how to understand this sentence? "There are not $n$ $S$-neighbours $y_1, \dots, y_n$ of $x$ with $C$ in $\mathcal{L}(y_i)$ and $y_i \not = y_j$ for $1 \leq i < j \leq n$."
If there are $n-1$ such $S$-neighbours, is that entailed by this sentence? Or this sentence only entails that there's no such $S$-neighbors?
 A: No. The statement only asserts (with absolute certainty) that given $n$ many $S$-neighborhood $y_1, ..., y_n$ such that $y_i \neq y_j$ for $i \neq j$, there must exists a $i$ such that $C \notin \mathcal{L}(y_i)$. 
It does not say anything about collections of less than $n$ or more than $n$ $S$-neighborhood. It is consistent with the statement that there are no collection of $S$-neighborhood with the above property. There could be some $k < n$ or $k > n$ for which property holds. The only thing you can gather from the statement is that the property does not hold for exactly $n$ many $S$-neighborhood. 

Also I have no idea what any of the symbols actually mean. If this is a concrete question from topology or another area of mathematics, then by using the definition of $S$-neighborhood and $\mathcal{L}(y_i)$, you may be able to get more information.I just interpreted the sentence using only its logical form. 
A: The sentence doesn't exclude the possibility that there are $n-1$ neighborhoods such that ...
So the sentence doesn't actually say anything about what does exist.
