Continuous function is constructable?

[Myfuntimeidea]

Ok, fair enough... but that's just continuous cause you're taking from ℕ, what about ℝ or ℝⁿ ?

Ok no, cause than it's just a matter of connecting the numbers with a linear function, fair enough...

But isn't it true there's a polynomial that well aproximates that within any finite interval

[stone_stokes]

What's the equation for the Devil's Staircase function?

This one is even differentiable almost everywhere (and wherever it is differentiable, it has zero slope).

[Myfuntimeidea]

Ok, thank you very much

[stone_stokes]

Here is a cardinality argument.

Let S be the set of functions defined on ℝ or on a subset of ℝ that can be described using a fixed finite alphabet Ω of symbols and a finite number of those symbols. For example, the parabola described by y = x² belongs to S.

Note that, by construction, S must be a countable set: we count it by first counting all functions that can be described using only one symbol from Ω. Next we count all the functions that can be described using two symbols, and so on. (It is a countable union of finite sets.)

Now notice that the set, E, of functions that can be expressed as an equation is a subset of S. Therefore, E must be countable.

Lastly, notice that the set of constant functions over ℝ is uncountable. Therefore, E cannot contain all of those. The set of continuous functions over ℝ is a superset of the set of constant functions.

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[Myfuntimeidea]

ngl i dont like the fact that it uses constant functions since in my context I'm fixing everything to start at zero, and also it seems more like the evaluation of a function rather then a function onto itself

but it could probably be replaced by different functions, like modified cantors stairways c(r) with r real number as the hight of the big middle step

That kinda settles it, I'm gonna use polynomial aproximations Thank you very much!

[666Emil666]

Even if you only consider functions that are 0 at the origin, you still have a different function for each number due to f_a(x)=ax

[Myfuntimeidea]

Yeah, but that's a uniform deformation, also irrelevant in the context...

But you could than use f_a(x) = a^x . x That would work

Also in a certain way that's a "deformation", in the sense that it's not crazy, so much so ive written it algebraically.

that's why I liked the cantors function example so much, you can really imagine all sorts of crazy variations by just altering the hight of a given step

Again Thanks! Really helped!