Supremum of Supremum over same variable The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Switching order of supremum for doubly indexed sequence?Supremum over a sequence with two indicesWhy is the expectation of essential supremum equal the supremum of expectationsInfimum and Supremum the sameDouble supremum over a function of two variablesApproximating supremum over countable set by suprema over finite setsTwo different supremum, are they the same?Convergence of a sequence over supremumSupremum of expected value over equivalent measuresessential supremum and supremum for stochastic processes.Supremum of a polynomial over reals
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Supremum of Supremum over same variable
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Switching order of supremum for doubly indexed sequence?Supremum over a sequence with two indicesWhy is the expectation of essential supremum equal the supremum of expectationsInfimum and Supremum the sameDouble supremum over a function of two variablesApproximating supremum over countable set by suprema over finite setsTwo different supremum, are they the same?Convergence of a sequence over supremumSupremum of expected value over equivalent measuresessential supremum and supremum for stochastic processes.Supremum of a polynomial over reals
$begingroup$
I am wondering if when taking the supremum twice of a function over the same variable, if the outer supremum is redundant/has no effect. I know in the case of two variables/indices that the nested supremum is meaningful (Switching order of supremum for doubly indexed sequence?), but I can't tell how to interpret this properly:
beginalign
sup_alpha in mathcalB Large[ sup_alpha in mathcalC f(alpha) Large]
endalign
I was thinking that the first supremum would make it so that the function value is evaluated, and thus the outer supremum is $sup_alpha in mathcalB [c] = c$, where $c$ is the constant that results from the evaluating the first supremum.
If the above is incorrect, then I'm wondering how the supremums could be manipulated, such as switching their orders or possibly leading to the supremum of the union of $mathcalB$ & $mathcalC$.
Thanks for the help!
real-analysis stochastic-processes
asked Mar 24 at 18:46
SladeSlade
83111
$endgroup$
add a comment |
$begingroup$
I am wondering if when taking the supremum twice of a function over the same variable, if the outer supremum is redundant/has no effect. I know in the case of two variables/indices that the nested supremum is meaningful (Switching order of supremum for doubly indexed sequence?), but I can't tell how to interpret this properly:
beginalign
sup_alpha in mathcalB Large[ sup_alpha in mathcalC f(alpha) Large]
endalign
I was thinking that the first supremum would make it so that the function value is evaluated, and thus the outer supremum is $sup_alpha in mathcalB [c] = c$, where $c$ is the constant that results from the evaluating the first supremum.
If the above is incorrect, then I'm wondering how the supremums could be manipulated, such as switching their orders or possibly leading to the supremum of the union of $mathcalB$ & $mathcalC$.
Thanks for the help!
real-analysis stochastic-processes
asked Mar 24 at 18:46
SladeSlade
83111
$endgroup$
$begingroup$
in $sup_alpha in mathcal C f(alpha), alpha$ is a dummy variable. The expression does not depend upon a value of $alpha$ at all. Thus when you take the outside supremum, that $alpha$ is a separate variable than the first, and you are just taking the supremum of a constant, just as you said. Your whole confusion here is another example why it is a very bad idea to use the same letter to represent two different variables in the same context.
$endgroup$
– Paul Sinclair
Mar 25 at 3:35
$begingroup$
Thanks for your comment! I think my question may be confusing. The function is a function of $alpha$, and the inner supremum is calculating the max/sup of the function over $alpha$ values in a particular domain $mathcalC$. The outer supremum is then attempting to calculate the supremum of the bracketed expression over $alpha$ values in some other domain $mathcalB$. I was wondering if this expression makes sense or if it's redundant since the bracketed expression is just a constant value.
