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Quick order preserving map question

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Let X and Y be ordered sets in the order topology.
    I want to show that a function f:X→Y is injective. We are given that f is surjective and preserves order.

    2. Relevant equations

    Definition of an order preserving map:

    If x≤y implies f(x)≤f(y)

    3. The attempt at a solution

    So if we assume that it is not injective then we are assuming that f(x)=f(y) and x[itex]\neq[/itex]y.

    Then either x>y or y<x. Assuming both, would clearly be a contradiction since it can only be one or the other in an ordered set.

    Now I need to show that x>y raises a contradiction and similarly x<y also raises a contradiction.


    My main question here is if x<y is it a contradiction if f(x)=f(y)?
    It seems like if order is preserved then if x<y then f(x)<f(y)
    but by the definition, it appears that if x<y then f(x)<f(y) or f(x)=f(y).

    Thank you for your time.
     
    Last edited: Dec 1, 2011
  2. jcsd
  3. Dec 1, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, you do. Use the fact that f is an order preserving map. If x< y, what can you say about f(x) and f(y)?


     
  4. Dec 1, 2011 #3
    Thank you, HallsofIvy. Actually, what you just said is exactly what my question is. My question was probably a little convoluted so let me restate it.

    I was trying to ask, if we have a function that preserves order
    and x<y

    1) do we conclude that f(x)<f(y)

    2)or do we conclude that f(x)≤f(y)

    The definition makes me think it is the second option but if I didn't have the definition I would have thought it was the first.
     
    Last edited: Dec 1, 2011
  5. Dec 1, 2011 #4
    Just to make it more clear what I am asking. If order is preserved does it mean

    if x<y implies f(x)<f(y)
    if x>y implies f(x)>f(y)
    if x=y imples f(x)=f(y)
    if x≤y imples f(x)≤f(y)
    if x≥y imples f(x)≥f(y)
     
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