[Topology]Counting and density

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SUMMARY

The discussion focuses on proving that the set of real numbers excluding a point \( p \) in the rational numbers, denoted as \( \mathbb{R} - \{p\} \), is dense in \( \mathbb{R} \). The argument presented establishes that \( \mathbb{R} - \{p\} \) can be expressed as a union of open sets, specifically \( (-\infty, p) \) and \( (p, \infty) \), leading to the conclusion that its closure is \( \mathbb{R} \). Additionally, the set of irrational numbers is shown to be a countable intersection of open and dense subsets of \( \mathbb{R} \), formulated as \( \text{Irr} = \bigcap_{i=0}^{\infty} (\mathbb{R} - \{q_i\}) \), where \( q_i \) are the rational numbers.

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  • Understanding of real numbers and rational numbers
  • Familiarity with concepts of density in topology
  • Knowledge of open and closed sets in metric spaces
  • Basic understanding of countable sets and intersections
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  • Study the properties of dense subsets in topology
  • Learn about the closure of sets in metric spaces
  • Explore the concept of countable intersections in set theory
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Mathematics students, particularly those studying topology and real analysis, as well as educators seeking to deepen their understanding of density and set theory concepts.

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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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