Show particular point topology, is a topology I'm teaching my self topology with the aid of a book. I'm trying to prove the following is a topology:

Let X be an infinite set, and $p$ be an arbitrary point in $X$. Show that $\mathscr{T}_3=\{U \subseteq X : U = \varnothing $ or $ p \in U \} $ is a topology.

My book calls this the particular point topology
Please let me know if my proof is valid.
Now to verify the definition.

(i) X and Ø are elements of $\mathscr{T}$.

Ø is given in the definition of $\mathscr{T}_3$, and $p \in X$, by definition. So (i) is "self-evident."

(ii) $\mathscr{T}$ is closed under finite intersections.

Let: $ A = \bigcap_{i=1}^n U_i ; $ where $U_i$ are sets that contain $p$ and $n \in \mathbb{N}$. I'm going to prove by contradiction. Assume $A \notin \mathscr{T}_3 $; that would imply that $p \notin A$, and by the nature of intersections, that would mean that AT LEAST ONE of the $U_i$'s does not contain $p$. However that contradicts the definition of $U_i$ (RAA).

(iii) $\mathscr{T}$ is closed under arbitrary unions.

Let: $A=\bigcup_{i \in I} U_i$ where $U_i$ are sets that contain $p$ and $n \in \mathbb{N}$. I'm going to prove by contradiction. Assume $A \notin \mathscr{T}_3 $;that would imply that $p \notin A$, and by the nature of unions, that would mean that ALL OF the $U_i$'s do not contain $p$. However that contradicts the definition of $U_i$ (RAA).
 A: Your proof is correct, but perhaps you will find this pair of comments useful:


*

*There is no need for $X$ to be infinite. It suffices to be non-empty. Note that you have never used that $X$ is infinite.

*Try to rewrite the two last proofs without using RAA. This is an extremely useful technique, but RAA make these particular proofs unnecessarily long and "overloaded".
A: @ajotatxe, here is my attempt to prove (ii) and (iii) without RAA.

(ii) $\mathscr{T}$ is closed under finite intersections.

Let: $ A = \bigcap_{i=1}^n U_i ; $ where $U_i$ are sets that contain $p$ and $n \in \mathbb{N}$. $A$ must be in $\mathscr{T}_3$ because intersections are the points that sets have in common, and every $U_i$ contains $p$. 

(iii) $\mathscr{T}$ is closed under arbitrary unions.

Let: $A=\bigcup_{i \in I} U_i$ where $U_i$ are sets that contain $p$ and $n \in \mathbb{N}$. $A$ will be in $\mathscr{T}_3$ if at least one of the $U_i$'s contain $p$. And, in fact, all the $U_i$'s contain $p$. So $A \in \mathscr{T}_3$.
