Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

"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?

share|improve this question
    
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

2 Answers 2

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.

share|improve this answer
    
"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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.