I'm completely stumped on this proof. I know that $E \subset X$ is a bounded set if $\exists R \in \mathbb{R}$ s.t. $E \subset N_R (x) $. I also know that finite unions of closed sets are close, and that arbitrary intersections of closed sets are closed.
I do have a rough idea for the finite union case. I know a set is bounded it is bounded above and below. I know there's the least upper bound property, but I can't wrap my head around how I would prove that.
Any guidance would be much appreciated.
Best Answer
For the finite union case, we will show that if $E_1 \subseteq X$ and $E_2 \subseteq X$ are bounded, then so is $E_1 \cup E_2$. That finite unions of bounded sets is bounded then follows by induction.
Since $E_1$ is bounded, there is some $x_1 \in X$ and $r_1 \in \mathbb{R}^+$ such that $E \subseteq B(x_1, r_1)$. Since $E_2$ is bounded, wer know that there is some $x_2 \in X$ and $r_2 \in \mathbb{R}^+$ such that $E_2 \subseteq B(x_2, r_2)$.
Let $r = r_1 + r_2 + d(x_1, x_2)$. Then if for any $x \in E_1 \cup E_2$, we have that either $x \in E_1$ or $x \in E_2$. If $x \in E_1$, then we know that $d(x, x_1) < r_1 < r$. If $x \in E_2$, then we have that $d(x, x_1) \leq d(x, x_2) + d(x_2, x_1) < r_2 + d(x_1, x_2) < r$ by the triangle inequality. Thus we have that $x \in B(x_1, r)$.
We see that $E_1 \cup E_2 \subseteq B(x_1, r)$, and hence $E_1 \cup E_2$ is bounded.
For the arbitrary intersection case, suppose that $\{E_i\}_{i \in I}$ is an arbitrary non-empty family of bounded sets. Let $E$ be their intersection. Then for any $i \in I$, we have that $B_i$ is bounded, and hence there is $x \in X$ and $r \in \mathbb{R}^+$ such that $E_i \subseteq B(x, r)$. But $E \subseteq B_i \subseteq B(x, r)$, and hence $E$ is also bounded.
Essentially, if you pick any set out of those that you're taking the intersection of, the intersection will be contained in that set. Since that set is bounded by assumption, so is the intersection.