Understanding Real analyticity I'm going to state my assumption of the definition of Real analyticity, and how I understand it based on my current understanding. Please tell me if they are correct or not and please help me perfecting them:
Definition:
$f$ is Real analytic at $x = c$, if f is equal to its Taylor Series not only at $x = c$ but also for all $x$ within the neighborhood of $x = c$ (neighborhood means the open interval with $x = c$ as center). In other words, $f$ is Real analytic at $x = c$ if $f$ is equal to its Taylor Series $\forall x \in (c - \epsilon, c + \epsilon): \epsilon > 0$ , therefore Taylor Series is:
$$ \sum_{n = 0}^\infty \frac{f^{(n)}(c)}{n!} (x-c)^n$$
My understanding of analyticity:
1/ It’s possible that Taylor Series doesn’t converge to $f$ for all $x$ within Interval of Convergence. Therefore, assume $x = c$ is being checked for analyticity, and $(c−R,c+R)$ is Interval of Convergence (R is Radius of Convergence). Then $(c - \epsilon, c + \epsilon) \subseteq (c - R, c + R)$
2/ $f$ only needs to be infinitely differentiable within $(c−R,c+R)$
3/ For any $x$ being checked of analyticity, Taylor Series has to be around that $x$, it means that the $x$ becomes a center of both the Taylor Series and the Interval of Convergence. For example, if $x = j$ is being checked of analyticity then Taylor Series would be ...
$$ \sum_{n = 0}^\infty \frac{f^{(n)}(j)}{n!} (x-j)^n$$
... , Interval of Convergence would then be $(j−R,j+R)$ and the interval where Taylor Series converges to $f$ is $(j - \epsilon, j + \epsilon)$, consequently $(j - \epsilon, j + \epsilon) \subseteq (j−R,j+R)$.
4/ There are 3 intervals, the largest interval being $(c−R,c+R)$ which is Interval of Convergence. The interval where Taylor Series converges to f $(c−M,c+M):M > 0$ . And the smallest one being $(c - \epsilon, c + \epsilon)$ . So, $(c - \epsilon, c + \epsilon) \subseteq (c−M,c+M) \subseteq (c−R,c+R)$
All what I wrote above are assumptions, nothing is confirmed, please help me fix them if they are false. Any help is appreciated.
Edit: Added the 4th point.
 A: Let me expand the comments above as real analyticity is a tricky concept; so consider $f$ an infinitely differentiable real function on some real interval $I$; for each $c \in I$ we can associate to $f$ its Taylor series centered there (here we consider only the case $c$ interior to $I$ as the case where $c$ is an end point is similar with the necessary changes to one-sideness):
$$T_c(x)=\sum_{n = 0}^\infty \frac{f^{(n)}(c)}{n!} (x-c)^n$$
Let $r_c$ the radius of convergence of $T_c$; there are three possibilities:
1: $r_c=0$ so the Taylor series is called divergent (it converges only at $c$ but not on any open neighborhood)
2: $r_c >0$ (including $r_c=\infty$) but there is no small open neighborhood of $c$ for which $T_c(x)=f(x)$ (here the usual bump function $f(x)=e^{-1/x^2}, x \ne 0, f(0)=0$ comes to mind as its Taylor series $T_0$ at zero is zero so has infinite radius of convergence but  $T_c(x) \ne f(x)$ for $x \ne 0$
3: $r_c >0$ and there is an $0<\epsilon_c \le r_c$ st $T_c(x)=f(x), |x-c|< \epsilon_c$ (while also we assume $|x-c| < \epsilon_c \subset I$); then $f$ is called real analytic at $c$ and by the usual theory of power series it follows that $f$ is real analytic on $(c-\epsilon_c, c+\epsilon_c)$ so $T_y(x)$ converges to $f(x)$ on a small interval around $y$ for any $y \in (c-\epsilon_c, c+\epsilon_c)$ and moreover we are guaranteed such convergence on any interval centered at $y$ and of radius at most $\epsilon_c-|c-y|$
Note that in case 3, it can happen that $\epsilon_c<r_c$ (take now the one sided bump $f(x)=e^{-1/x^2}, x >0, f(x)=0, x \le 0$ and note that for every $c<0$ the radius of $T_c$ is infinite but $T_c$ converges to $f$ only on $(-\infty,0]$ so $\epsilon_c=|c|$)
By the above real analyticity is an open and local property (so the points $c$ where $f$ is analytic form a (relatively) open subset of $I$ and analyticity depends only on the values of $f$ in a small neighborhood of $c$)
There are at least two important results due to Boas that should be known (and are not that hard to prove, see linked reference).
Assume $f$ infinitely differentiable on $I$ st $r_c>0$ for all $c \in I$ (so we are only in cases 2,3); then the set of points $E$ where $f$ is not real analytic is closed and nowhere dense.
If in addition there is $\delta>0$ st $r_c \ge \delta>0$ for all $c \in I$ then $E$ is empty and $f$ is real analytic on $I$
