1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof in relation with totally and partially ordered sets

  1. Aug 24, 2013 #1
    1. The problem statement, all variables and given/known data.

    Let A be a chain and B a partially ordered set. Now let f be an injective function from A to B and suppose that if a,b are elements of A and a≤b, then f(a)≤f(b). Prove that f(a)≤f(b) implies a≤b.

    3. The attempt at a solution.

    I want to check if this proof by contradiction is correct:

    Let a,b be elements in A such that f(a)≤f(b). Since A is a chain, a and b must satisfy a≤b or a>b. Suppose a>b, then the statement a≥b is also true, so, by hypothesis, f(a)≥f(b). But then, f(a)<=f(b)<=f(a). B is a partially ordered set, so by the antisymmetric property of posets f(a)≤f(b) and f(a)≥f(b) imply f(a)=f(b). The function f is injective and, by definition of injectivity, f(a)=f(b) implies a=b, which contradicts the assumption a>b. Therefore, it must be a≤b.
     
  2. jcsd
  3. Aug 25, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks good to me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Proof in relation with totally and partially ordered sets
  1. Partially ordered set (Replies: 4)

Loading...