How many continuous functions are there ?

[@blackpenredpen]

You forgot oreo(x)

[@drpeyam]

Discontinuous everywhere 😂

[@IoT_]

@@drpeyam like the Dirichlet function?)

[@VibingMath]

bprp(x) and drpim(x) are definitely two real, unique and unbounded functions 😎

[@blackpenredpen]

Awwww thank you!!!

[@VibingMath]

@@blackpenredpen Yay coz the thumbmail has bprp(x)😂

[@chirayu_jain]

BPRP(x) is really a continous function with a range and domain of (-infinite, infinite).

[@blackpenredpen]

nice!!!1!!

[@chirayu_jain]

@@blackpenredpen thanks 😃

[@seanfraser3125]

Isn’t it?

[@cadereinberger3315]

Just in case anyone is interested in just how surprising Cantor-Shroeder-Bernstein is, note that just because there exists an injective homomorphism from A to B and from B to A, with A and B groups, does not imply that A and B are isomorphic. If you'v never seen this, look for a counterexample, it's pretty great.

[@willnewman9783]

But how many Riemann Integrable functions are there?

[@drpeyam]

I feel there must be as many as continuous functions, because of the Darboux Integrability Criterion

[@ExplosiveBrohoof]

@@drpeyam Isn't the set of Riemann integrable functions a subset of the set of Lebesgue integrable functions, which is a subset of the set of measurable functions? The latter has cardinality 2^(aleph_0), so the cardinality is equal to the cardinality of continuous functions.

[@cosminaalex]

@@ExplosiveBrohoof well, N is a subset of R, sub N has a smaller cardinality of R rather, what's the density of the Riemann integrable functions in the set of measurable functions?

[@ExplosiveBrohoof]

@@cosminaalex I don't know what you mean by density. There's no canonical measure that I know of on the set of real valued functions.

[@cosminaalex]

@@ExplosiveBrohoof this en.m.wikipedia.org/wiki/Dense_set

[@maxpercer7119]

I would like to quibble a bit here. At 2:49 it seems misleading to say there are more continuous functions than reals, better to say there are at least as many continuous functions as reals (and, conversely, at most as many continuous functions as reals). Towards the end, you're rushing through very delicate points. But this video got me really interested in the result.

[@dougr.2398]

The first part of the proof is indeed easy & simple (done by/at 3:27). I’’ll have to view and re-view the second part a few times, carefully and perhaps take notes to follow it diligently and with full comprehension. A proof like this pushes the boundaries of believability, since one has already established the existence of more continuous functions than real numbers..... how to lump the extras in? Countable infinities of the extras? Did Cantor first study and elucidate this? Thanks for providing what is missing in my mathematical studies, real Real Analysis!!! (Of course I am reminded that perhaps if I had taken Real Analysis, I would never have needed so many years of Fake Analysis!! (Psychology department!) ). is that Oreo (bunny rabbit) in the lower right inset? Must be!

[@rb1471]

"since one has already established the existence of more continuous functions than real numbers." Another example could be *N* has the same cardinality as *N* . We know an easy mapping could be: 1 -> 1 2 -> 2 3 -> 3 N -> N We could also use: 1 -> 3 2 -> 6 3 -> 9 N -> 3N So we've shown a similar result that *N* has more numbers than *N* when in reality they are the same size. But this is just a property of infinite sets which is why it seems like there are "more continuous functions than real numbers" from the first part of the proof.

[@maxpercer7119]

The uploader did not show that there are *more* continuous functions than real numbers. If he did that, then clearly the cardinalities of the two sets would not be equal. What he did show is that the cardinality of the set of continuous functions from R to R is at least the cardinality of the set of real numbers, and vice a versa (the cardinality of real numbers is at least the cardinality of the set of continuous functions from Reals to Reals).

[@snipergranola6359]

BPRP(x) Wow what a function continues in you tube also

[@newtonnewtonnewton1587]

D peyam is a very wonderful continuos function

[@drpeyam]

❤️

[@Keithfert490]

This is really fascinating! Could you give a proof that if two continuous functions have equal values at the rational numbers, then they are equal? Does it have something to do with rational numbers being dense? What does that mean precisely?

