I'm reading Burago, Burago and Ivanov's book, and I'm on the section about Strainers. The authors say that it is obvious that the set of $(m,\varepsilon)$-strained points for any fixed natural number $m$ and $\varepsilon >0$ is open. I'm failing to see why is it so obvious. 


As I understand, I have to prove that for any $(m,\varepsilon)$-strained point $p$ there is a radius $r$ such that all the metric ball $B\_r(p)$ is composed of $(m,\varepsilon)$-strained points. I'm trying to build strainers for a point $q \in B\_r(p)$ from the strainers at $p$ but I've had no luck so far. Am I missing something very obvious here?



The set of strained points in an Alexandrov space is open
Note that this is an open condition, which depends on the distances $|a\_ia\_j|$, 
$|a\_ib\_j|$, $|b\_ib\_j|$, $|pa\_i|$ and $|pb\_i|$, hence the result.


