# Non-Hausdorff topology on the germs of holomorphic functions

Let $\mathcal{O}_{z}$ be the space of germs of holomorphic functions at $z$, defined as a direct limit of a system $$\mathcal{O}_{z} = \lim_{\rightarrow}{\mathcal{O}(U_{\frac{1}{n}}(z))}$$ where $U_{\frac{1}{n}}(z)$ stands for the unit disk centered at $z$.

Each $\mathcal{O}(U_{\frac{1}{n}}(z))$ is endowed with the topology of pointwise convergence and $\mathcal{O}_{z}$ inherits the corresponding inductive locally convex topology.

$\mathcal{O}_{z}$ can be decomposed as $$\mathcal{O}_{z} = \mathbb{C}1 \oplus \mathcal{O}_{z}^{0}$$ where the latter stands for $$\mathcal{O}_{z}^{0} = \{ f \in \mathcal{O}_{z} | f(z) = 0 \}$$ and the decomposition is as follows $$f(z) = a_{0} + (z-a)(a_{1}+a_{2}(z-a)+a_{3}(z-a)^{2} + \ldots)$$ where $a_{k}$ stands for the $k$-th coefficient in the Taylor series expansion at $z = a$.

How to prove that the topology on $\mathcal{O}_{z}^{0}$ induced from the $\mathcal{O}(z)$ is trivial? In fact, this result would imply that the chosen inductive topology is non-Hausdorff.

For each $n$, let $\rho_n \colon \mathcal{O}(U_{\frac{1}{n}}(z)) \to \mathcal{O}_z$ the natural map. A neighbourhood basis of $0$ in $\mathcal{O}_z$ is given by the family of absolutely convex (i.e. convex and balanced) $V \subset \mathcal{O}_z$ such that $\rho_n^{-1}(V)$ is, for every $n$, a neighbourhood of $0$ in $\mathcal{O}(U_{\frac{1}{n}}(z))$. Let $V$ be any such set. By definition of the topology of pointwise convergence, for every $n$ there is a finite $F_n\subset U_{\frac{1}{n}}(z)$ and a $\varepsilon_n > 0$ such that $\lvert f(w)\rvert < \varepsilon_n$ for all $w\in F_n$ implies $\rho_n(f) \in V$.

Let $\delta = \min\:\bigl\{ \lvert w-z\rvert : w \in F_1\setminus \{z\}\bigr\}$ and let $n > 1/\delta$. Consider an arbitrary $f \in \mathcal{O}(U_{\frac{1}{n}}(z))$ with $\lvert f(z)\rvert < \varepsilon_1/2$. There is an entire $g$ with $g(w) = 2f(w)$ for $w \in F_n$ and $g(w) = 0$ for $w \in F_1\setminus \{z\}$. Then $g \in \rho_1^{-1}(V)$, and therefore $\tilde{g} = g\rvert_{U_{1/n}(z)}\in \rho_n^{-1}(V)$. Also, $2f - \tilde{g}$ vanishes on $F_n$, so $2f-\tilde{g} \in \rho_n^{-1}(V)$. By convexity, it follows that

$$f = \frac{1}{2}\bigl(2 f - \tilde{g}\bigr) + \frac{1}{2}\tilde{g} \in \rho_n^{-1}(V)\,.$$

Thus

$$V \supset \bigl\{ f \in \mathcal{O}_z : \lvert f(z)\rvert < \varepsilon_1/2\bigr\}\,.$$

Conversely, for every $r > 0$ the set

$$V_r = \bigl\{ f \in \mathcal{O}_z : \lvert f(z)\rvert < r\bigr\}$$

is an absolutely convex set such that $\rho_n^{-1}(V_r)$ is a neighbourhood of $0$ in $\mathcal{O}(U_{\frac{1}{n}}(z))$ for every $n$, hence this family is a neighbourhood basis of $0$ in $\mathcal{O}_z$.

In other words, the inductive limit topology on $\mathcal{O}(z)$ is induced by the seminorm $p \colon f \mapsto \lvert f(z)\rvert$ whose kernel is $\mathcal{O}_z^0$. Hence the subspace topology on $\mathcal{O}_z^0$ is the trivial topology.

• @DanielFisher Thanks! I do admit that i was really far from the final solution. – hyperkahler Jan 28 '18 at 17:06