Theorem 2.14 (One-Parameter Subgroups) If $A(\cdot)$ is a one-parameter subgroup of $\text{GL}(n;\mathbb{C})$, there exists a unique $n\times n$ complex matrix $X$ such that $A(t)=\mathrm{e}^{tX}$
In the proof concerning existence, it is mentioned that $U$ is a open set in $\text{GL}(n,\mathbb{C})$ ($U:=\exp(B_{\epsilon/2}$) with $B_{\epsilon/2}$ the ball of radius $\epsilon/2$ around the origin in $M_{n}(\mathbb{C}),\, \epsilon/2<\log(2))$. The continuity of $A$ guarantees that there exists $t_0>0$ such that $A(t)\in U$ for all $t$ with $|t|\leq t_0$.
Question 1. Why $U$ is open?
My attempt. It is necessary to prove that for all $\mathrm{e}^{B}\in U$, exists $r>0$ such that $B(\mathrm{e}^{B};r)\subset U$. Let $Y\in B(\mathrm{e}^{B};r)$ then $\left\|Y-\mathrm{e}^B\right\|<r$, so $\left\|Y\right\|\leq \left\|\mathrm{e}^{B}\right\|+r\leq \mathrm{e}^{\left\|B\right\|}+r\leq \mathrm{e}^{\epsilon/2}+r\leq \mathrm{e}^{\log(2)}+r=2+r$
Question 2. From what I understand, if there exists $t_0>0$ then by continuity, (being close to $t_0$) we have that $A(t)\in U$ for all $|t|\leq t_0$. But, Why there exists $t_0>0$ such that $A(t_0)\in U$?