[@drpeyam]

Yep, it has to do with density. And that’s because f(x) = lim f(xn) where xn is any sequence converging to x by def of continuity

[@Keithfert490]

@@drpeyam Oh ok, I had never seen this definition of continuity ( en.wikipedia.org/wiki/Continuous_function#Definition_in_terms_of_limits_of_sequences ). Now it makes sense! Thanks!

[@michaelz2270]

Alright! Thanks for posting!

[@vanessakitty8867]

Ah cardinalities are fun.

[@pierreabbat6157]

But do they have the same oriolity?

[@thedoublehelix5661]

Is this what analysis is like? Seems really interesting

[@drpeyam]

It’s more set theory, but yeah :)

[@duckymomo7935]

No, this isn’t real analysis Analysis is It’s more like let {x(n) | x(n) = sqrt(x(n-1)+2), x(1) = 1} be a sequence Prove that it is bounded, monotone and then use monotone convergence to prove that its limit, L, is 2

[@duckymomo7935]

What he’s doing is set theory or meta function theory really as this result while interesting isn’t of much help to solve problems

[@thinkingpi]

Awesome.

[@JBJHPL]

when you say that R is in bijection with the number of functions from N to {0,1}, it seems to me that you are claiming that the continuum hypothesis holds

[@MikeRosoftJH]

That proposition doesn't depend on the continuous hypothesis. That there are as many real numbers as functions from N to {0,1} can be seen from that every real number has its binary representation. (Yes, binary representations don't exactly correspond to real numbers - some numbers have two different representations, like [bin]0.1000...=[bin]0.0111... . But of these are countably many. Of course, an interval of reals, and reals themselves, can be mapped one-to-one.)

[@max.caimits]

My idea would be to use a Fourier series to show that |C(ℝ)| = |C(−π .. π)| = |ℕ → ℝ|.

[@duckymomo7935]

Most functions are very and highly discontinuous, it’s only that continuous functions are also uncountable too

[@GIFPES]

Bijective functions, so invertible also, are continuous ones.

[@drpeyam]

Not true, try f(x) = 1/x with f(0) = 0

[@duncanw9901]

What's your definition of continuous? Cauchy-sequences of polynomials converge to a continuous function in some norm--this can't be true for all, because some norms allow countable discontinuity.

[@drpeyam]

The usual epsilon delta definition of continuity from R to R

[@ruffifuffler8711]

How about the lack of alternatives going anywhere else other then while on the established path?

[@duncanw9901]

@@ruffifuffler8711 about as non-rigorous as it gets. Wierstrass function is continuous but fractally sinusoidal, so there is no 'direction' or 'path' that is well-defined

[@ruffifuffler8711]

@@duncanw9901 Hmmm, ...I could be in a trap thinking digitally, ...probably still strung out on one of those non-homotopic chunks left over from euler's formula before presumptions were notarized.

[@duncanw9901]

@@ruffifuffler8711 not meaningless technobabble bro.... look up the wierstrass function if you zoom in on a section of it it's wavy, and if you zoom in on a section of these waves they are also wavy, and so it continues. Your def doesn't work for this one.

[@kevinfung6697]

Wait for a moment.....bprp(x)????

[@ThreePointOneFou]

Pretty sure that's a joke referring to BlackPenRedPen.

[@dgrandlapinblanc]

Not evident but interesting.

