GatorPower
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Homework Statement
Let p be a point in the rational numbers. Show that R (real numbers) - {p} is dense in R.
Show that the set of irrational numbers may be expressed as a countable intersection of open and dense subsets of R.
Homework Equations
Definitions of density,..
The Attempt at a Solution
On the first question I am not sure if what I'm thinking is sufficient..
R - {p} is a union of open sets (-inf, p) and (p, inf), and hence we need to close it. Closure of the set is the intersection of all closed sets containing the set, but the only closed set containing R - {p} is R itself, and hence the closure equals R. Then R - {p} is dense in R.
Second part. Let us write Irr = R - Q for the irrationals. We know that a open set R - {p|p in Q} is dense, so let us spin a bit on that... Rewriting the irrationals we get
Irr = R - Q = (R - q1) intersected with (R - {q| q in Q, q != q1}). Keep doing this until one gets Irr = intersection{R - {qi}} i from 0 to inf. Then you get the irrational numbers expressed as an intersection of dense and open subsets of R, and since Q is countable we only have so many rationals to choose, so the intersection is also countable.
Find any flaws in this? Got a better approach?
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