5

Given a topological space $(X, \tau)$ and a basis $B$ for $X$, every open set $A \in \tau$ can be written as $A = \cup B_\alpha$, where $B_\alpha \in B$.

The definition of a local basis is supposed to be similar to this: If $B$ is a local basis for $x \in X$, then every open set containing $x$ (say $A_x$) contains at least one local basis element.

However, in the latter case, we may not be able to write $A_x$ as a union of some local basis element of $x$. How can we modify the definition of a local basis so that it allows us to write every open neighbourhood of $x$ as a union of local basis elements? Will this definition be inferior to the current one?

Jos van Nieuwman
  • 2,084
shivamvats
  • 88

2 Answers2

13

The traditional definition:

$\mathcal{B}$ is a local base for $(X,\mathcal{T})$ at $x$ iff all $B \in \mathcal{B}$ are open ($\mathcal{B} \subseteq \mathcal{T}$) and also contain $x$, and moreover for all $O \in \mathcal{T}$ with $x \in O$ there exists some $B \in \mathcal{B}$ such that $B \subseteq O$.

This corresponds nicely to a localisation of this reformulation of the definition of a base (it amount to the same notion as the "every open set is a union of base elements"):

$\mathcal{B}$ is a base for $(X,\mathcal{T})$ iff all members of $\mathcal{B}$ are open (or $\mathcal{B} \subseteq \mathcal{T}$) and for every $O \in \mathcal{T}$ and every $x \in O$ there exists a $B \in \mathcal{B}$ such that $x \in B \subseteq O$.

This makes it almost immediate that if $\mathcal{B}$ is a base for $(X,\mathcal{T})$ then for every $x \in X$, $\mathcal{B}_x:= \{B \in \mathcal{B}: x \in B\}$ is a local base for $(X,\mathcal{T})$ at $x$. And vice versa, if, for every $x \in X$, $\mathcal{B}_x$ is a local base for $(X,\mathcal{T})$ at $x$, then $\mathcal{B}:= \bigcup \{ \mathcal{B}_x: x \in X\}$ is a base for $(X,\mathcal{T})$.

Also with this traditional definition of local base, a countable local base is special, in the sense that sequences can be used to describe the topology around such a point. (A first countable space is Fréchet-Urysohn).

If you would propose a definition like "$\mathcal{B}$ is a local base at $x$ iff every open set $O$ that contains $x$ is a union of members of $\mathcal{B}$" (which is what you seem to propose), then you can check that the Sorgenfrey line is first countable in the traditional sense (using sets $[x,x+\frac{1}{n})$ for $n \in \mathbb{N}$), but not in this alternative sense (little puzzle for you). So this alternative definition is too strong, and would miss out on such generalisations as we can use for traditionally first countable spaces. Then having a countable local base is almost as strong as having a countable base. So this is a less useful definition.

Other generalisations: if we leave out the $x \in B$ part of the correct local base definition we get the definition of a so-called local $\pi$-base at $x$, e.g. If we leave out $\mathcal{B} \subseteq \mathcal{T}$ in the definition of a base as given above, we get the definition of a network for $(X,\mathcal{T})$. Spaces with a countable local $\pi$-base everywhere, or a countable network have also been studied in general topology.

Henno Brandsma
  • 242,131
0

Disclaimer: This does not answer the question you asked, but perhaps it clarifies some things.

I think the point is that the local basis does not form a global one, but the union of local bases for each point in your space does. Indeed, any open set $U$ contains a local basic open set $U_x$ for all $x\in U$, hence we find $U=\bigcup_{x\in U} U_x$. Conversely, any global basis is a local basis for each point in your space, essentially by definition.

asdq
  • 3,900