Let A lessthanorequalto a and B lessthanorequalto b be order

Let (A, lessthanorequalto a) and (B, lessthanorequalto b) be order-isomorphic partially ordered sets. Let y and x isinv A. Then y is the immediate successor (resp. immediate predecessor) of x in A if and only if f(y) is the immediate successor (resp. immediate predecessor) of f(x) in B. Let

Solution

What properties f possess? Is f order isomorphism, i.,e ., f is bijective and order preserving? In that this is true, Here s the proof:

Suppose, y is immediate successor of x in A, that means:

x< y and there is no z in A such that x<z<y.

We want to prove, f(y) is immediate successor of f(x).

Since f is order preserving so x<y implies f(x) < f(y)

and if there exist b in B such that f(x) < b < f(y) then we get a contradiction.

Indeed, since f is bijective so there exist c in A such that f(c) = b, we get, f(x) < f(c) < f(y), taking inverse we get,

x<c<y, a contradiction, since y is immediate successor of f(x).

Thus, if y is immediate successor of x then f(y) is immediate successor of f(x).

Conversely,

Suppose, f(y) is immediate successor of f(x) then to prove y is immediate successor of x.

That also follows, same as above.

Since, f(x) < f(y),so taking inverse of f we get, x< y , ( This should happen since f is order preserving).

Also, there does not exist any m in A such that x < m < y, since this will give us,

f(x) < f(m)<f(y), since f is order preserving,, which is a contradiction, as f(y) is immediate successor of f(x).

Hence, y is immediate successor of x IFF f(y) is immediate successor of f(x).

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Exactly same proof work for immediate predecessor. Just replace the role of x and y. Give a try, its exactly the same, in above proof just replace successor by predecessor and you are done.

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