# Infimum and supremum of the empty set

Let $E$ be an empty set. Then, $\sup(E) = -\infty$ 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.

• What is the question?
– Did
Commented Jun 29, 2013 at 11:51
• @Did He's asking if the thought what he thought. Commented Jun 29, 2013 at 11:58
• It's because every real number is an upper bound/lower bound as its empty. Of course, in this definition we're using the extended reals.
– user70962
Commented Jun 29, 2013 at 11:59
• @Alexy Can you provide more context? Are you considering $E$ as a subset of a strictly larger poset? Is so, what poset is that? Where are you getting this from? I'm guessing it's a definition. It's not possible to prove a definition. Commented Jun 29, 2013 at 12:04
• @Git Gud: I read this from a textbook on Analysis some time ago( I can't remember the name). Can I see it this way: sup(E) = the least x such that x is greater than or equal to all elements in E. since E is empty, any x would be an upper bound for E, so just choose the least one; similarly, inf(E) = the greatest x such that x is smaller than or equal to all elements of E. Since E is empty, any x will be a lower bound for E, so just choose the greatest one. Commented Jun 29, 2013 at 12:18

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.

When you say "meaningful" you actually mean to say something "$\sup(E)$ exists". But what does it mean to exist? In the context of the real numbers it means that there exists a real number $r$ such that $r=\sup(E)$.

Of course $\pm\infty$ are not real numbers, but they still contain information for us. When we say that $\sup(E)=+\infty$ then we say that $E$ does not have a supermum in the real numbers, or equivalently for every $r\in\Bbb R$ we can find $e\in E$ such that $r<e$.

So in fact we can extend the definition of $\sup$ and $\inf$ to include $\pm\infty$ as possible values by adding two formal symbols for these "points at infinity".

Now, what was the definition of $\sup(E)$ anyway? It was the least point which is not smaller than any of the points in $E$. That is, $\sup(E)=x\iff\forall r\in E:r\leq x\land\forall z(\forall r\in E(r\leq z)\rightarrow x\leq z)$.

Plug this to the empty set, then $\forall r\in\varnothing$ is vacuously true, so every $r\in\Bbb R$ satisfies this condition. It follows that $-\infty$ also satisfies this condition. Therefore $\sup(\varnothing)=-\infty$.

If grasping this is somewhat difficult, here is an equivalent way of looking at this. Consider only $[0,1]$, and now define $\sup^*$ and $\inf^*$ as before but for subsets of $[0,1]$ and requiring that the $\sup^*$ and $\inf^*$ are from $[0,1]$ as well. What is $\sup^*(\varnothing)$?

• Thanks for your instructive comments! May I know how $-\infty$ satisfy this condition? Commented Jun 29, 2013 at 13:10
• Alexy, because every number does, so no real number is an upper bound. On the other hand, $-\infty$ is still vacuously larger than all the members of the empty set, but it is smaller than all the other values! Commented Jun 29, 2013 at 13:12
• Hi Asaf, since every real number satisfy this condition, this means every real number is sup(E).( Isn't this contradictory?) Hence, in order for sup(E) to be unique, we define sup(E) = $-\infty$ ? Commented Jun 29, 2013 at 13:23
• No, I meant to say that every number is an upper bound. The $\sup$ is the least possible value which is still an upper bound. Commented Jun 29, 2013 at 13:26

Let $\Bbb{R}$ be the set of real numbers with the usual order and let $\overline{\Bbb{R}}=\Bbb{R}\cup\{+\infty;-\infty\}$ with the following rules:$$-\infty< x< +\infty~~\forall x\in\Bbb{R}$$ Let $E=\emptyset$ be a subset of $\overline{\Bbb{R}}$. $U$ is an upper bound of $E$ if: $$\exists U\in\Bbb{R};~x\leq U~~~\forall x\in E$$ and $E$ doesn't have an upper bound in $\Bbb{R}$ if:$$\forall U\in\Bbb{R},~\exists x\in E;~~~x>U$$ which is clearly false since $E$ doesn't contain any element, so $E$ doesn't have an upper bound in $\Bbb{R}$. Let us define $\sup(E)=-\infty$ (you'll see why in a moment).

We have the relation $\sup(A\cup B)=\max\{\sup(A);\ \sup(B)\}$. If $B=\emptyset$, we expect $\sup(A\cup B)$ to be equal to $\sup(A)$ which is only verified if $\sup(\emptyset)=-\infty$

It seems you are aware of the following:

Fact: For any non-empty set of real numbers $E$ with an upper (lower) bound in $\Bbb R$, there is a least (greatest) upper (lower) bound. We call this $\sup E$ ($\inf E$).

