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
- Show that $\mathcal{T} \subset \mathcal{D}$ and $\mathcal{D} \subset \mathcal{T}$, and thus that $\mathcal{T} = \mathcal{D}$, by showing that each $U \in \mathcal{T} = U \in \mathcal{D}$
- 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?
Best Answer
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 family 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 topology 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 subcollection $\mathcal{K}$ 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 $\mathcal{D} = \mathcal{T}$. $\ \square$