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}_4=\{U \subseteq X : U = X $ or $ p \notin U \} $ is a topology.
My book calls this the excluded point topology
Please let me know if my proof is valid.
Now to verify the definition.
(i) X and Ø are elements of $\mathscr{T}$.
X is given in the definition of $\mathscr{T}_4$, and $p \notin \varnothing$, 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 do not contain $p$ and $n \in \mathbb{N}$. $A$ must be in $\mathscr{T}_4$ because intersections are the points that sets have in common, and none of the $U_i$'s contains $p$.
special case: $A \cap X$. This is still in $\mathscr{T}_4$ because $A \cap X = A$ .
(iii) $\mathscr{T}$ is closed under arbitrary unions.
Let: $A=\bigcup_{i \in I} U_i$ where $U_i$ are sets that do not contain $p$. $A$ will be in $\mathscr{T}_4$ because unions take all the elements of their constitutive parts, and none of the $U_i$'s contains $p$.
special case: $A \cup X$. This is still in $\mathscr{T}_4$ because $A \cap X = X$ .
Best Answer
In (ii) and (iii) you should also account for intersections and unions of families that include the open set $X$. For (ii), for instance, let $\mathscr{U}$ be a finite family of open sets. If there is a $U\in\mathscr{U}$ such that $p\notin U$, then clearly $p\notin\bigcap\mathscr{U}\subseteq U$. If not, then $\mathscr{U}=\{X\}$, and $\bigcup\mathscr{U}=X\in\mathscr{T}$. For (iii) let $\mathscr{U}$ be an arbitrary family of open sets. If $X\in\mathscr{U}$, then $\bigcup\mathscr{U}=X\in\mathscr{T}$, and if not, then $p\notin U$ for each $U\in\mathscr{U}$, so $p\notin\bigcup\mathscr{U}\in\mathscr{T}$.
The special cases involving $X$ are even more obvious than the cases that you did cover, but they should still be dealt with.