$endgroup$
– Slade
Mar 25 at 15:22
$begingroup$
No, I understood your question perfectly, but apparently you didn't understand my answer . No, the expression does not make sense, Yes, taking the second supremum does not change anything (though "redundant" is not an accurate description - "pointless" would be better). The reason the expression does not make sense is that it misleads you into thinking you are taking the supremum twice over the same variable, when in fact, you are not. $sup_alpha in mathcal C f(alpha)$ does not in any way depend on a variable named $alpha$.
$endgroup$
– Paul Sinclair
Mar 25 at 23:45
add a comment |
$begingroup$
I am wondering if when taking the supremum twice of a function over the same variable, if the outer supremum is redundant/has no effect. I know in the case of two variables/indices that the nested supremum is meaningful (Switching order of supremum for doubly indexed sequence?), but I can't tell how to interpret this properly:
beginalign
sup_alpha in mathcalB Large[ sup_alpha in mathcalC f(alpha) Large]
endalign
I was thinking that the first supremum would make it so that the function value is evaluated, and thus the outer supremum is $sup_alpha in mathcalB [c] = c$, where $c$ is the constant that results from the evaluating the first supremum.
If the above is incorrect, then I'm wondering how the supremums could be manipulated, such as switching their orders or possibly leading to the supremum of the union of $mathcalB$ & $mathcalC$.
Thanks for the help!
real-analysis stochastic-processes
asked Mar 24 at 18:46
SladeSlade
83111
$endgroup$
I am wondering if when taking the supremum twice of a function over the same variable, if the outer supremum is redundant/has no effect. I know in the case of two variables/indices that the nested supremum is meaningful (Switching order of supremum for doubly indexed sequence?), but I can't tell how to interpret this properly:
beginalign
sup_alpha in mathcalB Large[ sup_alpha in mathcalC f(alpha) Large]
endalign
I was thinking that the first supremum would make it so that the function value is evaluated, and thus the outer supremum is $sup_alpha in mathcalB [c] = c$, where $c$ is the constant that results from the evaluating the first supremum.
If the above is incorrect, then I'm wondering how the supremums could be manipulated, such as switching their orders or possibly leading to the supremum of the union of $mathcalB$ & $mathcalC$.
Thanks for the help!
real-analysis stochastic-processes
real-analysis stochastic-processes
asked Mar 24 at 18:46
SladeSlade
83111
asked Mar 24 at 18:46
SladeSlade
83111
asked Mar 24 at 18:46
SladeSlade
83111
asked Mar 24 at 18:46
SladeSlade
83111
asked Mar 24 at 18:46
SladeSlade
83111
83111
$begingroup$
in $sup_alpha in mathcal C f(alpha), alpha$ is a dummy variable. The expression does not depend upon a value of $alpha$ at all. Thus when you take the outside supremum, that $alpha$ is a separate variable than the first, and you are just taking the supremum of a constant, just as you said. Your whole confusion here is another example why it is a very bad idea to use the same letter to represent two different variables in the same context.
$endgroup$
– Paul Sinclair
Mar 25 at 3:35
$begingroup$
Thanks for your comment! I think my question may be confusing. The function is a function of $alpha$, and the inner supremum is calculating the max/sup of the function over $alpha$ values in a particular domain $mathcalC$. The outer supremum is then attempting to calculate the supremum of the bracketed expression over $alpha$ values in some other domain $mathcalB$. I was wondering if this expression makes sense or if it's redundant since the bracketed expression is just a constant value.
$endgroup$
– Slade
Mar 25 at 15:22
$begingroup$
No, I understood your question perfectly, but apparently you didn't understand my answer . No, the expression does not make sense, Yes, taking the second supremum does not change anything (though "redundant" is not an accurate description - "pointless" would be better). The reason the expression does not make sense is that it misleads you into thinking you are taking the supremum twice over the same variable, when in fact, you are not. $sup_alpha in mathcal C f(alpha)$ does not in any way depend on a variable named $alpha$.
