What is mean of "open condition"?

Question

We want to deform $I$ such that it is a almost complex structure on smooth manifold,for small $t$ . let $A(t):T^{0,1}M\longrightarrow T^{1,0}M $ is given, and define for small $t$ $$ T_{t}^{0,1}M:= (id + A(t))(T^{0,1}M). $$ The condition $t$ be a 'small' has to be imposed in order to ensure that with this definition $T_{t}^{0,1}M$ has right dimension. More precisely, $t$ small means that $T_{t}^{0,1}M\subset T_{\mathbb{C}} \longrightarrow T^{0,1}M$ is an isomorphism, which is certainly an open condition, at least on a compact manifold.

I don't understand "open condition" and it related to compact manifold. Can you explain me this condition?

Answer 1

It means that it defines an open set, like an inequation or an inequality for a continuous function. $\{x:f(x)>0\}$ and $\{x:f(x)\ne 0\}$ are open sets, thus one can call $f(x)>0$, $f(x)\ne 0$ "open conditions".

The restriction to compact manifolds is to ensure that the underlying function is continuous. Here one could define the condition as $$ f(t)=\inf_{x\in M}|\det(I+A(t,x))|>0. $$

---

An easy counter-example for the non-compact situation is $$ f(t)=\inf_{x\in\Bbb R}|1-tx| $$ where $f(0)=1$ while $f(t)=0$ for $t>0$, a jump discontinuity.

Answer 2

A condition is open if whenever it holds at a point $p$ then it also holds on a small neighbourhood of $p$ (in this case I think it means open with respect to the parameter $t$). Afraid I don't know enough about manifolds to answer why compactness makes this an open condition in this case though.