0

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:

\begin{align} \sup_{\alpha \in \mathcal{B} } {\Large[} \sup_{\alpha \in \mathcal{C}} f(\alpha) {\Large]} \end{align}

I was thinking that the first supremum would make it so that the function value is evaluated, and thus the outer supremum is $\sup_{\alpha \in \mathcal{B} } [c] = c$, 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 $\mathcal{B}$ & $\mathcal{C}$.

Thanks for the help!

Slade
  • 719
  • 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. – Paul Sinclair Mar 25 '19 at 03:35
  • 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 $\mathcal{C}$. The outer supremum is then attempting to calculate the supremum of the bracketed expression over $\alpha$ values in some other domain $\mathcal{B}$. I was wondering if this expression makes sense or if it's redundant since the bracketed expression is just a constant value. – Slade Mar 25 '19 at 15:22
  • 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 $\alpha$. – Paul Sinclair Mar 25 '19 at 23:45

1 Answers1

1

$\sup_{\alpha \in \mathcal C} f(\alpha)$ is by definition the least upper bound of the set $\{f(\alpha) | \alpha \in \mathcal C\}$. Note that

$$\{f(\alpha) | \alpha \in \mathcal C\} = \{f(\beta) | \beta \in \mathcal C\} = \{f(x) | x \in \mathcal C\} = \{f(\frak g) | \frak g \in \mathcal C\}$$

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 $f(\mathcal C)$:

$$f(\mathcal C) := \{f(\alpha) | \alpha \in \mathcal C\}$$

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 $\sup_{\alpha \in \mathcal B}\left[\sup_{\alpha \in \mathcal C} f(\alpha)\right] = \sup_{\alpha \in \mathcal B} \sup f(\mathcal C)$, which is, by definition $$\sup \{\sup f(\mathcal C) | \alpha \in \mathcal B\}$$ 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.

Paul Sinclair
  • 43,643