Let $ A \subseteq \mathbf{R}^{n} $ be a closed set and let $ B $ be a closed unit normal bundle over $ A $ ( that means for every $ a \in A $ we have closed subset $ B\_{a} \subseteq \mathbf{S}^{n-1} $ and $ B = \{ (a,u) : u \in B\_{a} \} $ is a closed set).


Consider the set $ F $ of functions $ f : A \rightarrow \mathbf{S}^{n-1} $ such that $ f(a) \in B\_{a} $ for every $ a \in A $. Is there in F at least one continuous function? 


Observe that if $ B\_{a} $ has only one element for every $ a \in A $ then the only element in $ F $ is continuous.



existence of continuous functions with values in the fiber of a closed bundle
Unless I'm missing something, the answer is no. 


Take $n=2$, $A=\mathbb{R}^2$, and let $a, b$ be distinct points on the unit circle. Now let $B$ be defined as follows:


* $B\_{(x, y)}=\{a\}$ if $x<0$
* $B\_{(x, y)}=\{b\}$ if $x>0$
* $B\_{(x, y)}=\{a, b\}$ if $x=0$.


It's clear that there is no continuous selector for $B$, and unless I'm missing something, $B$ defines a closed subset of $\mathbb{R}^2\times S^1$.


Or am I missing something?


