Set of analyticity of a function is an open set Let be $f$ a real function of real variable, prove that the set in which $f$ is analytic is an open set.
Any help please?
 A: I take "set of analyticity" to mean this: $f$ is analytic at $x_0$ if there is a power series
$$
f(x) = \sum_{n=0}^\infty \frac{a_n}{n!} (x-x_0)^n
$$
for which the radius of convergence is not $0$ and for which the "equals" sign means the series actually converges to $f(x)$ for each value of $x$ whose distance from $x_0$ is less than the radius of convergence.
Now I am tempted to say that the fact that the radius of convergence is positive finishes it off: there is an open interval about $x_0$ within which the series converges to $f$; that interval is a subset of the set of analyticity.  Then since every point in the set is surrounded by an open interval included in the set, it follows that the set is open.
However: What we need to prove is that if $x_1$ is some other point in that open interval then then there is a power series
$$
f(x)=\sum_{n=0}^\infty \frac{b_n}{n!} (x-x_1)^n
$$
with non-zero radius of convergence, for which again "equals" means the series actually converges to the right thing as long as $x$ is close enough to $x_1$.
At this point I realize the only way I know of doing this that comes to mind right away involves complex numbers.  Radii of convergence are one example of something whose theory is simpler with complex numbers than with real numbers only.
So before finishing the answer I will ask: Must this for some reason be done without complex numbers?
