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Theorem:

Let X be a set and let B be a basis for a topology on X. Then T equals the collection of all unions of elements of B.

Proof:

X is a non-empty set and let B be a basis for a topology on X.

Recall: A topology T has a countable basis IFF there is at least one basis B that generates T and has only countably many elements.

Indeed, there is at least one basis B that generates a topology T.

But what about B having countably many elements?

Hints are appreciated.

Thanks in advance.

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  • $\begingroup$ Your question has nothing to do with countability. In fact, your question doesn't even require a proof, since it is true just by definition. $\endgroup$ – Alex M. Aug 20 '16 at 10:22
  • $\begingroup$ see lemma 13.1 of munkres topology $\endgroup$ – Javier Dec 7 '18 at 18:09
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Let $\mathbf T$ be a topology on a set $X$ and $\mathbf B$ be an arbitrary basis of the topology $T$, consisting of its open subsets. Since a union of any family of open sets is open, the union of any family of members of $\mathbf B$ belongs to $\mathbf T$. Conversely, let $U\in\mathbf T$ be an arbitrary open set. By the definition of a basis, for any point $x\in U$ there exists a set $U_x\in\mathbf B$ such that $x\in U_x\subset T$. Then $U=\bigcup_{x\in U} U_x$ is the union of a family of members of $\mathbf B$.

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  • $\begingroup$ Should $U_x$ be chosen such that $U_x \subset U$? Also for my own curiosity/understanding: does this proof rely on the axiom choice to choose a single $U_x$ for each $x$ when there may in fact be (infinitely) many such sets $V$ satisfying $x \in V \subset U$? $\endgroup$ – Jagerber48 Jul 22 '18 at 6:19

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