Regard Q, the set of all rational numbers, as a metric space,

Click For Summary
SUMMARY

The discussion centers on the metric space of rational numbers, Q, defined by the metric d(p, q) = |p − q|. The set E, defined as all p ∈ Q such that 2 < p² < 3, is confirmed to be closed and bounded within Q, but it is not compact. The participants clarify that neighborhoods around points in the complement of E cannot contain irrational numbers, as they do not exist in the rational metric space. The conclusion emphasizes that the set defined by the inequalities 2 < x² < 3 and 2 ≤ x² ≤ 3 is equivalent in the context of rational numbers.

PREREQUISITES
  • Understanding of metric spaces and the definition of distance metrics.
  • Familiarity with the concepts of closed sets, bounded sets, and compactness in topology.
  • Knowledge of rational numbers and their properties.
  • Ability to analyze neighborhoods and limit points within a metric space.
NEXT STEPS
  • Study the properties of closed and bounded sets in metric spaces.
  • Learn about compactness in different types of spaces, particularly in Euclidean spaces.
  • Explore the implications of neighborhoods in metric spaces, focusing on rational numbers.
  • Review the definitions and examples of interior points and limit points in topology.
USEFUL FOR

Mathematics students, particularly those studying real analysis or topology, as well as educators looking to deepen their understanding of metric spaces and rational number properties.

Jamin2112
Messages
973
Reaction score
12

Homework Statement



Regard Q, the set of all rational numbers, as a metric space, with d(p, q) = |p − q|. Let E
be the set of all p ∈ Q such that 2 < p2 < 3. Show that E is closed and bounded in Q, but that E is not compact. Is E open in Q?

Homework Equations




Definition of interior point, limit point, open set, closed set, and compact set.


The Attempt at a Solution



Obviously E is bounded, but I'm not convinced that E is closed. Take a point in Ec. Supposedly there is a neighborhood, about that point, containing only members of E. But that can't be true, since any neighborhood about that point will also contain irrational numbers (Right?).
 
Physics news on Phys.org
NO. This is in the space of rational numbers. There are NO irrational numbers in that space. For the purposes of this problem, irrational numbers do not exist.

Notice that since there is no rational number whose square is 2 or 3, "for the purposes of this problem", again, the set of all x such that "2&lt; x^2&lt; 3" is exactly the same the set of all x such that "2\le x^2\le 3".
 
I suspected that might be the case. Now take a look, if you don't mind, at the last paragraph of the solution to this problem, found here: http://www.math.ucla.edu/~elewis/Math230PDFs/Math%20230a%20HW%203%20Extra%20Credit.pdf . It seems flawed to me, since they are putting using the given metric |p - q| and plugging irrational numbers into it, and it's only defined for rational numbers. nomsayin'?
 
Last edited by a moderator:

Similar threads

Exploring Open, Closed, Bounded, and Compact Sets in R with a Unique Metric
  • · Replies 1 ·
Replies
1
Views
2K
Is a Complete Subspace Necessarily Closed in a Metric Space?
  • · Replies 5 ·
Replies
5
Views
2K
Closure of Open Ball in Normed Vector Spaces
  • · Replies 7 ·
Replies
7
Views
2K
Convergence of sequence in metric space proof
  • · Replies 4 ·
Replies
4
Views
2K
Is Every Connected Metric Space Compact?
  • · Replies 1 ·
Replies
1
Views
2K
MHB Correctness of Statements about Set A in Metric Space X
  • · Replies 32 ·
2
Replies
32
Views
3K
Closed and Open Subsets of a Metric Space
  • · Replies 4 ·
Replies
4
Views
2K
Bounded Subsets of a Metric Space
  • · Replies 4 ·
Replies
4
Views
2K
Proof for convergent sequences, limits, and closed sets?
  • · Replies 11 ·
Replies
11
Views
3K
Proving Closedness of a Set in a Metric Space
  • · Replies 14 ·
Replies
14
Views
7K