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As a definition of " similarity mapping" I read in Lipschutz, Set Theory :

the mapping f from A to B ( A and B being ordered sets) is a similarity mapping iff, (a) f is a bijection and (b) for any elements a and a' belonging to A :

                   a < a'  iff f(a) < f(a')

The definition is expressed in terms of < : "strictly precedes".

Would f also be a similarity mapping in case < were replaced by " precedes or is equal to" ?

What I do not understand is that (1) the author has defined ordered sets in terms of " precedes or is equal to " and (2) defines an " order preserving function" in terms of " strictly precedes".

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Yes, the two notations are equivalent because $f$ is also required to be a bijection. This condition ensures that $f(a) = f(a') \iff a = a'$.

It follows that the strict and non-strict variants are equivalent.

As to the use of "strictly": one could argue it would have been better for the author to phrase the definition of "ordered set" in terms of "strictly precedes or is equal to" (which we would then abbreviate as "precedes").

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The given condition implies $a\le a'\implies f(a) \le f(a')$, but not the other way around, as any constant function is a counterexample.

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  • $\begingroup$ $f$ is also a bijection. $\endgroup$ – Lord_Farin May 4 at 12:31

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