mahler1
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1. Homework Statement .
Let A be a chain and B a partially ordered set. Now let f be an injective function from A to B and suppose that if a,b are elements of A and a≤b, then f(a)≤f(b). Prove that f(a)≤f(b) implies a≤b.
3. The Attempt at a Solution .
I want to check if this proof by contradiction is correct:
Let a,b be elements in A such that f(a)≤f(b). Since A is a chain, a and b must satisfy a≤b or a>b. Suppose a>b, then the statement a≥b is also true, so, by hypothesis, f(a)≥f(b). But then, f(a)<=f(b)<=f(a). B is a partially ordered set, so by the antisymmetric property of posets f(a)≤f(b) and f(a)≥f(b) imply f(a)=f(b). The function f is injective and, by definition of injectivity, f(a)=f(b) implies a=b, which contradicts the assumption a>b. Therefore, it must be a≤b.
Let A be a chain and B a partially ordered set. Now let f be an injective function from A to B and suppose that if a,b are elements of A and a≤b, then f(a)≤f(b). Prove that f(a)≤f(b) implies a≤b.
3. The Attempt at a Solution .
I want to check if this proof by contradiction is correct:
Let a,b be elements in A such that f(a)≤f(b). Since A is a chain, a and b must satisfy a≤b or a>b. Suppose a>b, then the statement a≥b is also true, so, by hypothesis, f(a)≥f(b). But then, f(a)<=f(b)<=f(a). B is a partially ordered set, so by the antisymmetric property of posets f(a)≤f(b) and f(a)≥f(b) imply f(a)=f(b). The function f is injective and, by definition of injectivity, f(a)=f(b) implies a=b, which contradicts the assumption a>b. Therefore, it must be a≤b.