Why is the empty set bounded?

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…

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Real Analysis

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