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Let (X,$\tau$) be a topological space and, for each point x$\in$ X, let $B_x$ be a local basis at x.

Show that B=$\bigcup${$B_x$:x$\in$ X} is a basis for the topology on X

My attempt (This is first proof dealing with this topic)

Let (X,$\tau$) be a topological space. Let x$\in$ X and x$\in $$\bigcup${$B_x$:x$\in$ X}

Then for some x$\in$ X, x$\in B_x$.For each B$\in\mathfrak{B_x}$ ,$B_x$ is an open set.

Let U $\in\tau$:x$\in$ B lt follows then x$\in$B$\subset$ U

So B is a union of open sets and that is an open set .

Thus B is a basis for X

Source:A First Course In Topology:Conover,R

I feel something is fishy Any help would be appeciated

Plotinus
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1 Answers1

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To check that $\mathcal{B}=\bigcup \{\mathcal{B}_x\mid x\in X\}$ is a base for the topology $\tau$, we need to check

$$\forall O \in \tau: \forall x \in O:\exists B \in \mathcal{B}: x \in B \subseteq O$$

and this immediately follows from the fact we can such a $B$ from $\mathcal{B}_x (\subseteq \mathcal{B})$ from the definition of a local base at $x$.

C'est tout.

Henno Brandsma
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