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could you answer my question?

Let $A \subset \mathbb{R}$ be a subset of the real numbers and $f$ a continuous functions. What happens the function $\sup_{t \in A} |f(t)|$ if $A$ is chosen to be the empty set?

I feel like it is undefined but can anyone confirm? Thank you!

PROB123
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  • I think the supremum of the empty set is sometimes taken to be $-\infty$ so that you get $\sup( A \cup B) = \max( \sup(A), \sup(B))$ – Asinomás May 18 '21 at 14:50
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    One common definition is $\sup\emptyset=-\infty$. A maximum obviously doesn't exist or make sense. – Benjamin May 18 '21 at 14:50
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    @Onir The same holds for other rules, such as $\sup(A\cap B) =\min{\sup A,\sup B}$. – Benjamin May 18 '21 at 14:52

1 Answers1

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Consider a non-empty subset $M \subset \mathbb R$. If $M$ is bounded above, then its supremum $\sup M$ is defined as the least upper bound of $M$. This is the unique number $\alpha \in \mathbb R$ with the following two properties:

  1. $\alpha$ is an upper bound for $M$.

  2. No $\beta < \alpha$ is an upper bound for $M$.

If $M \subset \mathbb R$ is unbounded above (which automatically implies that $M \ne \emptyset$), then one defines $\sup M = +\infty$.

$\sup \emptyset$ has not yet been defined. What would be adequate in this case?

Clearly each $\beta \in \mathbb R$ is an upper bound for $\emptyset$, simply because $x \le \beta$ for all $x \in \emptyset$. In other words,the set of upper bounds of $\emptyset$ is $\mathbb R$. This set does not have a least element and therefore one defines $$\sup \emptyset = -\infty .$$ Note that no $\alpha \in \mathbb R$ has the above properties 1. and 2.

Paul Frost
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