Intersection of Rationals and (0 to Infinity)?

[IntroAnalysis]

Homework Statement

Let A = [Q\bigcap(0,\infty)] \bigcup {-1} \bigcup(-3, -2]

Homework Equations

So A = (0,\infty) \bigcup{-1} \bigcup(-3,-2]

The Attempt at a Solution

I understand that the Rational numbers are cardinally equivalent to (0,\infty),

but why isn't the intersection of Rationals and (0,\infty) =>(0,\infty)\Irrationals ?

 

[sunjin09]

I understand that the Rational numbers are cardinally equivalent to (0,\infty),

Quite the contrary, the Irrationals are cardinally equivalent to (0,\infty)

 

[HallsofIvy]

The set of all rational numbers is countable, unlike [0, \infty).

 

[IntroAnalysis]

Then back to my original question why is the intersection of rationals and (0,∞) = (0,∞)

in other words, why don't irrationals come out of this intersection?

 

[clamtrox]

Then back to my original question why is the intersection of rationals and (0,∞) = (0,∞)

in other words, why don't irrationals come out of this intersection?

If x belongs to the intersection of A and B, then x belongs to A and x belongs to B. The intersection of rationals and (0,∞) is the set of numbers which are both rationals and positive real numbers.

 

[HallsofIvy]

Then back to my original question why is the intersection of rationals and (0,∞) = (0,∞)

in other words, why don't irrationals come out of this intersection?

It isn't. The intersection of the set of all rational numbers with all positive real numbers is all positive rational numbers.