$endgroup$
– Paul Sinclair
Mar 25 at 23:45
add a comment |
$begingroup$
in $sup_alpha in mathcal C f(alpha), alpha$ is a dummy variable. The expression does not depend upon a value of $alpha$ at all. Thus when you take the outside supremum, that $alpha$ is a separate variable than the first, and you are just taking the supremum of a constant, just as you said. Your whole confusion here is another example why it is a very bad idea to use the same letter to represent two different variables in the same context.
$endgroup$
– Paul Sinclair
Mar 25 at 3:35
$begingroup$
Thanks for your comment! I think my question may be confusing. The function is a function of $alpha$, and the inner supremum is calculating the max/sup of the function over $alpha$ values in a particular domain $mathcalC$. The outer supremum is then attempting to calculate the supremum of the bracketed expression over $alpha$ values in some other domain $mathcalB$. I was wondering if this expression makes sense or if it's redundant since the bracketed expression is just a constant value.
$endgroup$
– Slade
Mar 25 at 15:22
$begingroup$
No, I understood your question perfectly, but apparently you didn't understand my answer . No, the expression does not make sense, Yes, taking the second supremum does not change anything (though "redundant" is not an accurate description - "pointless" would be better). The reason the expression does not make sense is that it misleads you into thinking you are taking the supremum twice over the same variable, when in fact, you are not. $sup_alpha in mathcal C f(alpha)$ does not in any way depend on a variable named $alpha$.
$endgroup$
– Paul Sinclair
Mar 25 at 23:45
$begingroup$
in $sup_alpha in mathcal C f(alpha), alpha$ is a dummy variable. The expression does not depend upon a value of $alpha$ at all. Thus when you take the outside supremum, that $alpha$ is a separate variable than the first, and you are just taking the supremum of a constant, just as you said. Your whole confusion here is another example why it is a very bad idea to use the same letter to represent two different variables in the same context.
$endgroup$
– Paul Sinclair
Mar 25 at 3:35
$begingroup$
in $sup_alpha in mathcal C f(alpha), alpha$ is a dummy variable. The expression does not depend upon a value of $alpha$ at all. Thus when you take the outside supremum, that $alpha$ is a separate variable than the first, and you are just taking the supremum of a constant, just as you said. Your whole confusion here is another example why it is a very bad idea to use the same letter to represent two different variables in the same context.
$endgroup$
– Paul Sinclair
Mar 25 at 3:35
$begingroup$
Thanks for your comment! I think my question may be confusing. The function is a function of $alpha$, and the inner supremum is calculating the max/sup of the function over $alpha$ values in a particular domain $mathcalC$. The outer supremum is then attempting to calculate the supremum of the bracketed expression over $alpha$ values in some other domain $mathcalB$. I was wondering if this expression makes sense or if it's redundant since the bracketed expression is just a constant value.
$endgroup$
– Slade
Mar 25 at 15:22
$begingroup$
Thanks for your comment! I think my question may be confusing. The function is a function of $alpha$, and the inner supremum is calculating the max/sup of the function over $alpha$ values in a particular domain $mathcalC$. The outer supremum is then attempting to calculate the supremum of the bracketed expression over $alpha$ values in some other domain $mathcalB$. I was wondering if this expression makes sense or if it's redundant since the bracketed expression is just a constant value.
$endgroup$
– Slade
Mar 25 at 15:22
$begingroup$
No, I understood your question perfectly, but apparently you didn't understand my answer . No, the expression does not make sense, Yes, taking the second supremum does not change anything (though "redundant" is not an accurate description - "pointless" would be better). The reason the expression does not make sense is that it misleads you into thinking you are taking the supremum twice over the same variable, when in fact, you are not. $sup_alpha in mathcal C f(alpha)$ does not in any way depend on a variable named $alpha$.
