Finite intersection of nonempty sets is not empty (Poincaré-Miranda theorem) I am trying to understand a proof of the Poincaré-Miranda theorem given in this article.
So, the theorem states that if we have a continuous map $f = (f_1, \ldots, f_n) : I^n \rightarrow {\rm I\!R}^n$ where $I^n$ is a $n$-dimensional cube $[-1, 1]^n$ that satisfies the following condition:
$$
f_i(x_1, \ldots, x_{i-1}, -1, x_{i+1}, \ldots, n) \leq 0 \\
f_i(x_1, \ldots, x_{i-1}, 1, x_{i+1}, \ldots, n) \geq 0
$$
for each $i = 1, \ldots, n$, there exists a point $c \in I^n$ such that $f(c) = 0$.
To prove the theorem authors defined sets
$$
H_i^- := f_i^{-1}(- \infty , 0] \\
H_i^+ := f_i^{-1}[0, \infty)
$$
Then they say that "using a compactness argument we infer that the intersection
$$
H := \bigcap \{ H_i^- \cap H_i^+ : i=1, \ldots, n\}
$$
is not empty."
I see that each $H_i^-$ and $H_i^+$ is not empty as it follows from the propositions, but why are $H_i^- \cap H_i^+$ nonempty and why is $H$ not empty?
Additionally, if $H$ is indeed not empty, it is obvious that for every $x \in H$ it follows $f(x) = 0$, so we found the zero. Why does not the proof just end here?
Thanks to anyone who could help me understand.
 A: Their proof is laid out backwards and is indeed confusing. Their general proof is that they can show their are an odd (and so at least 1) simplices of each given diameter that intersect both $H_i^{-}$ and $H_i^{+}$ for all $i$. They then effectively wave their hands and say that's sufficient to prove that $H$ is nonempty by an unstated compactness argument before going back and proving the existence of such simplices.
Here's one way of showing the compactness part. Let $S_k$ be a simplex with diameter $1/k$ which intersect $H_i^-$ and $H_i^+$ for all $i$. Then for all $k$ there are exist $a_k$  in $S_k \cap H_i^-$ and $b_k$ in $S_k \cap H_i^+$ for all i. Then, by compactness, there is some convergent subsequence of the $a_k$. Let that subsequence be $a_{k_i}$ and let its limit be $a$. Note that $a\in H_i^-$ for all $i$ since they are closed. Since the diameter of $S_{k_i}$ is $1/k_i$, the distance $d(a_{k_i},b_{k_i})<1/k_i$ goes to 0, so $a=\lim a_{k_i}=\lim b_{k_i}$. Since $b_{k_i}$ are in the closed set $H_i^+$ for all $i$, the limit $a$ is in $H_i^+$ for all $i$, so $a\in H$.
