Proof about ordering-preserving function











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Let $f: C_1 to C_2$ be an order preserving function. Assume that for $A subset C_1$ there exist $Sup(A) in C_1$ and $Sup(f(A)) in C_2$. Prove that $Sup(f(A)) leq f(Sup(A))$.



The statement seems obvious, but how would one go about proving this?










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  • Show (using the "order-preserving" assumption) that $f(sup(A))$ is an upper bound for $f(A)$. Then use that $sup(f(A))$ is the least upper bound of the same set $f(A)$.
    – Andreas Blass
    2 days ago

















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Let $f: C_1 to C_2$ be an order preserving function. Assume that for $A subset C_1$ there exist $Sup(A) in C_1$ and $Sup(f(A)) in C_2$. Prove that $Sup(f(A)) leq f(Sup(A))$.



The statement seems obvious, but how would one go about proving this?










share|cite|improve this question






















  • Show (using the "order-preserving" assumption) that $f(sup(A))$ is an upper bound for $f(A)$. Then use that $sup(f(A))$ is the least upper bound of the same set $f(A)$.
    – Andreas Blass
    2 days ago















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $f: C_1 to C_2$ be an order preserving function. Assume that for $A subset C_1$ there exist $Sup(A) in C_1$ and $Sup(f(A)) in C_2$. Prove that $Sup(f(A)) leq f(Sup(A))$.



The statement seems obvious, but how would one go about proving this?










share|cite|improve this question













Let $f: C_1 to C_2$ be an order preserving function. Assume that for $A subset C_1$ there exist $Sup(A) in C_1$ and $Sup(f(A)) in C_2$. Prove that $Sup(f(A)) leq f(Sup(A))$.



The statement seems obvious, but how would one go about proving this?







analysis functions upper-lower-bounds






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asked 2 days ago









user14513462563

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  • Show (using the "order-preserving" assumption) that $f(sup(A))$ is an upper bound for $f(A)$. Then use that $sup(f(A))$ is the least upper bound of the same set $f(A)$.
    – Andreas Blass
    2 days ago




















  • Show (using the "order-preserving" assumption) that $f(sup(A))$ is an upper bound for $f(A)$. Then use that $sup(f(A))$ is the least upper bound of the same set $f(A)$.
    – Andreas Blass
    2 days ago


















Show (using the "order-preserving" assumption) that $f(sup(A))$ is an upper bound for $f(A)$. Then use that $sup(f(A))$ is the least upper bound of the same set $f(A)$.
– Andreas Blass
2 days ago






Show (using the "order-preserving" assumption) that $f(sup(A))$ is an upper bound for $f(A)$. Then use that $sup(f(A))$ is the least upper bound of the same set $f(A)$.
– Andreas Blass
2 days ago

















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