Show that T is a not a topology on X

Question

Let $X=\{a, b, c, d, e\}$ and $T_2=\{X,\emptyset, \{a\},\{c,d\}, \{a,c,e\}, \{b,c,d\}\}$.

I know that $T_2$ is not a topology on X as the union of two members of $T_2$ does not belong to $T_2$, but is it also because of the intersection. For example, $ \{c,d\} \cap \{a, c, e\}=\{c\}$ which is not a member of $T_2$. I just want to make sure I fully understand how to prove that $T_2$ is or is not a topology on X.

Answer 1

Yes, you are correct. Recall the definition of a topology:

Let $\tau\subseteq\mathcal{P}(X)$. Then, $\tau$ is a topology if

1.- $\emptyset\in\tau$ and $X\in\tau$

2.- For all $U,V\in\tau$ we have that $U\cap V\in\tau$

3.- For all $F\subseteq\tau$ we have that $\displaystyle\bigcup F\in\tau$.

Then, one form to disprove that some set isn't a topology is found two sets such that their intersection isn't in $\tau$ or a collection such that their union isn't in $\tau$. Your proof is correct.

Answer 2

Yes, exhibiting two sets in $T_2$ whose intersection is not an set in $T_2$ proves that it doesn't define the open sets of a topology, since the collection of open sets of a topology is closed under finite intersections.

(Friendly advice: this is the kind of question (checking against definitions) you must get comfortable and confident answering on your own. It will just slow down your learning otherwise, and also going over a proposed answer and checking it carefully helps you to learn the topic better.)