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Show that if $\mathscr{A}$ is a basis for a topology on $X$, then the topology generated by $\mathscr{A}$ equals the intersection of all topologies on $X$ that contain $\mathscr{A}$.

This is what I've managed. However, the following stumps me:

Prove the same if $\mathscr{A}$ is a subbasis.

That is, how to show that the topology generated by $\mathscr{A}$ as a subbasis is the same as the intersection of all topologies on $X$ that contain $\mathscr{A}$?

Let $\mathscr{T}$ be the topology that consists of all unions of finite intersections of sets in $\mathscr{A}$, and let $\mathscr{T}^\prime$ be the topology that is the intersection of all topologies that contain $\mathscr{A}$.

Now $\mathscr{A}$ is a collection of subsets of $X$ whose union is all of $X$, and each set $A \in \mathscr{A}$ is in both $\mathscr{T}$ and $\mathscr{T}^\prime$.

Am I right so far?

And if so, then what next?

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First, we show that $\mathscr{T}'\subseteq \mathscr{T}$.

Suppose $U \in \mathscr{T}'$. Then $U$ is an element of every topology containing $\mathscr{A}$, including $\mathscr{T}$, which is certainly one such topology. So $U \in \mathscr{T}$, which shows $\mathscr{T}'\subseteq \mathscr{T}$.

To see that $\mathscr{T} \subseteq \mathscr{T}'$:

Let $W \in \mathscr{T}$. Then $W = \bigcup W_i$ with each $W_i$ a finite intersection of sets in $\mathscr{A}$, say $W_i = W_{i1} \cap \dots \cap W_{in}$, with $W_{ij} \in \mathscr{A}$. Since $W_{ij} \in \mathscr{A}$, and $\mathscr{A} \subseteq \mathscr{T}'$, and because $\mathscr{T}'$ is a topology (and thus closed under finite intersections), $W_i \in \mathscr{T}'$, and thus (since $\mathscr{T}'$ is also closed under arbitrary unions) $W \in \mathscr{T}'$, so $\mathscr{T} \subseteq \mathscr{T}'$.

Hence, we see that $\mathscr{T} = \mathscr{T}'$.

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Since $\scr A$ is contained in $\scr T$ and $\scr T$ is a topology, it follows that $\scr T^\prime\subseteq \scr T$ (as $\scr T^\prime$ is the intersection of all topologies on $X$ that contains $\scr A$). For the other direction, if $V\notin \scr T^\prime$ then there exist a topology $\cal T$ on $X$ with $\scr A\subseteq \cal T$ such that $V\notin \cal T$. Since any arbitrary union of finite intersections of sets in $\scr A$ belongs to $\cal T$ (because $\cal T$ is a topology and $A\in \scr A\implies A\in \cal T$), it follows that $V$ cannot be a union of finite intersections of sets in $\scr A$ and hence $V\notin \scr T$. Hence $\scr T\subseteq \scr T^\prime$. So both the topologies are equal.

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