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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

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

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  • $\begingroup$ 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

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