f implies continuous

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Every function is continuous! Don't believe it? Then watch this video!

Dr Peyam

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This is what a mathematician thinks clickbait is

seanfraser

As soon as I saw the thumbnail I thought "discrete metric"

ProfOmarMath

The flip side of this coin: every function is also continuous when we put the _indiscrete_ topology on the target space.

EpicMathTime

Love the beginning :D

"Every function is continuous"
*stares*
"ᵖʳᵒᵛᶦᵈᵉᵈ ᵗʰᵃᵗ ʸᵒᵘ ᵘˢᵉ ʷʰᵃᵗˢ ᶜᵃˡˡᵉᵈ ᵃ ᵈᶦˢᶜʳᵉᵗᵉ ᵐᵉᵗʳᶦᶜ"

imnotarobot

Yes thats what I have been always advocating for

curiosityzero

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.

iabervon

This proof is like the cracks in the edifice of real numbers

txikitofandango

f^-1 (U) -> U reminds of covering spaces

Jaylooker

YAY, I never have to prove continuity again! Every function is continuous. lol. Way over my head! :)

albertaraujo

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

MrNygiz

Every function is continuous!!!... provided they are to an INdiscrete space

f-th

The real numbers are a countable union of countable sets...

Provided you forget about the axiom of choice 👍🏻

factsheet

i remember we had an exercise to prove that in Analysis I

pythoncake

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.

mikhailmikhailov

You're number one you're number one everyone's number one

mrmathman

f implies continuous

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