Exercise 13.4

Proof. First, clearly since $\text{∅}$ and $X$ are in every $T_{\alpha }$ since they are topologies, they are both in $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ so that property (1). Now suppose that $A$ is a subcollection of $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$. Consider any $T_{\beta }$ and any $A\in A$. Then $A$ is also in $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ since $A\subset \bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$. It then follows that $A$ is in our specific $T_{\beta }$. Since $A$ was arbitrary it follows that $A$ is a subcollection of $T_{\beta }$ so that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->A\in T_{\beta }$ also since $T_{\beta }$ is a topology. Since $T_{\beta }$ was also arbitrary it follows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->A\in \bigcap T_{\alpha }$. Lastly, since the subcollection $A$ was arbitrary, this shows property (2) for $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$.

Finally, suppose that $U_1,\dots ,U_n$ are sets in $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$. Consider any $T_{\beta }$ so that clearly then $U_i\in T_{\beta }$ for every $i\in \left\{1,\dots ,n\right\}$. It then follows that ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{U}_i\in T_{\beta }$ since $T_{\beta }$ is a topology. Since $T_{\beta }$ was arbitrary, this shows that ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{U}_i\in \bigcap T_{\alpha }$, which shows property (3) for $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$. This completes the proof that $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ is a topology on $X$ since all three properties have been shown. □

Proof. As a counterexample consider the set $X=\left\{a,b,c\right\}$, the topologies $T_1=\left\{\varnothing ,X,\left\{a\right\}\right\}$ and $T_2=\left\{\varnothing ,X,\left\{b\right\}\right\}$, and the collection of topologies $C=\left\{T_1,T_2\right\}$. Then we clearly have that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->C=T_1\cup T_2=\left\{\varnothing ,X,\left\{a\right\},\left\{b\right\}\right\}$, which is not a topology since $A=\left\{\left\{a\right\},\left\{b\right\}\right\}$ is a subcollection of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->C$ but $\bigcup <!--\; FUNCTION\; APPLICATION\; -->A=\left\{a,b\right\}$ is not in $\bigcup <!--\; FUNCTION\; APPLICATION\; -->C$. □

Proof. It was proven in part (a) that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ is not necessarily a topology. However, it is clearly always a subbasis for a topology since clearly $X\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ since it is in each $T_{\alpha }$ since they are topologies. Hence obviously then $\bigcup <!--\; FUNCTION\; APPLICATION\; -->\left(\bigcup T_{\alpha }\right)=X$ so that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ is a subbasis by definition. Then let $T_s$ be the topology generated by the subbasis $\bigcup <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$. We claim that $T_s$ is then the smallest topology that contains all the $T_{\alpha }$ as subsets.

First, from the proof following the definition of a subbasis, we know that the set $B$ of finite intersections of elements of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ is a basis for the topology $T_s$, and that $T_s$ is the set of all unions of subcollections of $B$.

We first show that every $T_{\alpha }$ is indeed contained as a subset of $T_s$. So consider any specific $T_{\beta }$ and any $U\in T_{\beta }$. Then clearly $U\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ so that $U\in B$ since $U=\bigcap <!--\; FUNCTION\; APPLICATION\; -->\left\{U\right\}$ is a finite intersection of elements of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$. It then follows that $U\in T_s$ since $U=\bigcup <!--\; FUNCTION\; APPLICATION\; -->\left\{U\right\}$ is the union of a subcollection of $B$. Since $U$ was arbitrary, this shows that $T_{\beta }\subset T_s$, which shows the result since $T_{\beta }$ was arbitrary.

Now we show that $T_s$ is the smallest such topology as ordered by $⊊$. So suppose that $T$ is a topology that contains every $T_{\alpha }$ as a subset. Consider any $U\in T_s$ so that $U=\bigcup <!--\; FUNCTION\; APPLICATION\; -->C$ for some subcollection $C\subset B$. Now consider any $Y\in C$ so that also $Y\in B$. Then $Y={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{Y}_i$ where each $Y_i\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$. Then each $Y_i$ is in some $T_{\beta }\subset T$ so that also $Y_i\in T$. Since $T$ is a topology, it follows that the finite intersection ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{Y}_i=Y$ is also in $T$. Since $Y$ was arbitrary, this shows that $C\subset T$ so that $C$ is a subcollection of $T$. It then follows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->C=U$ is also in $T$ since $T$ is a topology. Since $U$ was arbitrary, we have that $T_s\subset T$, which shows that $T_s$ is the smallest topology since $T$ was arbitrary.

It is easy to see that $T_s$ is unique since, if both $T_1$ and $T_s$ are the smallest topologies that contain each $T_{\alpha }$ as subsets, then we would have that both $T_1\subset T_2$ and $T_2\subset T_1$ so that $T_1=T_2$. Really this follows from the more general fact that the smallest elements in any order are always unique. □

Proof. It was shown in part (a) that $T_l=\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ is a topology on $X$. We claim that in fact, this is the unique largest topology contained in all $T_{\alpha }$. First, clearly $T_l=\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }$ is contained in each $T_{\alpha }$ since the intersection of a collection of sets is always a subset of every set in the collection. Now suppose that $T$ is a topology that is contained in every $T_{\alpha }$, i.e. $T\subset T_{\alpha }$ for every $T_{\alpha }$. Then clearly for any $U\in T$ we have that $U\in T_{\alpha }$ for every $T_{\alpha }$ so that $U\in \bigcap <!--\; FUNCTION\; APPLICATION\; -->T_{\alpha }=T_l$. Thus $T\subset T_l$ since $U$ was arbitrary. This shows that $T_l$ is the largest such topology since $T$ was arbitrary.

Clearly also $T_l$ is unique since, if $T_1$ and $T_2$ are two such largest topologies that are contained in every $T_{\alpha }$. Then we would have $T_1\subset T_2$ and $T_2\subset T_1$ so that $T_1=T_2$. This also follows from the fact that the largest element in any ordered set (or collection of sets in this case) is unique. □