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My question is really a basic (and, I dare say, naive) question about the theory of ordinary differential equations and is likely answered in a book on functional analysis or the like, so any pointers to such a book with the answer would be appreciated.

It is well-known that $\mathcal{C}^n(I)$ with the compact-open topology for any open interval $I$ of positive width is homeomorphic to $\mathcal{C}^n((0,1))$ with the compact-open topology, essentially via the unique orientation-preserving affine diffeomorphism of $I$ with $(0,1)$. In this post and this post, it is outlined that $\mathcal{C}^n((0,1))$ with the compact-open topology is metrizable by a metric that is complete, that is to say, that $\mathcal{C}^n((0,1))$ with the compact-open topology is topologically complete.

My questions is, are the elementary functions dense in $\mathcal{C}^n((0,1))$ with the compact-open topology?

If this is true, then given any $n^\text{th}$-order linear IVP with all coefficient functions continuous on $I$ and leading coefficient function $a_n(t)$ having no roots in $I$ has a Cauchy sequence (in any of the metrics outlined above) of elementary functions which converges to the IVP's unique solution $x(t)$ (which may be non-elementary but which will be $n$ times continuously differentiable on $I$ as $\mathcal{C}^n(I)$ with the compact-open topology is topologically complete). This fact would be useful in motivating differential equations students about some of the strategies for "solving", or approximating solutions to, IVPs.

Rob Arthan
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Jeffrey Rolland
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The polynomials are dense by the Stone-Weierstrass theorem.

Robert Israel
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    Isn't that only true for closed, bounded intervals $I$? https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem (I had thought of that but seemed to recall the compactness of the domain was an important hypothesis.) – Jeffrey Rolland Apr 13 '17 at 21:19
  • It is true for open intervals in the compact-open topology; the proof is pretty much follow-your-nose. (https://www.researchgate.net/post/Can_be_replaced_closed_interval_a_b_by_open_interval_a_b_in_Weierstrass_Approximation_Theorem2 See Rogier Brussee's comment) – Jeffrey Rolland Apr 13 '17 at 22:01
  • Compact-open topology: so a basic neighbourhood of $f$ is defined in terms of the values on a compact subset $K \subset (0,1)$. And Stone-Weierstrass says the polynomials are dense in the continuous functions on that compact set. – Robert Israel Apr 13 '17 at 23:11