Continuous function injective over a compact set, and locally injective on each point of the set Suppose we have a function $F: \mathbb R^n \rightarrow \mathbb R^k$ continuous over some open set $U \in \mathbb R^n$, and let compact set $K \subset U$. $F$ satisfies the following properties:
1) F is injective over K
2) For every $x \in K$, there is some open set $U_x$ such that F is injective on $U_x$. 
Show that there exists an $\epsilon > 0$ such that F is injective over the set $\{x\in \mathbb R^n | dist(x,K) < \epsilon \}$.
My attempt:
I can form an open cover of $K$ by forming  $W = \cup(U_x \cap U)$, then consider the boundary of this open cover $W$. The boundary is certainly compact, so $dist(x,K)$ over it takes on a minimum (it's easy to show this value is greater than 0), we may call it $\epsilon$.
But now I am stuck. I am not sure how to show that F is injective on all of $W$...
 A: This is a very nice problem indeed. Thank you.
Let's call $B(K,\epsilon)$ the set $\{x\in \mathbb{R} : \mathrm{dist}(x,K)
<\epsilon\}$. It is easy to see that for any $\epsilon$,
$\overline{B(K,\epsilon)}$ is compact, and that there exists an
$n\in\mathbb{N}$ such that $B(K,\frac{1}{n})\subseteq U$. We may assume without loss of
generality that $\overline{B(K,1)}\subseteq U$ (otherwise, shift
subindices below). Also, it can be shown that 
$$
K=\bigcap_{n\geq 1} B(K,\tfrac{1}{n})\qquad\qquad (*).
$$
Now assume that the assertion is false. That is, for every
$n\in\mathbb{N}$ there exist $x_n,y_n\in B(K,\frac{1}{n})$ such that
$x_n\neq y_n$  $(**)$ and 
$F(x_n)=F(y_n)$. Since $\{x_n\}_n$ and $\{y_n\}_n$ are sequences in
the compact $\overline{B(K,1)}$, there are convergent subsequences
$\{x_{n_j}\}_j$  and $\{y_{n_j}\}_j$ with limits $x$ and $y$, respectively.  By
$(*)$, $x,y\in K$. But then $x=y$, since $F$ is injective on $K$. Now,
for sufficiently large $j$, $x_{n_j},y_{n_j}\in U_x$. Since $F$ is
injective on $U_x$, $x_{n_j}=y_{n_j}$ for all such $j$. But this
contradicts $(**)$.
A: Getting a finite open cover $K$ consisting of metric balls seems to be a red herring. 
Your assumptions imply there is an open $U$ containing $K$ over which $f$ is injective. Let  $\epsilon=d(K,\mathbb R ^n \setminus U)$. All you have to show is $\epsilon>0$, because clearly $\{x\in \mathbb R^n : d(x,K) < \epsilon \}\subseteq U$.
Suppose $\epsilon=0$. There is a sequence $(k_i)$ of points in $K$ and a sequence $(x_i)$ of points in $\mathbb R ^n\setminus U$ such that $d(k_i,x_i)\to 0$ as $i\to\infty$. Because $K$ is compact, the sequence $(k_i)$ has a subsequence that converges to $k\in K$. But then you can get a subsequence of $(x_i)$ which also converges to $k$. This contradicts that $\mathbb R ^n\setminus U$ is closed.
