Infimum and supremum of the empty set

Let E be an empty set. Then, sup and \inf(E)=+\infty. I thought it is only meaningful to talk about \inf(E) and \sup(E) if E is non-empty and bounded?

Thank you.

Answer

There might be different preferences to how one should define this. I am not sure that I understand exactly what you are asking, but maybe the following can be helpful.

If we consider subsets of the real numbers, then it is customary to define the infimum of the empty set as being \infty. This makes sense since the infimum is the greatest lower bound and every real number is a lower bound. So \infty could be thought of as the greatest such.

The supremum of the empty set is -\infty. Again this makes sense since the supremum is the least upper bound. Any real number is an upper bound, so -\infty would be the least.

Note that when talking about supremum and infimum, one has to start with a partially ordered set (P, \leq). That (P, \leq) is partial ordered means that \leq is reflexive, antisymmetric, and transitive.

So let P = \mathbb{R} \cup \{-\infty, \infty\}. Define \leq the “obvious way”, so that a\leq \infty for all a\in \mathbb{R} and -\infty \leq a for all a\in \mathbb{R}.

With this definition you have a partial order and it this setup the infimum and the supremum are as mentioned above. So you don’t need non-empty.

Attribution
Source : Link , Question Author : Alexy Vincenzo , Answer Author : Thomas

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