$endgroup$
– Paul Sinclair
Mar 25 at 23:45
$begingroup$
No, I understood your question perfectly, but apparently you didn't understand my answer . No, the expression does not make sense, Yes, taking the second supremum does not change anything (though "redundant" is not an accurate description - "pointless" would be better). The reason the expression does not make sense is that it misleads you into thinking you are taking the supremum twice over the same variable, when in fact, you are not. $sup_alpha in mathcal C f(alpha)$ does not in any way depend on a variable named $alpha$.
$endgroup$
– Paul Sinclair
Mar 25 at 23:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$sup_alpha in mathcal C f(alpha)$ is by definition the least upper bound of the set $f(alpha) $. Note that
$$f(alpha) = f(beta) = x in mathcal C = f(frak g) $$
The variable used the set builder notation is not something the set depends on. It is only there to assist us in understanding the set definition. So it doesn't matter what we call it. That variable only exists within the set brackets. Outside, there is no variable. There is just some set. That is what a dummy variable is: a variable used within some expression that has no definition outside of that expression.
That set is usually denoted by $f(mathcal C)$:
$$f(mathcal C) := f(alpha) $$
Now the expression "$f(mathcal C)$" doesn't even meantion $alpha$, which is fine, because it does not depend on $alpha$ in any way.
And $sup_alpha in mathcal C f(alpha) = sup,f(mathcal C)$. The right side explicitly does not depend on $alpha$.
So, in the expression $sup_alpha in mathcal Bleft[sup_alpha in mathcal C f(alpha)right] = sup_alpha in mathcal B sup f(mathcal C)$, which is, by definition $$sup sup f(mathcal C) $$
Since $sup f(mathcal C)$ is independent of $alpha$, the set above has only one element, and therefore the supremum is that single element again.
You have been thinking of $sup_alpha in mathcal B sup_alpha in mathcal C f(alpha)$ as taking two supremums over the same variable. This is false, as the outside supremum does not even see any variable. The definition of the variable $alpha$ occurring in $f(alpha)$ ends with the inside supremum. All the outside supremum sees is a single value that does not depend on its own (separate) variable $alpha$ at all.
answered Mar 26 at 0:15
Paul SinclairPaul Sinclair
20.8k21543
$endgroup$
$begingroup$
Thanks a lot for the detailed answer. I get what you were saying in the comments now!
$endgroup$
– Slade
Mar 26 at 15:31
add a comment |
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$begingroup$
$sup_alpha in mathcal C f(alpha)$ is by definition the least upper bound of the set $f(alpha) $. Note that
$$f(alpha) = f(beta) = x in mathcal C = f(frak g) $$
The variable used the set builder notation is not something the set depends on. It is only there to assist us in understanding the set definition. So it doesn't matter what we call it. That variable only exists within the set brackets. Outside, there is no variable. There is just some set. That is what a dummy variable is: a variable used within some expression that has no definition outside of that expression.
That set is usually denoted by $f(mathcal C)$:
$$f(mathcal C) := f(alpha) $$
Now the expression "$f(mathcal C)$" doesn't even meantion $alpha$, which is fine, because it does not depend on $alpha$ in any way.
And $sup_alpha in mathcal C f(alpha) = sup,f(mathcal C)$. The right side explicitly does not depend on $alpha$.
So, in the expression $sup_alpha in mathcal Bleft[sup_alpha in mathcal C f(alpha)right] = sup_alpha in mathcal B sup f(mathcal C)$, which is, by definition $$sup sup f(mathcal C) $$
Since $sup f(mathcal C)$ is independent of $alpha$, the set above has only one element, and therefore the supremum is that single element again.
You have been thinking of $sup_alpha in mathcal B sup_alpha in mathcal C f(alpha)$ as taking two supremums over the same variable. This is false, as the outside supremum does not even see any variable. The definition of the variable $alpha$ occurring in $f(alpha)$ ends with the inside supremum. All the outside supremum sees is a single value that does not depend on its own (separate) variable $alpha$ at all.
answered Mar 26 at 0:15
Paul SinclairPaul Sinclair
20.8k21543
$endgroup$
$begingroup$
Thanks a lot for the detailed answer. I get what you were saying in the comments now!
