Understanding Double Quantifiers and Sets with Epsilon

  • Thread starter Thread starter eclayj
  • Start date Start date
  • Tags Tags
    Sets
Click For Summary
SUMMARY

The discussion focuses on the evaluation of sets A, B, C, and D against the proposition p: "for all ε > 0, ∃ x ∈ S such that x < ε." Set A, defined as {1/n : n ∈ Z+}, satisfies the proposition as its elements approach 0. Set B, defined as {n : n ∈ Z+}, does not satisfy the proposition since all its elements are greater than or equal to 1. Set C, being the union of A and B, satisfies the proposition due to the inclusion of set A. Set D, containing only the element -1, trivially satisfies the proposition as it is less than any positive ε.

PREREQUISITES
  • Understanding of double quantifiers in mathematical logic
  • Familiarity with the concept of limits in calculus
  • Knowledge of set theory and union of sets
  • Basic understanding of positive and negative integers
NEXT STEPS
  • Study the formal definition of limits in calculus
  • Learn about the properties of sets and their unions
  • Explore mathematical logic, focusing on quantifiers and their implications
  • Review examples of propositions involving ε and their applications in analysis
USEFUL FOR

Students of mathematics, particularly those studying calculus and set theory, as well as educators seeking to clarify concepts related to limits and quantifiers.

eclayj
Messages
20
Reaction score
0

Homework Statement



Determine if the sets A, B, C, and D satisfy the following proposition (p) for the set S:

Homework Equations



p: for all ε > 0, ∃ x \in S such that x < ε


A = {1/n : n \in Z+}
B = {n : n ε Z+}
C = A \cup B
D = {-1}


The Attempt at a Solution



I am looking for a little help in reading/interpreting the mathematical statements with double quantifiers and "ε," I have always had terrible trouble understanding what these mean (I have never quite understood the formal definition of a limit for example).

Let me attempt to explain how I understand proposition p, for example, and how it would relate to set A, B, C, and D. Then please tell me if I am off the mark.

For p, after I read it a few times I interpreted it as a condition requiring the set to contain at least one element, x, that is less than some number ε, where that number ε can be made arbitrarily close to 0. The only set I could visualize that would allow for this is a set that contains elements which get arbitrarily close to 0 (if the set does not contain negative numbers), or a set which contains negative numbers.

A satisfies this condition because the members of its set approach 0 as n approaches infinity. So, no matter what epsilon you choose, you can always find a smaller x value in set A

B does not satisfy this condition, because you can choose 0 < ε < 1, but the elements of this set are restricted to whole numbers, and therefore all elements in set B \geq 1

A \cup B satisfy p b/c it includes set A, the elements of which approach 0 as n approaches infinity (i.e., get infinitely close to 0).

Finally set D trivially satisfies condition p b/c condition p restricts the choice of ε > 0, and the only element of set D is < 0
 
Physics news on Phys.org
eclayj said:

Homework Statement



Determine if the sets A, B, C, and D satisfy the following proposition (p) for the set S:

Homework Equations



p: for all ε > 0, ∃ x \in S such that x < ε


A = {1/n : n \in Z+}
B = {n : n ε Z+}
C = A \cup B
D = {-1}


The Attempt at a Solution



I am looking for a little help in reading/interpreting the mathematical statements with double quantifiers and "ε," I have always had terrible trouble understanding what these mean (I have never quite understood the formal definition of a limit for example).

Let me attempt to explain how I understand proposition p, for example, and how it would relate to set A, B, C, and D. Then please tell me if I am off the mark.

For p, after I read it a few times I interpreted it as a condition requiring the set to contain at least one element, x, that is less than some number ε, where that number ε can be made arbitrarily close to 0. The only set I could visualize that would allow for this is a set that contains elements which get arbitrarily close to 0 (if the set does not contain negative numbers), or a set which contains negative numbers.

A satisfies this condition because the members of its set approach 0 as n approaches infinity. So, no matter what epsilon you choose, you can always find a smaller x value in set A

B does not satisfy this condition, because you can choose 0 < ε < 1, but the elements of this set are restricted to whole numbers, and therefore all elements in set B \geq 1

A \cup B satisfy p b/c it includes set A, the elements of which approach 0 as n approaches infinity (i.e., get infinitely close to 0).

Finally set D trivially satisfies condition p b/c condition p restricts the choice of ε > 0, and the only element of set D is < 0

That sounds just fine.
 

Similar threads

Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K