If we consider N as a subset of the usual metric space of real numbers R, we can think in a fundamental system of neighbourhoods (FSN) of N. I need to prove that does not exist a countable FSN, and as a suggestion Dieudonne (The autor of the book Modern analysis) says: Use contradiction. If ( ( a n m )) is a doble sequence of positive numbers, the sequence ( b n ) = a n n 2 is such that for no integer m is valid the inequality b n ≥ a m n for all the integers n.How can I use this fact to prove it?

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If we consider N as a subset of the usual metric space of real numbers R, we can think in a fundamental system of neighbourhoods (FSN) of N. I need to prove that does not exist a countable FSN, and as a suggestion Dieudonne (The autor of the book Modern analysis) says: Use contradiction. If (  (      a          n      m        )) is a doble sequence of positive numbers, the sequence   (      b    n    )  =            a              n        n              2   is such that for no integer   m is valid the inequality       b    n    ≥      a          m      n       for all the integers   n.How can I use this fact to prove it?

Mateo Barajas · Accepted Answer

Suppose that   N has a countable system of neighbourhoods       U    n    ,  n  ∈  N, which means that for any open set   O that contains   N we know that for some   k,      U    k    ⊂  O.As all       U    n   are open sets of   R, we know that after maybe shrinking each       U    n   (which does not affect the FSN property) each       U    n   can be assumed to be of the form       U    k    =      ∪    n    (  n  −      r          n      ,      k        ,  n  +      r          n      ,      k        )  ,      r          n      ,      k        &amp;lt;      1    2  . (As each   n has such a neighbourhood inside       U    k  , and we collect those).Now we have a double sequence       r          n      ,      k       and we apply the idea you mentioned:Define       s    n    =            r              n        ,        n              2  , we halve the &quot;diagonal&quot;. Define   O  =      ∪    n    (  n  −      s    n    ,  n  +      s    n    ), which is an open set that contains   N. So if we had a FSN for   N, for some   m,       U    m    ⊂  O. But the interval around   m in       U    m   equals   (  m  −      r          m      ,      m        ,  m  +      r          m      ,      m        ) and in   O it&#039;s   (  m  −      s    m    ,  m  +      s    m    ), which is strictly smaller, so the inclusion fails, contradiction.Note that this fact is often used to show that   R  /  N, the quotient space where   N gets identified to a &quot;point&quot;, is not first countable at this new point. A countable neighbourhood base for the class of   N, would, pulled back, be a countable FSN for the set   N. Fun fact: compact sets in metric spaces do have a countable FSN.

If we consider N as a subset of the usual metric space of real numbers R, we can think in a fundamen

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