# The topology generated by open intervals of rational numbers

Mathematics Asked by roslavets on January 7, 2022

Let
$$B = { mathbb{R} } cup { (a,b) capmathbb {Q} , alt b , a,b inmathbb{Q}}$$

Thus, a set $$V in B$$ if it is either equal to $$mathbb{R}$$ or if it is in the intersection of $$mathbb{Q}$$ with an open interval with rational endpoints.

a) Show that $$B$$ forms a base for a topology $$T$$ on $$mathbb{R}$$.

b) Show that this topology does not contain nor be contained in the usual topology of $$mathbb{R}$$.

c) Is $$(mathbb{R}, T)$$ compact? Is it separable? Is it connected?

I think have solved a)

All irrational numbers are contained in the set $$mathbb{R}$$ and all the other numbers are contained in some intersection of an open set with $$mathbb{Q}$$. So each point $$x$$, is contained in some set $$v in B$$. Also if a point $$x$$ belongs to the intersection of two base sets then I can find some set containing $$x$$ that is contained in that intersection.

Let $$B = mathcal{I} cup mathcal{J}$$ where $$mathcal{I} = left{ mathbb{R} right}$$, $$mathcal{J} = left{ (a,b) cap mathbb{Q} | a. Suppose that $$U,V in B$$. If $$U,V in mathcal{I}$$, then clearly $$U cap V in mathcal{I}$$. If one of $$U,V$$ is in $$mathcal{I}$$ whereas the other is in $$mathcal{J}$$, then clearly $$U cap V in mathcal{J}$$. If both $$U$$ and $$V$$ are in $$mathcal{J}$$, then $$U cap V in mathcal{J}$$ from the basic properties of open intervals. Hence $$B$$ forms a basis.

Let $$mathcal{T}_B$$ be the topology generated by $$B$$ and $$mathcal{T}$$ be the usual topology. Since the interval $$(0,1) in mathcal{T}$$ contains irrational numbers $$(0,1) not in mathcal{T}_B$$. Therefore $$mathcal{T}_B$$ does not contains $$mathcal{T}$$. Since the interval $$(0,1) cap mathbb{Q} in mathcal{T}_B$$ is not open in the usual topology, we have $$mathcal{T}$$ does not contains $$mathcal{T}_B$$.

The space $$X = (mathbb{R}, mathcal{T}_B)$$ is compact. To see this let $$mathcal{O}$$ be an open cover of $$X$$. Since $$sqrt{2} in U in mathcal{O}$$ implies $$U = mathbb{R}$$, we have a finite subcover $$I subseteq mathcal{O}$$.

The space $$X$$ is separable. To see this, consider $$D = left{ 0 right}$$. For every irrational $$r$$, there is unique neighborhood $$mathbb{R}$$ showing that the closure of $$D$$ is just $$X$$. Hence $$X$$ is separable.

The space $$X$$ is connected. To see this, suppose that $$U$$ and $$V$$ are proper non empty open subsets of $$X$$. Then they cannot cover $$X$$ simply because they do not have any irrational numbers.

Answered by seoneo on January 7, 2022

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