Why Zeros not isolated in $ \mathbb{R}$

Question

My question is about two theorems
1)isolated zeros theorem
Counter example in $\mathbb{R}$
Let $$f(x) = \left\{ \begin{gathered} {e^{ - \frac{1}{{{x^2}}}}}.\sin \left( {\frac{1}{x}} \right)\hspace{2cm} x \in \mathbb{R}\backslash \{ 0\} \hfill \\ f(0) = 0 \hfill \\ \end{gathered} \right.$$

the function satisfied that $C^\infty$ and $\mathbb{R}$ is Connected space but $0$ not isolated because there is a Sequence of zeros $U_n=\frac{1}{n\pi }$ , $n \in \mathbb{N^*}$
2)Liouville's theorem in complex analysis $$ g(x) = \sin (x)$$ $g$ is $C^\infty$ and Bounded but $g$ not constant
My question is What the secret that make theorems true for the holomorphic function in $\mathbb{C}$ and not true in differentiable function in $\mathbb{R}$? is the secret Topology and Neighbourhood ?

Answer 1

These functions are $C^\infty$ and bounded on $\mathbb{R}$, not on $\mathbb{C}$. The key difference is that differentiability in $\mathbb{R}$ is a much weaker condition that in $\mathbb{C}$. A geometric way of seeing how much weaker the differentiability on $\mathbb{R}$ is than on $\mathbb{C}$ is to consider that on $\mathbb{R}$ you have a two limiting processes to consider, $\lim_{h\to0^+} (f(x+h)-f(x))/h$ and $\lim_{h\to0^-} (f(x+h)-f(x))/h$ while on $\mathbb{C}$ you have limits coming from uncountably many directions. My professors always highlighted this fact to show why $C^\infty(\mathbb{C})=C^1(\mathbb{C})\subset C^1(\mathbb{R})$.

Answer 2

We could say that one of the main reasons holomorphic functions have more rigidity than $C^\infty$ functions is that holomorphic functions, if seen as functions from $U\subseteq \Bbb R^2$ to $\Bbb C=\Bbb R^2$, are solutions of a PDE.