TITLE: isomorphism Neighbourhood system and basis
QUESTION [0 upvotes]: Given a basis and a point, choosing all elements including that point build a Neighbourhood system for that point.
Given a Neighbourhood system for each point, taking the union will give a basis for the topology.
Is there an isomorphism between basis and Neighbourhood systems?
Is then one of the notions redundant (although of course useful)?

REPLY [3 votes]: Neither of them is redundant, I'd say. Yes, if we have a base $\mathcal{B}$ for the topology $(X,\mathcal{T})$, then $\mathcal{B}_x=\{B \in \mathcal{B}\mid x \in B\}$ is a neighbourhood base for $x$ for $(X,\mathcal{T})$.
And if $\mathcal{B}_x$ is an open neighbourhood base for $x$, for each $x \in X$, then $\mathcal{B} = \bigcup_{x \in x} \mathcal{B}_x$ is a base for $(X,\mathcal{T})$.
So both are alternative ways to describe a topology on a set: you can specify a base for it, or a neighbourhood system for each point. Both are common and useful. And there are quite a few ways to describe properties of spaces from either view point, and translate between them in similar ways as we did above. You can also "just" define the set $\mathcal{T}$ in some other way.
In this "equivalent description" sense there can be said to be an "isomorphism", but usually that term is reserved for other purposes (like in algebraic contexts).
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