If S is a bounded set, then inf(S) and sup(s) belong to the closure of S

Question

My proof:

Since $\bar S = S\cup bdy(S)$ $\Rightarrow S\subset \bar{S}$

and since S is bounded $\Rightarrow inf(S)$ and $sup(S)$ exist

Then is it safe to assume that since inf(S) and sup(S) exists, then inf(S) and sup(s) is already a subset of $\bar S$?

Answer 1

The definition of $\sup(S)$ is the real number $M$ such that $x \leq M$ for all $x \in S$, and such that if $M'$ also satisfies this inequality, then $M' \geq M$.

So let $M = \sup(S)$, and for all $n \in \mathbb{N}$, let $\delta_n = \frac{1}{n}$.

Now, for each $n$, let $a_n \in S$ be such that $\left| M - a_n \right| < \delta_n$. This exists by the definition of $\sup$, for if we could not find such an $a_n$, then we would have that $x \leq M - \delta_n$ for all $x \in S$, contradicting that $M$ is the $\sup$.

Now, by construction, we have $a_n \in S$ for all $n$ and $a_n \rightarrow M$ as $n \rightarrow \infty$. So by definition of closure, $M \in \bar{S}$.

The proof for $\inf$ is similar.

Answer 2

It isn't entirely safe. It's better to mention that $\inf(S)$, $\sup{S}$ are limit points of $S$, i.e. there exist sequences $S\supset(x_n)\longrightarrow\inf(S)$ and $S\supset(y_n)\longrightarrow\sup(S)$, and that the closure contains all limit points, which follows easily from any possible definition.

But it is not true in general that $\inf(S)$, $\sup(S)$ belong to $S$! Note e.g. $S=(0, 1)$.