuous and not locally constant. Then there exists a point s in S such that f is not constant on any U-open containing s. Now take the set U=f^-(f(s)) this must be open by continuity of f and contains s. as f is also constant on this set we have a contradiction
The definition of a function requires it to be defined for all elements in the domain so for 1/x to be a function we would have to exclude 0 from its domain.
Every subset is open, hence every function is continuous. meh, its a stupid example. i guess for n point discrete metric spwces you can isometrically embed them to space to R^n, so there is that. It would be interesting if you can get infinite dimensional "simplices" in some Hilbert space somehow and embed R or Z with the discrete topology in them.
As soon as I saw the thumbnail I thought "discrete metric"
The power of metrics... Also i love your channel, Professor Omar!!
Nice!!!
This is what a mathematician thinks clickbait is
I mean, it worked on me 😅
The flip side of this coin: every function is also continuous when we put the _indiscrete_ topology on the target space.
Hey Epic. Your video on the group of matrices with all entries equal was really nice. It's a great example to really focus on understand the definition of a group. Good stuff.
@@ProfOmarMath Thanks, glad you liked it!
Love ur videos man
Love the beginning :D
"Every function is continuous"
*stares*
"ᵖʳᵒᵛᶦᵈᵉᵈ ᵗʰᵃᵗ ʸᵒᵘ ᵘˢᵉ ʷʰᵃᵗˢ ᶜᵃˡˡᵉᵈ ᵃ ᵈᶦˢᶜʳᵉᵗᵉ ᵐᵉᵗʳᶦᶜ"
Yes thats what I have been always advocating for
As for the combination of both spaces having the discrete metric, the thing is that all functions whose domain uses that metric are uninterestingly locally constant, since they only locally have one input.
let f map to a discrete space and assume f is continuous and not locally constant.
Then there exists a point s in S such that f is not constant on any U-open containing s.
Now take the set U=f^-(f(s)) this must be open by continuity of f and contains s. as f is also constant on this set we have a contradiction
Thank you Dr. Peyam
The real numbers are a countable union of countable sets...
Provided you forget about the axiom of choice 👍🏻
f^-1 (U) -> U reminds of covering spaces
Every function is continuous!!!... provided they are to an INdiscrete space
This proof is like the cracks in the edifice of real numbers
Even 1/x at 0?
You can’t strictly speaking talk about continuity at points where it’s undefined
The definition of a function requires it to be defined for all elements in the domain so for 1/x to be a function we would have to exclude 0 from its domain.
i remember we had an exercise to prove that in Analysis I
YAY, I never have to prove continuity again! Every function is continuous. lol. Way over my head! :)
Weird. Thank you very much.
Every subset is open, hence every function is continuous.
meh, its a stupid example. i guess for n point discrete metric spwces you can isometrically embed them to space to R^n, so there is that. It would be interesting if you can get infinite dimensional "simplices" in some Hilbert space somehow and embed R or Z with the discrete topology in them.
You're number one you're number one everyone's number one
Me failing my calc 1 exam