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"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?

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If anything, the sentence states that if there are $m$ such $S$-neighbours then $m<n$ (since there aren't $n$, and if there are $m>n$ then plainly there are $n$ (just take a subset of size $n$)). It doesn't state that any number of them do exist, though. – Clive Newstead Aug 29 '12 at 16:52

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.

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"exactly $n$ many" sounds weird. – xando Aug 29 '12 at 17:04

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.

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