# I heard that the empty set $\emptyset$ is bounded. But I think this statement is not correct.

In Rudin's "Principles of Mathematical Analysis", there is the following definition of bounded.

Definition: Suppose $$S$$ is an ordered set, and $$E \subset S$$. If there exists a $$\beta \in S$$ such that $$x \leq \beta$$ for every $$x \in E$$, we say that $$E$$ is bounded above, and call $$\beta$$ an upper bound of $$E$$. Lower bounds are defined in the same way(with $$\geq$$ in place of $$\leq$$).

Let $$S = \emptyset$$.
Let $$E = \emptyset$$.
Then, there exists no $$\beta \in S$$ such that $$x \leq \beta$$ for every $$x \in E$$.
So, in this case, I think $$E (\subset S)$$ is not bounded above.

Am I correct or not?

Isn't $$\emptyset$$ an orderd set?

• If $S$ is empty then indeed it has no subsets that are bounded above, But IMV that is no reason to say that $S$ is not an ordered set. There are also non-empty sets endowed with an order that have no subsets that are bounded above. – drhab Jan 19 at 10:20

Being bounded is relative to a given ordered set. If $$S$$ is empty, then indeed it is not bounded, which is just one more reason to agree that in general the empty set shouldn't be counted as a partial order. But if $$S$$ is non-empty, then the empty set is bounded, and every element is an upper and lower bound.
There is an analogy here with $$(0,1)$$. As a subset of itself, it is not bounded. There is no maximal element. But as a subset of $$[0,1]$$ or $$\Bbb R$$ it is certainly bounded by $$0$$ and $$1$$.