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Topic: Order embedding
Replies: 8   Last Post: Sep 16, 2013 10:01 PM

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William Elliot

Posts: 1,551
Registered: 1/8/12
Re: Order embedding
Posted: Sep 16, 2013 10:01 PM
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On Mon, 16 Sep 2013, William Elliot wrote:

I continue the summary of basic order map properties
to include order maps with linear domains.

> Let X,Y be (partially) ordered sets. Are these definitions correct?
>
> f:X -> Y is order preserving when
> for all x,y, (x <= y implies f(x) <= f(y).
>
> f:X -> Y is an order embedding when
> for all x,y, (x <= y iff f(x) <= f(y)).
>
> f:X -> Y is an order isomorphism when f is surjective
> and for all x,y, (x <= y iff f(x) <= f(y)).
>
> The following are immediate consequences.
>
> Order embedding maps and order isomorphisms are injections.
> If f:X -> Y is an order embedding,
> then f:X -> f(X) is an order isomorphism.
>
> Furthermore the composition of two order preserving, order
> embedding or order isomorphic maps is again resp., order
> preserving, order embedding or order isomorphic.
>
> Finally, the inverse of an order isomorphism is an order isomorphism.
>
> That all is the basics of order maps, is it not?
> Or is the more to be included?


Well now that we all agree upon the definitions including the easy to prove
facts, as would be expected, that order embeddings and order isomorphisms,
are injections. There are a couple of additional basic properties which
are easy to prove.

If X is a linear order, Y an order and f:X -> Y is an
order preserving injection, then f is an order embedding.

If X is a linear order, Y an order and f:X -> Y is an
order preserving bijection, then f is an order isomorphism.

Problem. Show by counterexample, that for those two propositions,
the linearity of X is a necessary premise for the conclusions.

Thusly in general, order embeddings and order isomorphisms cannot
be defined as injective and bijective order preserving maps.





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