(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Quick order preserving map question

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