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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.

tnt235711
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    What do you mean by "$A$ generates a topology over $\mathbb{R}$"? – asdq Apr 13 '18 at 18:35
  • Do you mean it is a topology? – saulspatz Apr 13 '18 at 18:36
  • This is the second time on MSE that I've seen the terminology "topology over $X$''; perhaps it simply means "topology on $X$". I wonder if this terminology has appeared in someone's book or course. The other occurence is https://math.stackexchange.com/questions/1892818/topology-over-mathbb-n?rq=1 – Lee Mosher Apr 13 '18 at 18:39
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    @LeeMosher it might be a literal translation from, for example, spanish, where we say "sobre" (literally translated as "over"). – Javi Apr 13 '18 at 18:41
  • I think what I have to do, is to show that A is base for a topology $\tau$ on $\mathbb{R}$...because A is not a topology on $\mathbb{R}$. – tnt235711 Apr 13 '18 at 19:04
  • A subset of $\Bbb Z$ is bounded iff it is finite. If $Y$ is the set of all finite subsets of a set $X$ then we may let $Y\cup {X}$ be the set of all closed subsets of $X.$ This is called the co-finite topology on $X.$ – DanielWainfleet Apr 16 '18 at 07:06

2 Answers2

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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 $S=U_1\cap U_2$.

Just dropped in
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  • The meaning of "$A$ generates the topology $T$ on $B$" is that $T$ is the $\subset$-smallest topology on $B$ that has $A$ as a subset.... This is the usual type of meaning of "generates" in mathematics....Any collection of subsets of $B$ generates a toplogy on $B$....... And $A$ is not necessarily a base (basis) for $T$. The set ${(-\infty,x):x\in \Bbb R}\cup {(y,\infty):y\in \Bbb R}$ generates the standard topology on $\Bbb R.$ – DanielWainfleet Apr 16 '18 at 07:18
  • I am familiar with the concept of a sub-basis. Since any collection of subsets generates a topology, there is nothing to prove in this approach, maybe the only thing worth mentioning is that $A$ is closed under intersection. – Just dropped in Apr 16 '18 at 07:58
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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.

William Elliot
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