Quick order preserving map question

[EV33]

Homework Statement

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.

Homework Equations

Definition of an order preserving map:

If x≤y implies f(x)≤f(y)

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\neqy.

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.

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

Definition of an order preserving map:

If x≤y implies f(x)≤f(y)

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\neqy.

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.

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)?

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).

What, exactly, is that definition? "if x< y the f(x)\le f(y)?

Thank you for your time.

 

[EV33]

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.

 

[EV33]

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)