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Let $\mathcal{B}$ and $\mathcal{B}'$ be bases of topologies $\tau$ and $\tau'$ on a space $X$ respectively. Show that if each $B \in \mathcal{B}$ is the union of some members of $\mathcal{B}'$, then $\tau \subset \tau'$.

I have two solutions for this and I'm wondering if they're both correct.

Firstly if $O \in \tau$, then I think that by definition there exists $B \in \mathcal{B}$ such that $B \subset O$. And since $B = B_1 \cup B_2 \cup \dots$ for $B_i$'s in $\mathcal{B}'$ we have that $O \in \tau'$.

Second is that if $O \in \tau$, then by definition $O$ is the union of the basis elements $B_i \in \mathcal{B}$. But since each $B_i$ is the union of elements in $\mathcal{B}'$ the set $O$ must also be an union of elements of $\mathcal{B}'$ and thus by definition $O \in \tau'$.

I'm convinced that the second one is correct, but I don't know if the first one is right? Isn't it true by definition that for any open set of $X$ there is a basis element contained in it?

Lakri tsa
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  • The first solution kind-of assumes that $O\in\tau \iff(\exists B\in\mathcal{B})B\subset O$, because it kind-of hints at $O\in\tau'$ because $O$ contains some of the $B\in\mathcal{B'}$. However, the above equivalence is incorrect. Just because a set contains a base set, it doesn't mean it is open. (Ask a closed disc, which contains the open disc inside!) Also: just because a set is open, it doesn't mean it contains a base set (ask the empty set; though this is the only counterexample.) –  Jan 15 '22 at 21:12
  • You must seek the solution in the fact that every open set is the union of all basis elements that are a subset of this set. In mathematical notation:$$O=\bigcup{B\in\mathcal B|B\subseteq O}$$I think this corresponds with your second solution. The first is at least incomplete. – drhab Jan 15 '22 at 21:26
  • Just passing by (so haven't looked carefully at your question), but my answer to Confusion Regarding Munkres's Definition of Basis for a Topology might be of interest. – Dave L. Renfro Jan 16 '22 at 07:47

1 Answers1

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The second one is correct (but could be more formalised), while the first has a gap and is handwavy about the union IMO.

Formalised second : Let $O \in \tau$. By the definition of a base, for some $\mathcal{O} \subseteq \mathcal{B}$ we have $\bigcup \mathcal O = O$. For each $B \in \mathcal{O}$ we know that (as it is in $\mathcal{B}$) we can write it as $B=\bigcup \mathcal O_B$ for some $\mathcal{O}_B \subseteq \mathcal B'$. So in fact $O=\bigcup \mathcal C$ where $\mathcal C = \bigcup \{\mathcal O_B\mid B \in \mathcal{O}\} \subseteq \mathcal B'$ and so $O$ is a union of sets from $\mathcal B'$ too and hence in $\tau'$.

It's however conceptually easier to just say: the assumption implies that $\mathcal B \subseteq \tau'$ (as $\mathcal B' \subseteq \tau'$ and a topology is closed under unions) and as $\tau$ is the minimal topology that contains $\mathcal B$ we have $\tau \subseteq \tau'$ QED. This is IMO by far the simplest and notation-minimal solution.

Henno Brandsma
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