$\mathcal{T}= \{\emptyset, X, A\}$ is a topology. Let $X$ and $A\subset X$ a set and pruve that $\mathcal{T}= \{\emptyset, X, A\}$ is a topology.
I've got this:
[$(i)$]$\emptyset\in\mathcal{T}_m$ y $X\in\mathcal{T}_m$ for definition. but the other one?
 A: Recall that a topology on a set $X$ is a subset $\mathcal T\subseteq\mathscr P(X)$ satisfying


*

*$\varnothing\in\mathcal T$

*$X\in\mathcal T$

*$U,V\in\mathcal T,\ U\neq V\Rightarrow U\cap V\in\mathcal T$

*$\mathcal U\subseteq\mathcal T\Rightarrow\bigcup_{U\in\mathcal U}U\in\mathcal T$


The first two conditions are clearly satisfied in your case. For the third condition, note that there are three ways to choose two distinct elements of $\mathcal T$:


*

*$\varnothing,A\in\mathcal T$ and $\varnothing\cap A=\varnothing\in\mathcal T$

*$\varnothing, X\in\mathcal T$ and $\varnothing\cap X=\varnothing\in\mathcal T$

*$A, X\in\mathcal T$ and $A\cap X=A\in\mathcal T$


Hence the third condition is satisfied. Can you prove the fourth condition?
A: Key idea:
Let $X$ be a set.
Let $I$ be a well-ordered index set with a minimum $m$ and a maximum $M$.
Finally, let $U_i \; : \; i \in I$ be subsets of $X$ such that


*

*$U_m = X$

*$U_i \supset U_j$ for $i < j$ (the subsets are "decreasing")

*$U_M = \varnothing$
Then $\mathcal{T} = \{ U_i \}$ forms a topology on $X$.
Proof. $\varnothing, X \in \mathcal{T}$.  Finite intersections are in $\mathcal{T}$ because every finite subset of the totally ordered set $I$ has a maximum.  Arbitrary unions are in $\mathcal{T}$ because every subset of the well-ordered set $I$ has a minimum.

In your case, let $U_0 = X$, $U_1 = A$, $U_2 = \varnothing$.  Then $\{0,1,2\}$ is well-ordered in the usual way, and $U_i$ are a decreasing sequence of sets, as required.
