Example of a set of real numbers $S$ where $\inf S = \sup S$?

Question

I am trying to come up with some examples for possible test questions where $\inf S = \sup S$,

The only thing I can think of is $S = \{1\}$ or any other $1$-element set, are there any others?

Answer 1

If $|S|>1$, there are $x, y \in S$ s.t. $x<y$. What does this tell about $\inf S$ and $\sup S$?

Answer 2

Considering nonempty bounded sets, the infimum and supremum are equal if and only if the set has exactly one element. (Suppose $a,b\in S$ with $a<b$. Then $\sup S\ge b$ and $\inf S\le a$.)

Answer 3

Your solution is correct; the only sets with this property are those with one element.
If $|S|>1$, then chose $x\ne y\in S$. Because $\mathbb R$ is ordered, either $x<y$ or $y<x$ must hold. WLOG assume $x<y$ and see by definition of $\inf$ and $\sup$ $$\inf_{x\in S}x \le x<y\le \sup_{y\in S}y$$ And therefor $\inf S \ne \sup S$. Now for $S=\emptyset$ it is defined $$\inf \emptyset = \infty$$ and $$\sup \emptyset = -\infty$$ and thus also $\inf S \ne \sup S$ by definition. This concludes the proof.