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

This what what I did before getting stuck

Proof:

The topology $\mathcal{T}$ on $X$ generated by the basis $\mathcal{A}$ is defined as follows:

$$\mathcal{T} = \{ U \subset X \ | \ \forall x \in U, \exists A \in \mathcal{A} \ \text{ such that } x \in A \ \text{ and } A \subset U \}$$ Lemma: If $\mathcal{B}$ is a basis and $\mathcal{K}$ is the topology generated by $\mathcal{B}$, then $\mathcal{B} \subset \mathcal{K}$

Therefore we have $\mathcal{A} \subset \mathcal{T}$.

Now let $\mathcal{F} = \{T_{\alpha}\}$ be the family of topologies on $X$ such that $\mathcal{A} \subset \mathcal{T_{\alpha}}$. Put $\mathcal{D} = \bigcap_{\alpha} \mathcal{T_{\alpha}}$, where $\mathcal{T_{\alpha}} \in \{\mathcal{T_{\alpha}}\}$. By a well known theorem $\mathcal{D}$ is a topology on $X$ (this can be easily proven but I've left it out here to keep the proof as short as possible) , and since $\mathcal{A} \subset \mathcal{T_{\alpha}}$ for every $\mathcal{T_{\alpha}} \in \{\mathcal{T_{\alpha}} \}$,we have $\mathcal{A} \subset \mathcal{D}$.


But that is as far as I've got before getting stuck. To show that $\mathcal{T} = \mathcal{D}$, I can do one of two things

  1. Show that $\mathcal{T} \subset \mathcal{D}$ and $\mathcal{D} \subset \mathcal{T}$, and thus that $\mathcal{T} = \mathcal{D}$, by showing that each $U \in \mathcal{T} = U \in \mathcal{D}$
  2. Show that $\mathcal{T}$ and $\mathcal{D}$ both have the same basis $\mathcal{A}$ and thus must be the same topology as every basis generates a unique topology.

But in either case I'm not sure how to proceed. For 2, I could assume that $\mathcal{D}$ has some underlying basis $\mathcal{O} \neq \mathcal{A}$, and then attempt to arrive at some contradiction, but I don't see how I would reach a contradiction here. For 1, I can show that $U \in \mathcal{T}$ can be expressed as the union of basis elements, but for $U \in \mathcal{D}$ all I know is that $U \in \mathcal{T_{\alpha}}$ for every $\alpha$, and that for one $\alpha$ we have $\mathcal{T_{\alpha}} = \mathcal{T}$.

How could I complete this proof using approach 1, or 2? Or is there a shorter easier way to complete this proof?

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Proof:

By a lemma, $\mathcal{T}$ equals the collection of all unions of elements of its basis $\mathcal{A}$, therefore we have

$$\mathcal{T} = \left\{\ \bigcup_{A \in \mathcal{K}} A\ \ | \ \mathcal{K} \subset \mathcal{A}\right\}$$

Now let $\mathcal{F} = \{\mathcal{T_{\alpha}}\}$ be the famiy of topologies on $X$ such that $\mathcal{A} \subset \mathcal{T_{\alpha}} \in \mathcal{F}$. Put $\mathcal{D} = \bigcap_{\alpha} \mathcal{T_{\alpha}}$, then $\mathcal{D}$ is a topoogy on $X$ as one can easily verify.

Since $\mathcal{A} \subset \mathcal{T_{\alpha}}$ for every $\mathcal{T_{\alpha}} \in \mathcal{F}$, we have $\mathcal{A} \subset \mathcal{D}$. Now since $\mathcal{D}$ is a topology, we take $\mathcal{K} \subset \mathcal{A}$ the union of all the elements of the subcolelction $\mathcal{K}$ is contained in $\mathcal{D}$, i.e. $\bigcup_{A \in \mathcal{K}}A \in \mathcal{D}$ for every possible $\mathcal{K}$. Therefore we have $\mathcal{T} = \left\{\ \bigcup_{A \in \mathcal{K}}A \ | \mathcal{K} \subset \mathcal{A}\right\} \subset \mathcal{D}$.

Conversely we also have $\mathcal{T} = \mathcal{T_{\alpha}}$, for some $\mathcal{T_{\alpha}} \in \mathcal{F}$, and since we have $\mathcal{T} \subset \mathcal{T_{\alpha}}$ for every $\mathcal{T_{\alpha}}$, this implies $\mathcal{D} = \mathcal{T}$. $\ \square$

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To show the reverse inclusion $\mathcal{D} \subseteq \mathcal{T}$, it's enough to see that $\mathcal{T}$ is a topology containing $\mathcal{A}$ (by definition) which means that $\mathcal{D}$ is an intersection over a family of topologies that contains $\mathcal{T}.$

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Reverse inclusion is already given. So I will prove $\mathcal{T}\subset \mathcal{D}$. Suppose $T_\alpha$ is a topology that is described above. Take $U\in \mathcal{T}$. Let $x\in U$, then we can write $x\in A_x\subset U$ for some $A_x\in\mathcal{A}$. Therefore, $U=\cup_{x} A_x $, but $A_x\in\mathcal{A}\subset T_{\alpha}$. Therefore, $A_x$ is open in $T_\alpha$ which implies $U\in T_\alpha$. This proves that $\mathcal{T}\subset \mathcal{T}_\alpha$, but since this is true for every such $\mathcal{T}_\alpha$, we have $\mathcal{T}\subset \mathcal{D}$

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You know that $\mathcal{A} \subseteq \mathcal{D}$, where the latter is the intersection topology. Now let $O \in \mathcal{T}$, the topology generated by $\mathcal{A}$. Then for every $x \in O$, pick $A_x \in \mathcal{A}$ such that $x \in A_x \subseteq O$. Then $O \subseteq \cup \{A_x: x \in O\}$ (as each $x \in O$ is in its "own" $A_x$) $\subseteq O$ (as all $A_x$ are subsets of $O$). So

$$ O = \cup \{A_x: x \in O\}$$

and as all $A_x \in \mathcal{D}$, $O \in \mathcal{D}$, as topologies are closed under unions. So $$\mathcal{T} \subseteq \mathcal{D}$$

On the other hand, $\mathcal{T}$ is a topology that contains $\mathcal{A}$ as a subset (because if $A \in \mathcal{A}$, we can pick $A$ itself for each $x \in A$ to see that $A \in \mathcal{T}$ by definition), and so it is one of the topologies that $\mathcal{D}$ is the intersection of (!), so clearly $$\mathcal{D} \subseteq \mathcal{T}$$ giving us equality.

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