Now, suppose we are given any set $E\subseteq R$. Define $U(E):=\{x\in\Bbb R:\forall y\in E(y\le x)\}$ to be the set of upper bounds of $E$ in $\Bbb R$. Similarly, let $L(E):=\{x\in\Bbb R:\forall y\in E(x\le y)\}$ be the set of lower bounds of $E$. You should be able to show the following:

• $U(E)$ is an upward-closed set--that is, if $x\in U(E)$ and $z\in\Bbb R$ with $x\le z,$ then $z\in U(E)$. Likewise, $L(E)$ is downward-closed.
• The only upward-closed subsets of $\Bbb R$ are $\emptyset,$ $\Bbb R,$ and the sets of the form $(a,\rightarrow):=\{x\in\Bbb R:a<x\}$ and $[a,\rightarrow):=\{x\in\Bbb R:a\le x\}$ for $a\in\Bbb R$. Similarly, the downward-closed subsets of $\Bbb R$ are $\emptyset,$ $\Bbb R,$ and the sets $(\leftarrow,b):=\{x\in\Bbb R:x<b\}$ and $(\leftarrow,b]:=\{x\in\Bbb R:x\le b\}.$
• $\Bbb R$ is the only upward-closed (downward-closed) subset of reals that is unbounded below (above).
• $E=\emptyset$ if and only if $U(E)=L(E)=\Bbb R.$

Now we can see the following:

Fact (rephrased): For any non-empty set of real numbers $E,$ either $U(E)$ is empty--meaning that $E$ has no upper bound in $\Bbb R$--or $U(E)$ has a least element--meaning that $U(E)=[a,\rightarrow)$ for some $a\in\Bbb R.$ This $a$ is uniquely determined by $E$, and we call it $\sup E$. Likewise, for any non-empty set of real numbers $E,$ either $L(E)$ is empty or $L(E)=(\leftarrow,b]$ for some unique $b\in\Bbb R$, which we call $\inf E$.

In the case that $E=\emptyset$, we have $U(E)=\Bbb R,$ which has no least element, so we can't say that there is any real number that we call $\sup E$ in this case. Accordingly, "$\sup E=-\infty$" should be interpreted (using my notation) to mean that $U(E)$ is unbounded below, which (by the discussion in the bullet points) is just another way of saying that $E$ is empty. Likewise, "$\inf E=+\infty$" means that $L(E)$ is unbounded above, just another way of saying that $E$ is empty.

Finally, note that if $E$ is non-empty and bounded above, then $\sup E=\inf U(E)$. In this light, we should interpret "$\sup E=+\infty$" in the same way as "$\inf U(E)=+\infty$," meaning $U(E)$ is empty (see previous paragraph), meaning that $E$ is unbounded above. Similarly, we should interpret "$\inf E=-\infty$" as saying that $E$ is unbounded below. These conventions are usually easier to grasp/swallow than that "$\sup E=-\infty$" and "$\inf E=+\infty$" both mean "$E=\emptyset$"--at least, that has been my experience--and I know you didn't directly ask about them, but I felt they were worth mentioning for elucidation.

If $$A\subset B\subset \mathbb R$$, then lower bound of $$B\leq$$ lower bound of $$A\leq$$ upper bound of $$B$$, and lower bound of $$B\leq$$ the upper bound of $$A\leq$$ upper bound of $$B$$.

Now, consider $$p\in \mathbb Z$$, where $$\mathbb Z$$ is the set of +ve and -ve integers. Since $$\emptyset \subset \{p\}$$, so lower bound of $$\{p\}\leq$$ lower bound of $$\emptyset\leq$$ upper bound of $$\{p\}$$, i.e. a lower bound of $$\emptyset$$ is $$p\in \mathbb Z$$. Hence, the greatest lower bound of $$\emptyset$$ is $$\infty$$.

Similarly, the least upper bound of $$\emptyset$$ is $$-\infty$$.

Notice that, $$\emptyset$$ has a nice/absurd property that it is a subset of any set, so it runs over the real line and so its upper and lower bound also run over the real line. While a nonempty set is fixed in its place, so it becomes unable to play such a game what $$\emptyset$$ plays.

We can prove it by contradiction. Let $$S = \varnothing$$ be the empty set and let $$S$$ does not have any upper bound i.e. there exists no real number $$M$$ such that $$M$$ is greater than or equal to every $$x \in S$$, which implies there exist at least one element $$x \in S$$ such that $$x > M$$, which is a contradiction since $$S$$ is an empty set and has no elements.

Similarly if $$S$$ does not have any lower bound then there exist no real number $$m$$ which is smaller than or equal to every element of $$S$$, implies there exist at least one element $$x \in S$$ such that $$x < m$$, again a contradiction since $$S$$ is an empty set and has no elements.

Hence our assumption is wrong and $$S$$ has both upper and lower bounds in fact every real number is an upper bound and every real number is a lower bound of an empty set $$S$$.

But an empty set has no least upper bound since we don't have any smallest real number, similarly an empty set has no greatest lower bound since there exists no greatest real number.