Supremum of the image of a monotonic function
1
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For a monotonic function $f:Dtomathbb{R}$ where $Dsubsetmathbb{R}$, and two sets $A$ and $B$ such that $sup A=sup B$, is it true that $sup f(A)=sup f(B)$, where $f(A)$ denotes the image of set $A$ under function $f$? If so, how to prove it, assuming $f(sup A)$ and $f(sup B)$ may be undefined?
real-analysis supremum-and-infimum
asked Jan 17 at 13:12
Yutong ZhangYutong Zhang
587
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add a comment |
1
$begingroup$
For a monotonic function $f:Dtomathbb{R}$ where $Dsubsetmathbb{R}$, and two sets $A$ and $B$ such that $sup A=sup B$, is it true that $sup f(A)=sup f(B)$, where $f(A)$ denotes the image of set $A$ under function $f$? If so, how to prove it, assuming $f(sup A)$ and $f(sup B)$ may be undefined?
real-analysis supremum-and-infimum
asked Jan 17 at 13:12
Yutong ZhangYutong Zhang
587
$endgroup$
add a comment |
1
1
1
$begingroup$
For a monotonic function $f:Dtomathbb{R}$ where $Dsubsetmathbb{R}$, and two sets $A$ and $B$ such that $sup A=sup B$, is it true that $sup f(A)=sup f(B)$, where $f(A)$ denotes the image of set $A$ under function $f$? If so, how to prove it, assuming $f(sup A)$ and $f(sup B)$ may be undefined?
real-analysis supremum-and-infimum
asked Jan 17 at 13:12
Yutong ZhangYutong Zhang
587
$endgroup$
For a monotonic function $f:Dtomathbb{R}$ where $Dsubsetmathbb{R}$, and two sets $A$ and $B$ such that $sup A=sup B$, is it true that $sup f(A)=sup f(B)$, where $f(A)$ denotes the image of set $A$ under function $f$? If so, how to prove it, assuming $f(sup A)$ and $f(sup B)$ may be undefined?
real-analysis supremum-and-infimum
real-analysis supremum-and-infimum
asked Jan 17 at 13:12
Yutong ZhangYutong Zhang
587
asked Jan 17 at 13:12
Yutong ZhangYutong Zhang
587
asked Jan 17 at 13:12
Yutong ZhangYutong Zhang
587
asked Jan 17 at 13:12
Yutong ZhangYutong Zhang
587
asked Jan 17 at 13:12
Yutong ZhangYutong Zhang
587
587
add a comment |
add a comment |
1 Answer
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No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?
answered Jan 17 at 13:26
Ravi SatputeRavi Satpute
267
$endgroup$
$begingroup$
What if $f$ is monotonically increasing?
$endgroup$
– Yutong Zhang
Jan 17 at 13:30
1
$begingroup$
@YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
$endgroup$
– user3482749
Jan 17 at 13:33
$begingroup$
For proving something we need proof. And to disprove things even counter example is sufficient.
$endgroup$
– Ravi Satpute
Jan 17 at 13:34
$begingroup$
But I'll write for increasing as well....
$endgroup$
– Ravi Satpute
Jan 17 at 13:35
1
$begingroup$
Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
$endgroup$
– Ravi Satpute
Jan 17 at 14:03
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?
answered Jan 17 at 13:26
Ravi SatputeRavi Satpute
267
$endgroup$
$begingroup$
What if $f$ is monotonically increasing?
$endgroup$
– Yutong Zhang
Jan 17 at 13:30
1
$begingroup$
@YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
$endgroup$
– user3482749
Jan 17 at 13:33
$begingroup$
For proving something we need proof. And to disprove things even counter example is sufficient.
$endgroup$
– Ravi Satpute
Jan 17 at 13:34
$begingroup$
But I'll write for increasing as well....
$endgroup$
– Ravi Satpute
Jan 17 at 13:35
1
$begingroup$
Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
$endgroup$
– Ravi Satpute
Jan 17 at 14:03
|
show 1 more comment
1
$begingroup$
No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?
answered Jan 17 at 13:26
Ravi SatputeRavi Satpute
267
$endgroup$
$begingroup$
What if $f$ is monotonically increasing?
$endgroup$
– Yutong Zhang
Jan 17 at 13:30
1
$begingroup$
@YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
$endgroup$
– user3482749
Jan 17 at 13:33
$begingroup$
For proving something we need proof. And to disprove things even counter example is sufficient.
$endgroup$
– Ravi Satpute
Jan 17 at 13:34
$begingroup$
But I'll write for increasing as well....
$endgroup$
– Ravi Satpute
Jan 17 at 13:35
1
$begingroup$
Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
$endgroup$
– Ravi Satpute
Jan 17 at 14:03
|
show 1 more comment
1
1
1
$begingroup$
No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?
answered Jan 17 at 13:26
Ravi SatputeRavi Satpute
267
$endgroup$
No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?
answered Jan 17 at 13:26
Ravi SatputeRavi Satpute
267
answered Jan 17 at 13:26
Ravi SatputeRavi Satpute
267
answered Jan 17 at 13:26
Ravi SatputeRavi Satpute
267
answered Jan 17 at 13:26
Ravi SatputeRavi Satpute
267
267
$begingroup$
What if $f$ is monotonically increasing?
$endgroup$
– Yutong Zhang
Jan 17 at 13:30
1
$begingroup$
@YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
$endgroup$
– user3482749
Jan 17 at 13:33
$begingroup$
For proving something we need proof. And to disprove things even counter example is sufficient.
$endgroup$
– Ravi Satpute
Jan 17 at 13:34
$begingroup$
But I'll write for increasing as well....
$endgroup$
– Ravi Satpute
Jan 17 at 13:35
1
$begingroup$
Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
$endgroup$
– Ravi Satpute
Jan 17 at 14:03
|
show 1 more comment
$begingroup$
What if $f$ is monotonically increasing?
$endgroup$
– Yutong Zhang
Jan 17 at 13:30
1
$begingroup$
@YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
$endgroup$
– user3482749
Jan 17 at 13:33
$begingroup$
For proving something we need proof. And to disprove things even counter example is sufficient.
$endgroup$
– Ravi Satpute
Jan 17 at 13:34
$begingroup$
But I'll write for increasing as well....
$endgroup$
– Ravi Satpute
Jan 17 at 13:35
1
$begingroup$
Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
$endgroup$
– Ravi Satpute
Jan 17 at 14:03
$begingroup$
What if $f$ is monotonically increasing?
$endgroup$
– Yutong Zhang
Jan 17 at 13:30
$begingroup$
What if $f$ is monotonically increasing?
$endgroup$
– Yutong Zhang
Jan 17 at 13:30
1
1
$begingroup$
@YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
$endgroup$
– user3482749
Jan 17 at 13:33
$begingroup$
@YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
$endgroup$
– user3482749
Jan 17 at 13:33
$begingroup$
For proving something we need proof. And to disprove things even counter example is sufficient.
$endgroup$
– Ravi Satpute
Jan 17 at 13:34
$begingroup$
For proving something we need proof. And to disprove things even counter example is sufficient.
$endgroup$
– Ravi Satpute
Jan 17 at 13:34
$begingroup$
But I'll write for increasing as well....
$endgroup$
– Ravi Satpute
Jan 17 at 13:35
$begingroup$
But I'll write for increasing as well....
$endgroup$
– Ravi Satpute
Jan 17 at 13:35
1
1
$begingroup$
Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
$endgroup$
– Ravi Satpute
Jan 17 at 14:03
$begingroup$
Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
$endgroup$
– Ravi Satpute
Jan 17 at 14:03
|
show 1 more comment
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