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There is a version of the theorem for continuous mappings $f$ from a compact metric space $X$ to real numbers $R$. Namely, if an algebra $A$ of continuous functions separates points and is non-vanishing, then any $f$ can be approximated arbitrarily well by an element $a$ in $A$.

I wonder whether the same exists for functions that map from a compact metric space $X$ to real numbers but are only piecewise continuous.

An example of $f$: Take a series of 100 numbers $\vec x=(x_1,x_2,\cdots,x_{100})$, each number in the interval $[0,1]$, and return $+1$ if they are sorted and $-1$ if they are not.

An example of $A$: an element $a\in A$ is defined by a tuple $a\equiv (i,j,v)$ where $i$ and $j$ are indexing elements of the sequence of 100, $v$ is any two-argument piecewise continuous function that takes $x_i$ and $x_j$, and returns a number. Thus the element $a$ would operate on $\vec x$ as follows: $a\circ\vec x \equiv (i,j,v)\circ\vec x=v(x_i,x_j)$.

Can I use $A$ to approximate $f$? I would like to use the Stone-W theorem to figure it out. However, the theorem only works for continuous functions. So unless there is a version for piecewise continuous functions such a problem cannot be solved.

Best

Zoran

p.s. This is an improved version of a similar question posted elsewhere. Based on the comments I’ve got there I realized that the present post is really the question I am interested in.

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What you seek is false, as I discovered thinking about my comment on Carl Schildkraut's answer.

Continuous functions are closed under uniform limits, so anything well-approximated by continuous functions is continuous. Since polynomials are continuous, Stone-Weierstrass can only apply to continuous functions.

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I believe that any piecewise continuous function (assuming the pieces are all of size $>\epsilon$) can be arbitrarily well-approximated by a continuous function. So, it can be approximated arbitrarily well by a polynomial. You can probably find some sort of reference for this.

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    $\begingroup$ Your claim ("piecewise continuous can be arbitrarily well-approximated by continuous") is false. Continuous functions are closed under uniform limits, so anything arbitrarily well-approximated by continuous functions must be continuous. $\endgroup$ – Jacob Manaker Feb 27 '17 at 18:35
  • $\begingroup$ @JacobManaker Can't the sign function ($-1$ for negative numbers, $0$ for $0$, and $1$ for positive numbers) be arbitrarily well-approximated by $\tanh(nx)$ as $n$ grows large? $\endgroup$ – Carl Schildkraut Feb 28 '17 at 4:45
  • $\begingroup$ Not uniformly. Pointwise, yes, but the point of Stone-Weierstrass is that it provides good approximations in the uniform norm. $\endgroup$ – Jacob Manaker Feb 28 '17 at 18:36

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