Show that $A = \{U\subset \mathbb{R}: \mathbb{R}-U$ is a bounded subset of $\mathbb{Z}\}$ generates a topology on $\mathbb{R}$.

Question

A is topology over $\mathbb{R}$ if:

1) $\emptyset$ and $\mathbb{R}$ belong to A. 2) Any union of sets in A, belongs to A. 3) Any finite intersection of sets in A, belongs to A.

but when I try to show that $\emptyset \in A$. I have to show that ${\emptyset}^\complement = \mathbb{R}$ is a bounded subset of $\mathbb{Z}$. I do not know what I am doing wrong.

Answer 1

Maybe "generates a topology" means it is a basis for a topology? We say that a class $\mathcal{B}$ is a basis for a topology $\mathcal{\tau}$ on a set $X$ when

1)$\bigcup\mathcal{B}=X$ and

2) For each $B_1, B_2 \in \mathcal{B}$ and for each $x\in B_1\cap B_2$ there exists a $B_3\in\mathcal{B}$ such that $x\in B_3\subset B_1\cap B_2$

In that sense, $A$ generates a topology on $\mathbb{R}$ indeed:

First, you have $\mathbb{R}\in A$, since $\mathbb{R}\setminus\mathbb{R}=\emptyset$ is a bounded subset of $\mathbb{Z}$, so $\bigcup A=\mathbb{R}$. Second, If $U_1, U_2\in A$ then $\mathbb{R}\setminus U_1$ and $\mathbb{R}\setminus U_2$ are both bounded subsets of $\mathbb{Z}$, from which one can conclude that $\mathbb{R}\setminus (U_1\cap U_2)$ is a bounded subset of $\mathbb{Z}$ (use De Morgan's law). If $x\in U_1\cap U_2$, then there does indeed exist a set $S$ in $A$ such that $x\in S\subset U_1\cap U_2$: simply put $S=U_1\cap U_2$.

Answer 2

Any collection of subsets A, generates a topology,
the smallest topology containg A. Is A a topology?

A bounded subset of Z only has a finite number of integers.
For R - U to be a bounded subset of Z, U has to be
(R-Z) $\cup$ K for some cofinite subset K, of Z.

Now if the empty set is added to A. A is a topology and the
proof of that is similar to showing all cofinite subsets of a
set (empty set added) is a topology for that set.