Closure of the Rational Numbers (Using Standard Topology)

[Fluffman4]

Prove that Cl(Q) = R in the standard topology
I'm really stuck on this problem, seeing as we haven't covered limit points yet in the text and are not able to use them for this proof. Can anybody provide me with help needed for this proof? Many thanks.

 

[Dick]

You should state the definition of 'Cl(Q)' you are using. Is it the intersection of all closed sets containing Q? If so, think about what the complement of a closed set containing Q must be.

 

[Fluffman4]

We defined the closure of a subset A of a topological space as the intersection of all closed sets containing A. We basically want to find the smallest closed set containing A for closure.

 

[Dick]

We defined the closure of a subset A of a topological space as the intersection of all closed sets containing A. We basically want to find the smallest closed set containing A for closure.

Ok. So describe a closed set A in R containing Q. You might find it easier to describe the complement of A.

 

[HallsofIvy]

Crucial point: given any interval of real numbers, there exist both rational and irrational numbers in that interval.