If $X$ is a non-empty subset of $\mathbf{R}$, show that $X$ is bounded if and only if $\inf(X)$ and $\sup(X)$ are finite.

Question

If $X$ is a non-empty subset of $\mathbf{R}$, show that $X$ is bounded if and only if $\inf(X)$ and $\sup(X)$ are finite.

MY ATTEMPT

Let us prove the implication $(\Rightarrow)$ first.

If $X$ is bounded, the $-M\leq x\leq M$ whenever $x\in X$.

Since $X$ is not empty, we conclude that $-M\leq\inf(X)\leq\sup(X)\leq M$.

Thus we conclude that $\inf(X)$ and $\sup(X)$ are finite.

Conversely, let us prove the implication $(\Leftarrow)$.

Once $-\infty < \inf(X) \leq x\leq \sup(X) < \infty$, we conclude that $X\subseteq[\inf(X),\sup(X)]$.

Consequently, $X$ is bounded.

I am a little bit worried about the formalism. Am I missing any steps?

Answer 1

There are only two things that I would add: one is the words "for all $x \in X$" after

Once $-\infty < \inf(X) \leq x \leq \sup(X) < \infty$

Secondly, it seems from your forward implication $(\implies)$ that you are taking the definition of a bounded set to be one which is contained in $[-M,M]$ for some real number $M \geq 0$. However, in the reverse implication $(\impliedby)$ you have "only" shown that $X \subseteq [\inf(X),\sup(X)]$. It remains for you to argue the following, in order to make your argument more complete:

Let $M = \max\{ \lvert \inf(X) \rvert, \lvert \sup(X) \rvert \}$. Then, $X \subseteq [-M,M]$, and hence $X$ is bounded.

---

Lastly, as other commenters have said, in the future try and specify what parts of your proof you are unsure about so that you can get more focussed feedback.