# Why is the empty set bounded?

Why is the empty set bounded below and bounded above? If it has no elements, how can you say that an upper or lower bound exists?

• Every number $b$ is both, an upper and a lower bound. For there is no $x \in \varnothing$ such that $x < b$ or $x > b$. Oct 19, 2013 at 16:25
• Note that $17$ is an upper bound. For it is true that every element of the empty set is $\lt 17$. Can you name one that isn't? Oct 19, 2013 at 16:25
• This is an instance of something being vacuously true. Oct 19, 2013 at 16:25
• Definition: a subset $S$ of $\mathbb{R}$ is blue if there exists $s\in S$ such that $17s+\sqrt{3}$ is rational. Proposition: the empty set is not blue. Oct 19, 2013 at 16:34

Recall that implication has the property that when the assumption is false, the implication is true. In other words, if $P$ is false, then $P\implies Q$ is true.

Let $S$ be a set of real numbers. Then $M$ is an upper bound for $S$ if the following implication holds, $$s\in S\implies s \leq M.$$ Now let us examine the case for the empty set, $\emptyset$.

Proposition Let $M$ be any real number. Then $M$ is an upper bound for the empty set (of real numbers).

Proof:

Since the statement, $s\in\emptyset$ is false, the implication, $$s\in\emptyset\implies s \leq M$$ is true.

Note: An almost identical proof works for the lower bound case. A nice little slogan to remember here is

All things are true about the MEMBERS of the empty set.

Argue by contradiction. Suppose $\emptyset$ is unbounded. Then for every $M > 0$ there is a point $x \in \emptyset$ such that $|x| > M$. But this contradicts that the empty set has no elements.

Alternatively, a set $S$ is bounded if there exist numbers $a$ and $b$ such that: $$S\subseteq[a,b]$$ Now, $\varnothing$ is a subset of every set, so…

A set in a metric space is bounded if, and only if, there exists a ball (of finite radius) containing it. Then trivially the empty set is contained in a ball (actually in every ball).

In mathematical analysis and related areas of mathematics, a set is called bounded, if it is, in a certain sense, of finite size. Conversely, a set which is not bounded is called unbounded. - Wikipedia definition.

Now there is only one "nothing" just like there is only one 1, or only one 2, or say only one zero. There may be 2 cats or 2 dogs or 2 giraffes but 2 is 2. there is only one 2 along the entire number-line. similarly there are only one "nothing".

so the empty set, the content (nothing) is finite in number, which is 1. so empty set should be finite.

please correct me i may make mistake im not math people.

• The quote you provide says that a set is bounded "if it is, in a certain sense, of finite size." This does not mean that the set is of finite cardinality. This is not the "certain sense" alluded to in the quote. Sep 1, 2019 at 18:55

Note: this is an informal explanation and not a rigorous proof. It's just meant to strengthen your intuition (I hope!)

It might help to think of upper and lower bounds as limits on the size of a set's elements. Sets are either unbounded or bounded.

Let's look first at an unbounded set, $$A$$. If $$A$$ is unbounded, its elements get arbitrarily large. Large positive and "large negative". They have no upper limit to their absolute value. They "go off to infinity, and to negative infinity."

Bounded sets are sets that don't do this.

A set $$S$$ has an upper bound $$b$$ if no element of $$S$$ is greater than $$b$$. $$S$$ doesn't "go off to infinity". It doesn't even get past $$b$$

The empty set certainly doesn't "go off to infinity or negative infinity". It doesn't have elements greater than $$b$$, no matter what number $$b$$ is. It doesn't have elements less than $$b$$ either.

Every number is both an upper and lower bound of the empty set, a size restriction so impossibly restrictive that no real numbers can obey.

The empty set is so tightly bound up, so tightly restricted, it doesn't even have any elements!

We can prove it by contradiction. Let S=∅ 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∈S, which implies there exist at least one element x∈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∈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.

• This is a really bad way of looking at it. The supremum is $-\infty$; this means that any extended real number is an upper bound. Similarly the infimum is $+\infty$, which means that any extended real number is a lower bound. No other subset of $\mathbb{R}$ has either of these properties. In this sense the empty set is "more bounded" than any other set.