[@ogorangeduck]

And people said Infinity War was the most ambitious crossover event...smh

[@ethancheung1676]

Wouldn’t the second part of the proof invalid because there cannot be enumeration of R. It is uncountable

[@drpeyam]

Not an enumeration, but there is still a bijection between R and sequences with 0 and 1, because of binary expansions

[@jekoddragon6227]

Dr Peyam! Can you cover the newly found fact in linear algebra - that you can derive eigenvectors directly from eigenvalues? Read the paper EIGENVECTORS FROM EIGENVALUES PETER B. DENTON, STEPHEN J. PARKE, TERENCE TAO, AND XINING ZHANG

[@peorakef]

What's the answer

[@drpeyam]

As many as real numbers

[@mike4ty4]

beth one

[@stanleychan3212]

When you consider how many continuous functions are differentiable, the result is even more suprising. Let X = C([0,1], R) be the space of real-valued continuous functions on [0,1], endowed with the supremum norm ||f|| = sup|f|. Then X is a Banach space and can be given a Borel probability measure on X, called the classical Wiener measure. A suprising result is the following: With respect to this probability measure on X, *the set of nowhere differentiable functions has probability measure 1*. This means that if you randomly pick a function f from X, then almost surely f is nowhere differentiable, and you have *zero chance to get a function that is differentiable at even 1 point*.

[@drpeyam]

Not that surprising :) Consider for instance f(x) = -x for x < 0. The only way to make f differentiable at 0 is to have f slope -1, so among all the possible directions f could have, only one direction would work, otherwise you have a kink

[@rb1471]

I don't quite understand why *R* ~ Functions from *N* to {0, 1}

[@drpeyam]

That’s because you can write any number as a binary expansion, like 100001.010100, so for each slot, you assign either 0 or 1, and that’s like defining a function from N to {0,1}

[@arthurreitz9540]

Plenty but not that much

[@yashovardhandubey5252]

BPRP is a function..... Lol never knew that.....

[@Austin101123]

What is incorrect with the following proof then, which shows there are more continuous functions than real numbers: Take each element of the reals and choose it or not. Create the polynomial with roots equal to the reals chosen. The number of ways to do this is the power set of the reals.

[@carl4578]

the power set contains infinite size sets of real numbers in it. there is no polynomial with infinite roots roots

[@Austin101123]

@@carl4578 Why can't it have infinite roots? Is it by definition or does it not work to have an infinite root polynomial?

[@MikeRosoftJH]

@@Austin101123 Every polynomial equation of the form a*x^n+b*x^(n-1)+...+u*x+w=0, where a, b, ..., w are n-many constants, and n is a positive integer, and a is not zero, has at most n real (or complex) solutions. You can surely make an infinite polynomial sum, but it's not guaranteed to be continuous, or even defined everywhere on reals. But that doesn't matter - there are only continuum many infinite sequences of reals. So there certainly are sets of reals which are not a set of roots of any infinite polynomial sum. For one, how would you get the Vitali set by this method? (Every function you get in this way is measurable.)

[@Austin101123]

@@MikeRosoftJH oh true true it probably wouldn"t be continuous, like p=0 odds its continuous. At all the roots it will be 0 and then I think p=1 to have p=1 almost everywhere else has infinite, or negative infinite values.

[@MikeRosoftJH]

@@Austin101123 There certainly exists a continuous function which is zero exactly on odd integers: a sinusoid with appropriate parameters. And that function can be expressed as an infinite polynomial sum (Taylor series). That's not the main problem. The real problem is, as I have said: there are continuum many infinite sequences of real numbers (the set of all sequences of reals can be mapped one-to-one with reals themselves). So there are sets of reals which can't be expressed as a set of zeroes of an infinite polynomial sum - there are more than continuum many sets of reals. (Likewise, not every set of reals is a set of zeroes of a continuous function; consider for example the set of all rational numbers. Again, it can be seen that this can't be the case, because there are continuum many continuous functions.)

[@AndDiracisHisProphet]

I think it is three

[@janivlevi2213]

He’s left-handed

[@tracyh5751]

Hmm... It seems like your argument holds for the set of continuous real valued functions on a separable metric space. Is this the case?

[@drpeyam]

I think so!

[@laugernberg4817]

Isnt the definition of being seperable exactly just that there exist a countable dense set? So yes

[@karinano1stan]

bprp(x) haha :)

[@CTJ2619]

There are an infinite number of continuous functions. Next question

[@zacharieetienne5784]

FHaE

[@jonathangrey6354]

Bprp(x)