inf e sup of empty set?

Question

If $S$ is an ordered set then the empty set is a subset of $S$. What are $\inf$ and $\sup$ of such set? To be honest I don't know what it could be. I'm not talking about of real numbers, but any ordered set. The definitions I have been given hold for any ordered set, but I don't know how to use such definitions to give an answer.

Answer 1

Every $x\in S$ is vacuously an upper bound for $\varnothing$: there is no $y\in\varnothing$ such that $x\not\le y$ (since there is no $x\in\varnothing$ at all). Thus, $\varnothing$ has a least upper bound if and only if $S$ has a minimum element: $\sup\varnothing=\min S$ if that exists, and otherwise $\sup\varnothing$ does not exist.

Similarly, if $x\in S$, there is no $y\in\varnothing$ such that $x\not\le y$ (since there is no $y\in\varnothing$ at all), so it’s vacuously true $x$ is a lower bound for $\varnothing$. Thus, $\varnothing$ has a greatest lower bound if and only if $S$ has a maximum element: $\inf\varnothing=\max S$ if that exists, and otherwise $\inf\varnothing$ does not exist.

Answer 2

If you append the elements $-\infty$ and $+\infty$ to your ordered set (if it does not already have them) to be such that $-\infty < x$ and $x<+\infty$ for all $x\in S$, then what you have is that $\inf\emptyset = +\infty$ and $\sup\emptyset = -\infty.$ The way to see this is that the $\inf$ is the largest lower bound. Every element in $S$ is a lower bound for $\emptyset$ vacuously. Similar logic applies in the case of $\sup$. It is very counter-intuitive at first but makes sense when phrased in the above manner.

Answer 3

In general, a subset of an ordered set may not have an infimum and/or supremum. So it's entirely possible that the empty set, in particular, doesn't.

By unfolding the definition we see that $\inf \varnothing$ should be the largest element of $S$ if a largest element exists and otherwise is not defined.

Similarly, $\sup\varnothing$ is the smallest element of $S$, if it exists.

Answer 4

The terms $\inf$ and $\sup$ define concepts which, like other kinds of limit, may not always exist.

For example, consider the real numbers $\mathbb{R}$ with the usual $\leq$ order. The lower bound $\inf \{x \in \mathbb{R} : x < 0\}$ does not exist, because there is no $a\in \mathbb{R}$ which satisfies the greatest lower bound property.

Sometimes for convenience, we append the "infinite" elements $-\infty$ and $\infty$ to our sets. You can introduce such infinite elements into any ordered set, not just the real numbers— they are elements defined by the property that $-\infty \leq x \leq \infty $ for each $x\in S$. They are the extreme minimum and extreme maximum elements of the set, respectively.

Whenever $S$ has such extreme elements, you will have by definition that:

$$\inf \varnothing = +\infty$$

$$\sup \varnothing = -\infty$$

Here is one intuitive explanation for why this should be true: when you add an element to a set, its infimum should always decrease or remain the same. As such, it makes sense for the infimum of the empty set to be the maximum possible value. Similarly, the supremum of the empty set should be the minimum possible value.

Answer 5

First off, if $S$ doesn't have the least-upper-bound property there is no reason to assume any bounded subset (which the empty set certainly is) has inf or sup.

So I'll assume $S$ has the least upper bound property. I scramble to page 4 of Rudin's Principals of Mathematical Analysis and look up the exact definition of least upper bound property. It sees "If $E\subset S, E$ is not empty...."

So empty set is exempt in the definition. Probably because of the obvious resulting conflicts: Although the empty is bounded above (for any $x$ then $x \ge y$ for all $y \in \emptyset$) but clearly there is clearly no $x$ such that $y < x$ means $y$ is not an upper bound as all $y$ would be upper bounds. (Unless S has a minimum element). So the empty set can not have a least upper bound (on sets that do not have a mimimum element-- in those cases, the least upper bound of the empty set would, paradoxically, be the minimum value of S[#]). To to make the "least upper bound property" meaningful we must make an exception to the empty set.

tl;dr -- the empty set is an exception and doesn't need to have a $\sup$ or $\inf$.

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[#]-- If $a = \min S$ then $ a \ge y \forall y \in \emptyset$ so $a$ is upperbound of $\emptyset$. If $y < a$ is a false hypothesis so it implies any conclusion. So $y < a\implies y$ is not an upper bound of $\emptyset$. (It also implies $y$ is an upper bound. And $y$ both is and is not a trumpet playing green elephant eating cheetohs.)

So $a = \sup \emptyset$.

Which is .... kinda dumb. But in the end it all works out....