# Subbase of a topology

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

• You might find my answer here useful too. – Henno Brandsma Oct 26 '18 at 13:32

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.

• Thanks for you answer @Ingix, it is very helpful and clear. I proved that $X \in \tau^\prime$ by assuming that $\cap_{i \in I} C_i = X$ if $I = \emptyset$. – F.Vitiello Oct 24 '18 at 11:35
• I would have said that this value is $\emptyset$, but of course definitions may vary. – Ingix Oct 24 '18 at 12:56
• @Ingix This is the standard convention. Like the empty product is $1$. See the Wikipedia entry on vacuous truth, e.g. The empty set we always get as any union over an empty index set. – Henno Brandsma Oct 24 '18 at 13:57