I am currently reading through Wade's introduction to analysis and am having some confusion with a lemma before the inverse function theorem. We have by hypothesis that f is $C^1$ on an open set $V$ in $R^n$ and that $\det(J_f(a)) \neq 0$ for some $a \in V$.
Then Wade defines $h:V^{(n)} \to R$ by
$$h(x_1, x_2, ..., x_n) = \det[\frac{\partial f_i}{\partial x_j}(x_i)]$$
I agree that h is continuous on $V^{(n)}$. But then it is claimed that since $h(a, a, ..., a) = \det J_{f}(a) \neq 0$ it follows that there is an $r > 0$ such that $B_r(a) \subset V$ and $h(c_1, ..., c_n) \neq 0$ for $c_i \in B_r(a).$
This is where I am confused, I know that by continuity there exists some $B \subset V^{(n)}$ such that $h(b) \neq 0 $ for all $b \in B$ but how then can we break this up and speak about each component of this b and say they must be in a ball close to a?
By $h(a,...,a)=J_f(a)\neq 0$ I will assume you mean $h(a,...,a)=det[J_f(a)]\neq 0$. As Paul Frost commented, $J_f(a)\neq0$ does not imply $det[J_f(a)]\neq 0$.
Continuity on $V^n$ means that we can find a ball in which $h(b)\neq 0$ for all $b\in B$. This is because we can make $||h(b)-h(a,...,a)||<\frac{||h(a,...,a)||}{2}$ so then by triangle inequality $||h(b)||\geq\frac{||h(a,...,a)||}{2}>0$ in $B$.
But we are using euclidean norm, so if $b=(b_1,...,b_n)\in B_r(a,...,a)$, that means we have $\left(\sum_{i=1}^n|b_i-a|^2\right)^\frac{1}{2}\leq r$ so in particular, each part of that sum has to be less than $r$ since we are summing positive terms. So $|b_i-a|\leq r$ for each $i$, but this is exactly saying $b_i\in B_r(a)$ for each $i$.