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a very simple question I find different answers to online: In topology, what is the definition of a neighbourhood basis for a set, and what is the definition of the topology generated by it?

M. Van
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  • Can you provide some links to those different answers? – Henno Brandsma May 09 '16 at 13:40
  • You want the definition for an indexed (by $X$) set of subsets to be a neighbourhood base system for a topology defined by it, I think? – Henno Brandsma May 09 '16 at 13:42
  • Yes, I think that is what I want. On the internet I can only find the definition of a neighbourhood basis of a point of a set on which a topology is already defined. – M. Van May 09 '16 at 15:41

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Suppose we have a set $X$ and for each $x \in X$ we have a collection of subsets of $x$ called $\mathscr{B}(x)$. Then we call this assignment from $x$ to $\mathscr{P}(\mathscr{P}(X))$ a system of neighbourhood bases for $X$ iff the following axioms hold:

  1. $\forall x \in X: \mathscr{B}(x) \neq \emptyset$.
  2. $\forall x \in X: \forall B \in \mathscr{B}(x): x \in B$.
  3. $\forall x \in X : \forall U,V \in \mathscr{B}(x): \exists W \in \mathscr{B}(x): W \subseteq U \cap V$.
  4. $\forall x \in X: \forall U \in \mathscr{B}(x): \exists V \in \mathscr{B}(x) : \forall y \in V: \exists W \in \mathscr{B}(y) : W \subseteq U$.

There is an alternative axiom 4, which one encounters too and which in the end gives the same topology, but which is less "pure" as it were, because it involves a subset that is not in some $\mathscr{B}(x)$:

4'. $\forall x \in X: \forall U \in \mathscr{B}(x): \exists V \subseteq X: (x \in V \subseteq U) \land (\forall y \in V: \exists W \in \mathscr{B}(y): W \subseteq V)$.

In either system we define $O \subseteq X$ to be open iff $\forall x \in O \exists B \in \mathscr{B}(x): B \subseteq O$. So axiom 4' can be reformulated (using the observation that the $V$ in it is open according to this definition) as "every neighbourhood $U$ of $x$ contains an open set $V$ containing $x$ that sits inside $U$".

One can easily show that this defines a topology (4 or 4' are not needed for that, even).

In this answer I check that the topology on $X$ so defined is unique with the property that the original $\mathscr{B}(x)$ form a neighbourhod base for that topology, based on the 4' axiom.

In this answer there is a similar check based on the 1-4 axioms.

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Henno Brandsma
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  • question: does every neighbourhood basis induce a basis and vice versa? – M. Van Jan 03 '17 at 16:02
  • @M.Van the collection of all open neighbourhoods of all points is a base. And vice versa, all sets in a base that contain $x$ form a local base for $x$. – Henno Brandsma Jan 04 '17 at 16:32