Consider the following topological space $(X, \tau)$ and let $\mathcal{C} \subseteq \tau$. Define
\begin{align}
\mathcal{B} &= \left\lbrace \cap_{i \in I} C_i \ \ | \ \ I \ \text{is a finite index set and} \ C_i \in \mathcal{C} \ \forall i \in I \right\rbrace \\
\tau^\prime &= \left\lbrace \cup_{j \in J} T_j \ \ | \ \ J \ \text{is an arbitrary index set and} \ T_j \in \mathcal{B} \ \forall j \in J \right\rbrace \\
\end{align}
I need to show that $\tau^\prime$ is a topology on $X$.
I know that in order to solve this question I need to show the following
- $\emptyset, X \in \tau^\prime$
- if $\{A_k\}_{k \in K} \subseteq \tau^\prime$ for some arbitrary $K$ then $\cup_{k \in K} A_k \in \tau^\prime$
- if $\{B_s\}_{s \in S} \subseteq \tau^\prime$ for some finite $S$ then $\cap_{s \in S} B_s \in \tau^\prime$
I have already proved the first bullet point. Anyone knows how I can prove the second and the third one? Thanks
Best Answer
How did you prove $X \in \tau'$ when this is not necessarily true? Take the extreme case that $\mathcal C = \{\emptyset\}$, then $\tau'=\{\emptyset\}$. Generally, any $x \in X$ that is not in any $C \in \mathcal C$ cannot be in any open set of $\tau'$.
So either you have to assume $\bigcup_{C \in \mathcal C} C = X$, or consider the underlying set of your topology to be $X'=\bigcup_{C \in \mathcal C} C$.
To your question (second bullet point): By definition $A_k=\bigcup_{j \in J_k} T_j, T_j \in \mathcal B$. If you set $I=\cup_{k \in K}J_k$, then you get
$$\bigcup_{k \in K} A_k = \bigcup_{k \in K} \cup_{j \in J_k} T_j = \bigcup_{i \in I} T_i$$
because the union of sets is associative and commutative. Since all $T_i$ are in $\mathcal B$, the last expression must be in $\tau'$.
The last bullet point is going into the same direction but a little more complicated, as you need to use the definition of $\mathcal B$ as well.