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

  • $\begingroup$ You might find my answer here useful too. $\endgroup$ – 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.

  • $\begingroup$ 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 $I = \emptyset$. $\endgroup$ – F.Vitiello Oct 24 '18 at 11:35
  • $\begingroup$ I would have said that this value is $\emptyset$, but of course definitions may vary. $\endgroup$ – Ingix Oct 24 '18 at 12:56
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Oct 24 '18 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.