Prove this is a topology in $\mathbb{Z}^+$

Question

Let $U\subset \mathbb{Z}^+$. Define $U$ to be open if it satisfies the condition: $n\in U\implies$ every divisor of $n$ belongs to $U$ Prove this is a topology in $\mathbb{Z}^+$

My attempt

Let $\tau=\{U_n\subset \mathbb{Z}^+ | n=x.q\,\,\, \forall x,q\in \mathbb{Z}^+\}$. We need prove $\tau$ is a topology, for this i need prove that:

1. $\bigcup_nU_n\in\tau$,
2. $ \bigcap_{i=1}^nU_i\in\tau$,
3. $ \mathbb{Z}^+,\emptyset\in\tau.$

1. Let $x\in\bigcup_nU_n$ then $x\in U_n$ and this implies for hypothesys for $n\in U_n$ exists $q$ such that $n=x.q$. As $x$ is arbitrary, then $ \bigcup_nU_n$ is a open set, and this implies $\bigcup_nU_n\in \tau$.

2. Let $x\in \bigcap_{i=1}^nU_i$ then $x\in U_i\forall i\in\mathbb{N}$ this implies for $n\in U_i $ exists $q$ such that $n=x.q$. Then, $\bigcap_{i=1}^nU_i\in \tau$.

I'm stuck trying to prove $\emptyset\in \tau$ and the other set.

Can someone help me with this exercise?

Answer 1

The sets $U_n=\{d:d|n\} $ form only a basis for the given topology, they are not closed under arbitrary union.
Alternatively, the empty set and $\Bbb Z^+$ clearly satisfy the original condition.

Answer 2

Firstly, a typical set $U$ as described in the question would be an element of:

$ \tau = \{ U \subset \mathbb{Z}^+ \mid$ if $x \in U$ and $y \in \mathbb{Z}^+$ such that $y \mid x$ then $y \in U$}

In this form, $\emptyset$ satisfies the condition trivially. Also the set $\mathbb{Z}^+$ contains every possible divisor of any element, so $\mathbb{Z}^+ \in \tau$ as well.

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As for the other conditions:

1. Let $\{ U_{\alpha} \} \subseteq \tau$. We would like to show that $\bigcup U_\alpha \in \tau$. Let $x \in \bigcup U_\alpha$. Then there exists $U_i \in \{ U_\alpha \}$ such that $x \in U_i$. As $x \in U_i$ and $U_i \in \tau$, if $y \in \mathbb{Z}^+$ and $y \mid x$ then $y \in U_i \subseteq \bigcup U_\alpha$. So $\bigcup U_\alpha \in \tau$
2. Let $\{ U_i \} \subseteq \tau$ be a finite subset of $\tau$ of order $n$. We would like to show that $\bigcap_n U_i \in \tau$. Let $x \in \bigcap_n U_i$. If $0 \leq i \leq n$, then $x \in U_i$ and $U_i \in \tau$ and hence all of the divisors of $x$ are in $U_i$. So all of the divisors of $x$ are in $\bigcap_n U_i$. So $\bigcap_n U_i \in \tau$

Answer 3

This answer provides a more general setting.

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Let $X$ be a set and let $f:X\to\wp(X)$ be a function.

Now sets $U\subseteq X$ can have the following property:$$\forall x[x\in U\implies f(x)\subseteq U]$$

and $\tau\subseteq\wp(X)$ can be defined as the collection of all subsets of $X$ that have this property.

It is obvious that $\varnothing$ and $X$ have this property so that $\varnothing,X\in\tau$.

Also it is obvious that $\tau$ is closed under intersections and unions.

So $\tau$ is a topology, special in the sense that it is closed even under arbitrary intersections.

This can be applied on $X=\mathbb Z^+$ and the function $f:\mathbb Z^+\to\wp(\mathbb Z^+)$ prescribed by $n\mapsto\{k\in\mathbb Z^+\mid k\text{ divides }n\}$