# Why is compactness needed for proving that Kahler forms are open

I am studying complex geometry from Huybrechts' Complex Geometry - An Introduction. In Corollary 3.1.8 he proves that:

The set of all Kahler forms on a compact complex manifold $X$ is an open convex cone in the linear space $\{\omega\in \pmb{A}^{1,1}(X)\cap \pmb{A}^2(X)|d\omega = 0\}$.

In his proof:

(Writing $(h_{ij})>0$ as a short hand for $(h_{ij})$ is positive definite.)

The positivity of an hermitian matrix $(h_{ij}(x))$ is an open property and, since $X$ is compact, the set of forms $\omega\in \pmb{A}^{1,1}(X)\cap \pmb{A}^2(X)$ that are locally of the form $\omega = \frac{i}{2} \sum h_{ij} dz_i \wedge d\bar{z}_j$ with $(h_{ij})$ positive definite at every point is open.

But I am confused in where he used the property of compactness in the proof. I thought that, if I could find a point $x\in X$ such that $(h_{ij}(x))>0$, then I could find a small open neighbourhood around this point, in which $(h_{ij})>0$ for every point in the neighbourhood. Then all the Kahler forms is a union of such neighbourhoods, and is thus open. So no compactness is required.

I realize that I had been sloppy in my thought. So a counter example is also welcome!

• @taper No, it's much simpler than that... for every chart $U$, there's an open set of forms $V_U$ that are positive on $U$. If you have a collection of charts $U_i$, then the set of forms positive on all of the $U_i$ is the set $\bigcap_i V_{U_i}$... – user98602 Apr 20 '16 at 2:18