Is there anything wrong with how the supremum of a set is written?

[Eclair_de_XII]

TL;DR

Let $A\subsetℝ$ be bounded from above. Let $f:A\rightarrow ℝ$. Then can I write the supremum of the image of $A$ through $f$ as $\sup f(A)$, or do I have to write it as $\sup\{f(x):x\in A\}$?

I'm just having random thoughts today, and I didn't know where to put this, since this isn't even a homework problem.

Anyway, is my way of writing the supremum of a set correct syntax-wise, or no?

 

[member 587159]

Both are the same thing: simply because $f(A) = \{f(a)\mid a \in A\}$ by definition. That A is bounded above does not imply that f(A) is bounded above, so the supremum may not exist as a real number.

 

[hilbert2]

I think having a continuous $f$ and compact $A$ would guarantee the existence of a supremum (probably even maximum).

 

[member 587159]

I think having a continuous $f$ and compact $A$ would guarantee the existence of a supremum (probably even maximum).

Yes, that's work. You are right that it is even a maximum because the image of a compact set under a continuous map is compact, so the image is in particular closed and bounded hence contains its supremum.

What we need here is the requirement that f is bounded above, but this is just a reformulation of f(A) is bounded above.

 

[Eclair_de_XII]

That A is bounded above does not imply that f(A) is bounded above

Oh, I should have caught that. My mista;e; thanks for the reply, besides.

 

[hilbert2]

A situation where that happens is the function (defined on a half-open interval)

$f:]0,1]\mapsto \mathbb{R}$ such that $f(x)=1/x$

which gets arbitrarily large values when approaching $x=0$.

If the domain is compact (closed and bounded), this can't happen because in conventional calculus you can't have a function that is defined as "infinite" at an endpoint of the domain.