$endgroup$
– Slade
Mar 26 at 15:31
add a comment |
$begingroup$
$sup_alpha in mathcal C f(alpha)$ is by definition the least upper bound of the set $f(alpha) $. Note that
$$f(alpha) = f(beta) = x in mathcal C = f(frak g) $$
The variable used the set builder notation is not something the set depends on. It is only there to assist us in understanding the set definition. So it doesn't matter what we call it. That variable only exists within the set brackets. Outside, there is no variable. There is just some set. That is what a dummy variable is: a variable used within some expression that has no definition outside of that expression.
That set is usually denoted by $f(mathcal C)$:
$$f(mathcal C) := f(alpha) $$
Now the expression "$f(mathcal C)$" doesn't even meantion $alpha$, which is fine, because it does not depend on $alpha$ in any way.
And $sup_alpha in mathcal C f(alpha) = sup,f(mathcal C)$. The right side explicitly does not depend on $alpha$.
So, in the expression $sup_alpha in mathcal Bleft[sup_alpha in mathcal C f(alpha)right] = sup_alpha in mathcal B sup f(mathcal C)$, which is, by definition $$sup sup f(mathcal C) $$
Since $sup f(mathcal C)$ is independent of $alpha$, the set above has only one element, and therefore the supremum is that single element again.
You have been thinking of $sup_alpha in mathcal B sup_alpha in mathcal C f(alpha)$ as taking two supremums over the same variable. This is false, as the outside supremum does not even see any variable. The definition of the variable $alpha$ occurring in $f(alpha)$ ends with the inside supremum. All the outside supremum sees is a single value that does not depend on its own (separate) variable $alpha$ at all.
answered Mar 26 at 0:15
Paul SinclairPaul Sinclair
20.8k21543
$endgroup$
$begingroup$
Thanks a lot for the detailed answer. I get what you were saying in the comments now!
$endgroup$
– Slade
Mar 26 at 15:31
add a comment |
$begingroup$
$sup_alpha in mathcal C f(alpha)$ is by definition the least upper bound of the set $f(alpha) $. Note that
$$f(alpha) = f(beta) = x in mathcal C = f(frak g) $$
The variable used the set builder notation is not something the set depends on. It is only there to assist us in understanding the set definition. So it doesn't matter what we call it. That variable only exists within the set brackets. Outside, there is no variable. There is just some set. That is what a dummy variable is: a variable used within some expression that has no definition outside of that expression.
That set is usually denoted by $f(mathcal C)$:
$$f(mathcal C) := f(alpha) $$
Now the expression "$f(mathcal C)$" doesn't even meantion $alpha$, which is fine, because it does not depend on $alpha$ in any way.
And $sup_alpha in mathcal C f(alpha) = sup,f(mathcal C)$. The right side explicitly does not depend on $alpha$.
So, in the expression $sup_alpha in mathcal Bleft[sup_alpha in mathcal C f(alpha)right] = sup_alpha in mathcal B sup f(mathcal C)$, which is, by definition $$sup sup f(mathcal C) $$
Since $sup f(mathcal C)$ is independent of $alpha$, the set above has only one element, and therefore the supremum is that single element again.
You have been thinking of $sup_alpha in mathcal B sup_alpha in mathcal C f(alpha)$ as taking two supremums over the same variable. This is false, as the outside supremum does not even see any variable. The definition of the variable $alpha$ occurring in $f(alpha)$ ends with the inside supremum. All the outside supremum sees is a single value that does not depend on its own (separate) variable $alpha$ at all.
answered Mar 26 at 0:15
Paul SinclairPaul Sinclair
20.8k21543
$endgroup$
$sup_alpha in mathcal C f(alpha)$ is by definition the least upper bound of the set $f(alpha) $. Note that
$$f(alpha) = f(beta) = x in mathcal C = f(frak g) $$
The variable used the set builder notation is not something the set depends on. It is only there to assist us in understanding the set definition. So it doesn't matter what we call it. That variable only exists within the set brackets. Outside, there is no variable. There is just some set. That is what a dummy variable is: a variable used within some expression that has no definition outside of that expression.
