denseness of polynomials in bounded borel measurable functions Let $K\subseteq \mathbb{R}$ be compact, consider $B(K)$ the set of all bounded borel measurable functions $f:K\to \mathbb{C}$ and endow $B(K)$ with the uniform norm, so you obtain a Banach space. My question is, is the set of all polynomials $p\in B(K)$ with $p(0)=0$ dense in $B(K)$ if you take a suitable closure? Which closure do we need here?
 A: Uniform closure, of course not. Closure in the topology of pointwise convergence yes, as has been pointed out, but this is trivial (immediate from the fact that a polynomial can take any values you want on any finite set) and I don't see how it's good for anything.
There's a closure for which the answer is yes, and which also actually allows one to prove things about Borel functions: what I call "bounded pointwise sequential" or bps closure: Say $X$ is bps closed if $f_1,f_2,\dots\in X$, $(f_n)$ uniformly bounded, $f_n(x)\to f(x)$ for all $x$ implies $f\in X$. Say the bps closure of a space of bounded functions is the smallest bps closed superset.
Then the bps closure of the polynomials restricted to $K$ is exactly $B(K)$. This is not quite trivial perhaps, but not hard to prove.
(If $K$ is a compact Hausdorff space then the bps closure of $C(K)$ is the set of bounded functions which are measurable with respect to the sigma-algebra generated by the compact $G_\delta$ sets.)
NOTE It's not true that $f\in B(K)$ implies there exists a sequence of polynomials $p_n$ that have bps limit $f$. To get the bps closure you need to throw in all the bps limits of sequences, then do that again, and again, a total of $\omega_1$ times.
A: All polynomials are dense in borel functions in pointwise convergence topology because polynomials are dense in continous functions in uniform convergence topology, and those are dense in borel in pointwise convergence topology. However if you restrict to polynomials vanishing at zero you will only get borel functions vanishing at zero.
