Conside a straight line $L$ in the plane $\mathbb{R}^{2}$. The filter of neighborhods of $L$ in $\mathbb{R}^{2}$ is the filter formed by the sets which contain an open set containing $L$. Prove that there is no basis of this filter which in countable.
First of all, I noticed that this family really defines a filter. Indeed, let
$$\mathscr{F}=\{A \subset \mathbb{R}^2: A \supset U \supset L, U=U^\circ\}$$
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$\emptyset \notin \mathscr{F}$ since this set cannot contain an open set containing L.
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Obviously the intersection of two sets, belonging to the family, also belongs to the family $\mathscr{F}$
- If $A \in \mathscr{F}$ then there exists an open subset of $\mathbb{R}^{2}$ such that $A \supset U \supset L$. Thus, if $B \supset A$ the same set $U$ satisfies the desired properties, that is, $B \supset U \supset L.$
Therefore, $\mathscr{F}$ is a filter.
However, I had no idea how to show that this filter has no countable basis.
Best Answer
We can assume $L=\mathbb{R}\times\{0\}$.
Let $(B_n)_{n\in\mathbb{N}}$ be a family of elements of $\mathscr{F}$. For $k\in\mathbb{N}$ and $n\in\mathbb{N}$, there must exist $\epsilon_{kn}>0$ such that $\{k\}\times[-\epsilon_{nk},\epsilon_{nk}]\subset B_n$. One can define a set $A$ which is a member of $\mathscr{F}$, but does not contain $(n,\epsilon_{nn})$ for any $n\in\mathbb{N}$. This set $A$ cannot contain any of the $B_n$.