That set is usually denoted by $f(mathcal C)$:
$$f(mathcal C) := f(alpha) $$
Now the expression "$f(mathcal C)$" doesn't even meantion $alpha$, which is fine, because it does not depend on $alpha$ in any way.
And $sup_alpha in mathcal C f(alpha) = sup,f(mathcal C)$. The right side explicitly does not depend on $alpha$.
So, in the expression $sup_alpha in mathcal Bleft[sup_alpha in mathcal C f(alpha)right] = sup_alpha in mathcal B sup f(mathcal C)$, which is, by definition $$sup sup f(mathcal C) $$
Since $sup f(mathcal C)$ is independent of $alpha$, the set above has only one element, and therefore the supremum is that single element again.
You have been thinking of $sup_alpha in mathcal B sup_alpha in mathcal C f(alpha)$ as taking two supremums over the same variable. This is false, as the outside supremum does not even see any variable. The definition of the variable $alpha$ occurring in $f(alpha)$ ends with the inside supremum. All the outside supremum sees is a single value that does not depend on its own (separate) variable $alpha$ at all.
answered Mar 26 at 0:15
Paul SinclairPaul Sinclair
20.8k21543
answered Mar 26 at 0:15
Paul SinclairPaul Sinclair
20.8k21543
answered Mar 26 at 0:15
Paul SinclairPaul Sinclair
20.8k21543
answered Mar 26 at 0:15
Paul SinclairPaul Sinclair
20.8k21543
20.8k21543
$begingroup$
Thanks a lot for the detailed answer. I get what you were saying in the comments now!
$endgroup$
– Slade
Mar 26 at 15:31
add a comment |
$begingroup$
Thanks a lot for the detailed answer. I get what you were saying in the comments now!
$endgroup$
– Slade
Mar 26 at 15:31
$begingroup$
Thanks a lot for the detailed answer. I get what you were saying in the comments now!
$endgroup$
– Slade
Mar 26 at 15:31
$begingroup$
Thanks a lot for the detailed answer. I get what you were saying in the comments now!
$endgroup$
– Slade
Mar 26 at 15:31
add a comment |
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$begingroup$
in $sup_alpha in mathcal C f(alpha), alpha$ is a dummy variable. The expression does not depend upon a value of $alpha$ at all. Thus when you take the outside supremum, that $alpha$ is a separate variable than the first, and you are just taking the supremum of a constant, just as you said. Your whole confusion here is another example why it is a very bad idea to use the same letter to represent two different variables in the same context.
$endgroup$
– Paul Sinclair
Mar 25 at 3:35
$begingroup$
Thanks for your comment! I think my question may be confusing. The function is a function of $alpha$, and the inner supremum is calculating the max/sup of the function over $alpha$ values in a particular domain $mathcalC$. The outer supremum is then attempting to calculate the supremum of the bracketed expression over $alpha$ values in some other domain $mathcalB$. I was wondering if this expression makes sense or if it's redundant since the bracketed expression is just a constant value.
$endgroup$
– Slade
Mar 25 at 15:22
$begingroup$
No, I understood your question perfectly, but apparently you didn't understand my answer . No, the expression does not make sense, Yes, taking the second supremum does not change anything (though "redundant" is not an accurate description - "pointless" would be better). The reason the expression does not make sense is that it misleads you into thinking you are taking the supremum twice over the same variable, when in fact, you are not. $sup_alpha in mathcal C f(alpha)$ does not in any way depend on a variable named $alpha$.
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– Paul Sinclair
Mar 25 at 23:45