Let $M$ be a metric space and let $\mathscr B$ be a family of open balls in $M$ whose radii are bounded. Assuming that $\mathscr B$ is totally ordered by inclusion, is the union of all members of $\mathscr B$ an open ball?
I believe this is false for the rational numbers, but what if $M$ is complete? Does it hold in every normed space? Every Banach space? What about $\mathbb R^n$?
My attempt at the case $M=\mathbb R^n$:
Let $R$ be the set of real numbers formed by the radii of the balls in $\mathfrak{B}$ and
let $r=\sup R$. If $r\in R$ then the answer is obvious, so let us suppose otherwise.
One may then find an increasing sequence
$\{r_i\}_i$ in $R$ converging to $r$. The centers $c_i$ of the corresponding balls $B_i$ must form a converging
sequence (this seems awful obvious, but might require a rather lengthy proof) so let $c$ be the limit.
I guess it should also be clear that $\bigcup \mathfrak B = \bigcup_i B_i$, and it seems reasonable to try to prove that this set coincides with
the open ball $B_r(c)$, centered at $c$, with radius $r$. Regarding the inclusion
$$
\bigcup_i B_i\subseteq B_r(c),
$$
let $x$ belong to the left-hand-side set, so there exists some $i_0$ such that $x\in B_{i_0}$ and, since the balls are
increasing, this should also hold for all $i>i_0$. In other words
$$
\|x-c_{i}\|<r_{i},\quad \forall i\geq i_0.
$$
Taking the limit as $i\to\infty$, one concludes that
$$
\|x-c_{i_0}\|\leq r_{i_0}.
$$
Among many, this is one of the main outstanding points! How to get "$<$" instead of "$\leq $"???
Best Answer
To answer your first question, it's not necessarily true in a complete metric space that the union of a chain of open balls is a ball. Here is a counterexample.
Let $M=\{a_i:i\in\mathbb N\}\cup\{b_i:i\in\mathbb N\}\cup\{c\}$ with the following metric:
$d(a_i,a_j)=1$ if $i\ne j$;
$d(b_i,b_j)=2$ if $i\ne j$;
$d(a_i,b_j)=1$ if $j\le i$;
$d(a_i,b_j)=2$ if $j\gt i$;
$d(a_i,c)=d(b_i,c)=2$.
The triangle inequality holds, since all nonzero distances are $1$ or $2$.
The metric is complete, since every Cauchy sequence is eventually constant.
Let $\mathscr B=\{B_n:n\in\mathbb N\}$ where
$B_n=\{x\in M:d(a_n,x)\le1\}=\{x\in M:d(a_n,x)\lt2\}=\{a_i:i\in\mathbb N\}\cup\{b_i:i\le n\}$.
S0 $\mathscr B$ is a chain of open balls and a chain of closed balls. The union $\bigcup\mathscr B=M\setminus\{c\}$ is not a ball because, for each point $x\ne c$, there is a point $y\ne c$ such that $d(x,y)=d(x,c)=2$.
Regarding your other questions. I'm going to guess that it's true for Banach spaces, false for incomplete normed spaces.