About lower upper bound of a subset of all topologies for a non empty set X

Question

How i may prove what is stated? That the topology generated using the theorem D as in the image, is equal to the lower upper bound of a non-empty family of topologies for a non empty set $X$.

Definitions: The lower upper bound of a non-empty family of topologies of $X$ is the topology whose arise from the intersection of all topologies stronger than each topology in the family. A topology T is stronger than R if R is cointained in T.

I tried to prove that one set is cointained in the other by proving that every open set from one is in the another, but i could not, in special because the union of the topologies. One there know how to prove it or have a hint? Thank you very much.

Answer 1

If you want to prove it in detail, all you have to show is that last statement:

the class of all unions of finite intersections forms a topology

which is to say, this class is closed under the formation of finite intersection and arbitrary unions.

Doesn't it follow by construction?
(Notice that $\bigcup \varnothing = \varnothing$ and $\bigcap \varnothing = X$.)

Answer 2

Suppose $T$ is a topology stronger than every topology in the family ${T_i}$. Let $T'$ be the topology generated by the open subbase $\cup_i T_i$. You need to prove that $T$ contains $T'$. If this is true then...

Edit: changed "$T$ is stronger than $T'$" to "$T$ contains $T'$". This should be more clear.