A fundamental system of neighbourhoods of $N $ (natural numbers)

Question

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_{nm}) $ is a doble sequence of positive numbers, the sequence $(b_n) = \frac{a_{nn}}{2}$ is such that for no integer $m $ is valid the inequality $b_n \ge a_{mn} $ for all the integers $n $

How can I use this fact to prove it?

Answer 1

Suppose that $\mathbb{N}$ has a countable system of neighbourhoods $U_n, n \in \mathbb{N}$, which means that for any open set $O$ that contains $\mathbb{N}$ we know that for some $k$ ,$U_k \subset O$.

As all $U_n$ are open sets of $\mathbb{R}$, 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 $U_k = \cup_n (n-r_{n,k}, n+r_{n,k}), r_{n,k} < \frac{1}{2}$. (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 $s_n = \frac{r_{n,n}}{2}$, we halve the "diagonal". Define $O = \cup_n (n - s_n, n + s_n)$, which is an open set that contains $\mathbb{N}$. So if we had a FSN for $\mathbb{N}$, for some $m$, $U_m \subset O$. But the interval around $m$ in $U_m$ equals $(m - r_{m,m} , m+r_{m,m})$ and in $O$ it'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 $\mathbb{R}/\mathbb{N}$, the quotient space where $\mathbb{N}$ gets identified to a "point", is not first countable at this new point. A countable neighbourhood base for the class of $\mathbb{N}$, would, pulled back, be a countable FSN for the set $\mathbb{N}$. Fun fact: compact sets in metric spaces do have a countable FSN.