I was told that $\sup\emptyset=-\infty$ and $\inf\emptyset=\infty$, where $\emptyset$ is the empty set. This seems paradoxical to me as to how supremum can be less than infimum. Is there any proof for this or is it taken for granted ?

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    $\begingroup$ Agreed, it seems odd. However, if you think about how to form $\sup$ and $\inf$, it works. For the supremum, pick a real number with the property that there does not exist an element of the set that exceeds it. Since the set is empty, any real number will do. Now start pushing the number lower and lower until the condition is violated. Since there is no element of the set to violate the condition, you can keep pushing it lower and lower indefinitely--so the supremum is the "smallest" possible value, $-\infty$. Similar reasoning justifies that the infimum is $+\infty$. This is purely heuristic. $\endgroup$
    – MPW
    Feb 14, 2015 at 5:17
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    $\begingroup$ BTW, the symbol "\varnothing" $\varnothing$ looks better than "\emptyset" $\emptyset$, at least to me. $\endgroup$
    – MPW
    Feb 14, 2015 at 5:21
  • $\begingroup$ I suppose you are working with $\emptyset$ as a subset of $\mathbb R$. Maybe it would be worth mentioning this explicitly in the post. (The same question makes sense in a complete lattice, in arbitrary poset, ...) $\endgroup$ Feb 14, 2015 at 7:04
  • $\begingroup$ The other question states whether you can define supremum or infimum for empty sets or not. My question is how is sup $\emptyset=-\infty$ and inf $\emptyset=\infty$. How this became duplicate ? $\endgroup$
    – creative
    Feb 15, 2015 at 6:02

3 Answers 3


In the context of the extended real number line (i.e. real numbers and $\pm \infty$), this is true. It's not hard to prove - the supremum of a set is defined as the least upper bound of the set. Every number* is an upper bound to the empty set (since "upper bound" means "greater than or equal to every element"). So, the least upper bound is the least number - which is $-\infty$. This acts similarly for the infimum.

I agree that it is counterintuitive - it is, indeed, the only case where the supremum is less than the infimum. However, it does follow from definition. One way to think about it is that the supremum of a set $S$ is what we get if we take a point, and drag it down from $\infty$ until it cannot go lower without hitting $S$ and the infimum is what happens if we take a point and drag it up from $-\infty$ until it hits $S$. That is, we sort of imagine $S$ like an impassable block of stuff, and the supremum and infimum are clamped to the sides of it. But if there is no $S$, then there is no block, and as we clamp these points together, they just pass through each other and keep going - they always had motion inwards, but now nothing stops them, so they end up at $-\infty$ and $\infty$ respectively, as far as they can go.

(*"Number" in the sense of "element of the extended real line")


Since every real number $x$ is an upper bound for $\emptyset$, $x \ge \sup \emptyset$ for all $x\in \Bbb R$. Therefore $\sup \emptyset = -\infty$. Similar reasoning gives $\inf \emptyset = +\infty$.


If we look at the definition of supremum:

We say that $x$ is the supremum of a set $S$ if $x$ is the least upper bound of $S$. I.e., $x \geq s$ for all $s \in S$ and $x \leq y$ for any $y$ that is an upper bound on $S$.

So if we consider $\emptyset$, every $x\in \mathbb{R}$ is an upper bound of $\emptyset$. So the supremum of $\emptyset$ must be the $\min(\mathbb{R})$, which is often $-\infty$.

We can reason similarly for the infemum.


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