Proving a topology

Question

Let \begin{equation}\Pi=\left\{\cdots ,\frac{1}{\pi^3},\frac{1}{\pi^2},\frac{1}{\pi},1,\pi,\pi^2,\pi^3,\cdots\right\}\end{equation} and $T$ be a collection of subsets of $\mathbb{R}$ consisting of $\emptyset,\mathbb{R}$ and those $U\subseteq\mathbb{R}$ such that the complement $\mathbb{R}\setminus U$ is contained inside $\Pi$. Prove that $T$ forms a topology on $\mathbb{R}$.

My attempt: The axioms for $T$ to be a topology on $X$ are:

(1) $\emptyset,X\in T$.

(2) for any $T'\subseteq T$ we have \begin{equation}\bigcup_{O\in T'}O\in T.\end{equation}

(3) for any $O,O'\in T$ we have $O\cap O'\in T$.

So our goal is to verify each axiom. Axiom (1) is immediate as it is given in the problem.

As for Axiom (3), Suppose $A,B\in T$. Then $\mathbb{R}\setminus A\in T$ and $\mathbb{R}\setminus B\in T$. I cannot see how to manipulate this to get what we want. I have no clue what to do for Axiom (2) and nothing jumps out to me as to how to approach it.

Answer 1

Actually if $Y$ is a (not empty) subset of the set $X$ then you can show that the collection $\mathcal T_Y$ given by $\emptyset$ and by any set $U$ such that $X\setminus U$ is in $Y$ is a topology. Indeed, if $U_1,U_2\in\mathcal T_Y$ then $$ X\setminus (U_1\cap U_2)=(X\setminus U_1)\cup(X\setminus U_2)\subseteq Y\cup Y=Y $$ which proves that $U_1\cap U_2\in\mathcal T_Y$; moreover if $U_i\in\mathcal T_Y$ for any $i\in I$ then $$ X\setminus\Big(\bigcup_{i\in I}U_i\Big)=\bigcap_{i\in I}X\setminus U_i\subseteq X\setminus U_j\subseteq Y $$ with $j\in I$ which proves that $\bigcup_{i\in I}U_i\in\mathcal T_Y$; finally, we observe that $$ X\setminus X=\emptyset\subseteq Y $$ which proves that $X\in\mathcal T_Y$. Moreover it is interesting to observe that the trivial topology $\{\emptyset, X\}$ is just $\mathcal T_\emptyset$.

Answer 2

Axiom (2). Let be $A,B \in T$. In consequence, one has $A^c = \mathbb{R} \setminus A \subset \Pi$ and $B^c = \mathbb{R} \setminus B \subset \Pi$ (which is not the same as $A^c,B^c \in T$ !), hence $A^c = \{\pi^k \,|\, k\in\mathscr{A}\}$ and $B^c = \{\pi^k \,|\, k\in\mathscr{B}\}$, with $\mathscr{A},\mathscr{B}\subset\mathbb{Z}$. Then, by de Morgan's law, we have $(A \cup B)^c = A^c \cap B^c = \{\pi^k \,|\, k\in\mathscr{A}\cap\mathscr{B}\} \subset \Pi$, since $\mathscr{A}\cap\mathscr{B}\subset\mathbb{Z}$, hence finally $A \cup B \in T$ and, by induction, $\displaystyle \bigcup_{A\in T'} A \in T \quad \forall T' \subset T$.

Axiom (3). Let $A,B$ be as before. Then, by the same type of argument, we have $(A \cap B)^c = A^c \cup B^c = \{\pi^k \,|\, k\in\mathscr{A}\cup\mathscr{B}\} \subset \Pi$, since $\mathscr{A}\cup\mathscr{B}\subset\mathbb{Z}$, hence $A \cap B \in T$.

Conclusion. As you said, the first axiom is obvious and thus we can conclude that $T$ forms a topology over $\mathbb{R}$. QED.