Studying topology on $\mathbb{Q}$ with an example

Question

I am studying mathematics on my own and today I am learning topology of rationals.

I want to analyze the topology through some examples.

Let $X=\mathbb{Q}$ be the set of rational numbers and let $d$ be standard Euclidean metric on $\mathbb{Q}$

Example 1: $A=\{r\in\mathbb{Q}: r^2<2\}$

Is $A$ closed?

A is closed in $\mathbb{Q}$ as $A=\mathbb{Q}\cap [-\sqrt{2},\sqrt{2}]$

Is $A$ open?

$A$ is open in $\mathbb{Q}$ as $A=\mathbb{Q}\cap(-\sqrt{2},\sqrt{2})$

Is $A$ compact and connected?

I am stuck here. I know the definitions but I am not able to proceed.

We say a topological space $X$ is connected if if it can be written as the disjoint union of two non empty open sets. A topological space $X$ is connected if for every open cover of $X$ there is a finite sub cover. @Arthur — Shweta Aggrawal, Sep 09 '18 at 08:00
I think your definition of "connected" is turned around. That's the definition of a disconnected space. — Arthur, Sep 09 '18 at 08:06
Q is totally disconnected: only the singeltons are connected subsets — , Sep 09 '18 at 08:22

score 2 · Answer 1 · answered Sep 09 '18 at 08:20

2

Not compact. Consider the open cover
{ $\mathbb{Q}\cap(-\sqrt{2},r)$: r in A }

Not connected. Let r be an irrational in $(-\sqrt{2},\sqrt{2})$ and consider $(-\sqrt{2},r), (r,\sqrt{2})$ both intersected by Q.

answered Sep 09 '18 at 08:20

William Elliot

drhab · Accepted Answer · 2018-09-10T09:21:08.200

2

Setting $A_s=\{r\in\mathbb Q\mid r^2<s\}=\mathbb Q\cap(-\sqrt s,\sqrt s)$ it is evident that the sets $A_s$ are open for $0<s<2$ and cover $A$.

Can you find a finite subcover?

We have $A=\{r\in\mathbb Q\mid r^2<\frac{1}{2}\}\cup \{r\in\mathbb Q\mid \frac{1}{2}<r^2<2\}$ since $\{r\in\mathbb Q\mid r^2=\frac12\}=\varnothing$.

The sets are open, not empty and disjoint.

edited Sep 10 '18 at 09:21

answered Sep 09 '18 at 08:23

drhab

2 Answers2