Existence of Least Upper Bounds (Proof Verification)

Question

I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary. I had previously put a proof for this prompt on stack exchange, but I deleted it as my answer was incorrect.

The Prompt:

Let $A$ be a nonempty subset of $\mathbb{R}$ and let $B = \{b : b$ is an upper bound of $A\}$. Assuming that $B$ is nonempty, show that $B$ is an interval of the form $B = [a,∞]$. Thus $a$ is the minimum of $B$ and is the definition of $sup A$.

My Proof:

For any $x∈A$, we know that $B⊆[x,∞]$, as $x\le{b}$ for all $b∈B$. Since $B$ is comprised of the upper bounds of all of X, we then know that $B = ⋂\{[x,∞]|x∈A\}$ and since the intersection of a collection of closed sets is closed, $B$ is closed. We also know that $B$ has no upper bound (as it is the intersection of sets which all have no upper bound), so since $B$ is closed on $\mathbb{R}$ it has the form $[a,∞]$ for some $a∈\mathbb{R}$. Thus $a\le{b}$ for all $b∈B$, so $a$ is the minimum of $B$ and is the definition of $sup A$.

Answer 1

More should be said about why $B = \bigcap \{ [x,\infty] \mid x \in A\}$.

Clearly $B \subseteq \bigcap \{ [x,\infty] \mid x \in A\}$, as you argued. You should show the other containment as follows:

Let $y \in \bigcap \{ [x,\infty] \mid x \in A\}$. Let $z \in A$. Then $y \in [z,\infty]$, so $y\geq z$. This holds for all $z \in A$, so $x$ an upper bound of $A$. Hence $y \in B$.

This shows the other containment.

You are correct to then deduce that $B$ is closed, but you haven't successfully argued $B$ is of the form $[a,\infty]$. Just because it doesn't have an upper bound, doesn't mean it is of this form. E.g. $[0,1] \cup [2,\infty]$ is a closed set without an upper bound but doesn't have this form.

Instead argue that since you are intersecting a nested collection of intervals, that $B$ is an interval. And argue that closed intervals with a lower bound and without an upper bound must be of the form $[a,\infty]$.

Final note: You are writing $[a,\infty]$ which is only correct if you are working over the extended real numbers, which from the prompt, it seems like you are.