Proof about ordering-preserving function
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
asked Nov 19 at 23:33
user14513462563
917
add a comment |
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
asked Nov 19 at 23:33
user14513462563
917
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
Nov 20 at 2:35
add a comment |
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
asked Nov 19 at 23:33
user14513462563
917
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
analysis functions upper-lower-bounds
asked Nov 19 at 23:33
user14513462563
917
asked Nov 19 at 23:33
user14513462563
917
asked Nov 19 at 23:33
user14513462563
917
asked Nov 19 at 23:33
user14513462563
917
asked Nov 19 at 23:33
user14513462563
917
917
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
Nov 20 at 2:35
add a comment |
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
Nov 20 at 2:35
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
Nov 20 at 2:35
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
Nov 20 at 2:35
add a comment |
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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
Nov 20 at 2:35