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Assume $M$ to be a compact $n$-dimensional manifold, endowed with a complete metric.

Let us consider the space $C^\infty(M)$ endowed with the standard $C^\infty$ topology, i.e. generated by the seminorms $$\sup_K\left|\frac{\partial}{\partial_x^\alpha}f\right|,$$ over the compact subsets of $M$. Now consider the following set: let $1\in C^\infty(M)$ be the constant function and $0<\varepsilon<<1$. Define $$A\doteq\left\{a\in C^\infty(M)\;\colon\;\|a-1\|_{C^1(M)}<\varepsilon\right\}\subset C^\infty(M).$$

It seems to me that Ascoli Arzelà applies here, in particular $A$ is relatively compact in $C(M)$.

The question is as follows:

is it possible to choose an open covering of $A$ such that all the balls are centered at smooth functions? The consequence of Ascoli Arzelà is that there exists these balls but in principle they may be centered just on continuous functions, not even differentiable, I was thinking about this stronger result because of the compactness of $M$ and (maybe) something along the lines of Stone-Weierstrass which allows to slightly modify the center of the balls of the covering so that they are indeed smooth functions. Is this possible?

Thanks for the attention.

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    $\begingroup$ I think maybe I am misunderstanding you because I don't see where the trouble is--view $A\subset C(M)$ and put a $\delta$-ball (in the $C(M)$ norm) around every point of $A$. These cover the closure of $A$ in $C(M)$, which is compact, hence finitely many balls suffice. But these are centered at points of $A$ which are all smooth functions to begin with. $\endgroup$
    – Mike Jury
    Commented Aug 28, 2014 at 13:42
  • $\begingroup$ @MikeJury no I was thinking the same.. so probably I was confusing myself for nothing. Or maybe we are both wrong :) just joking.. thank you very much $\endgroup$
    – guido giuliani
    Commented Aug 28, 2014 at 13:50
  • $\begingroup$ You wanted to ask about small balls, surely: as stated, for any $f \in A$, the (sup norm) ball centered at $f$ of radius $2 \epsilon$ contains all of $A$. $\endgroup$
    – Nate Eldredge
    Commented Aug 28, 2014 at 14:48
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    $\begingroup$ Given a ball of radius $r$ centred at $a$, if you can find smooth $b$ at distance less than $\epsilon$ from $a$, the ball centred at $b$ with radius $r+\epsilon$ contains your original ball. $\endgroup$
    – Robert Israel
    Commented Aug 28, 2014 at 16:06

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