I am studying analysis on my own and need some help verifying the solution to the above exercise found in Bartle's Elements of Real Analysis. I know there are other posts answering the same question but I need some criticism of my attempt if someone is kind enough to read through it for me. I also am in need of some advice on whether my terminology is alright. I have also chosen to interpret an open set in $\Bbb R ^ p$ to mean a set such that to each of its points there is an open ball centred at that point entirely contained in the original set. Is this fine??
And also I found it very difficult to come up with a proof of this exercise - Took me a good couple of days but did do it entirely on my own. Should I learn a first course on Analysis described in Cartesian space or should I begin with something like Abbott or Strichartz which focus just on the real line first??
Any help is appreciated. Thanks in advance.
Here is my proof. The absolute value sign "$ \left| {x} \right|$" stands for the standard Euclidean norm.
Proof:
Let $J \subseteq \Bbb R^p$ be an open ball. Then there exists $r \gt 0$ such that $J = \{ y \; | \; \left| { y - x} \right| \lt r\}$ for some point $x \in J$. For every point $ y \in J$, $ \; r_y = r - \left| { y - x} \right| \gt 0 $. Consider the open ball $J_y = \{ z \; | \; \left| { z - y} \right| \lt r_y \}$ centered at $y$. $ \; z \in J_y \implies \left| { z - y} \right| \lt r_y \implies \left|{(z - x) - (y - x)}\right| \lt r_y$
$\implies \left|{z - x}\right| - \left|{ y -x}\right| \lt r - \left| { y - x} \right| \implies z \in J \implies $ there is an open ball centered at every point $y \in J_y$ entirely contained in $J_y$.
Therefore an open ball is an open set. And since the union of any collection of open sets is open in $\Bbb R ^ p$, the union of countably many open balls is an open set in $\Bbb R ^ p$.
Let $G$ be an open set. The set $\{ a_n \}$ of all points in $G$ with rational coordinates is countable. Since each $a_n \in G$ there corresponds an open ball centred at $a_n $entirely contained in G. Let $B_n$ be the largest of these open balls. We shall prove that $$G = \bigcup_{n \in \Bbb N} B_n$$
Since $B_n \subseteq G \;\;\; \forall n \in \Bbb N$, the union is trivially contained in $G$.
Let $x = (x_1, x_2, .., x_p) \in G$. Then there is an open ball $B_x = \{ y \in \Bbb R ^ p \ | \ \left| {y - x} \right| \lt r_x (\gt 0) \}$ centred at $x$ entirely contained in $G$.
If $ a_t = (t_1, t_2,.., t_p)$ where $t_i$ is a rational number in the open interval $(x_i - \frac {r_x}{3\sqrt p}, x_i + \frac {r_x}{3\sqrt p}) $, then since; $$\left| {a_t - x} \right| = \sqrt { \sum_{i=1}^p (t_i - x_i)^2} \lt \sqrt { \sum_{i=1}^p \frac {(r_x)^2}{ 9p}} \lt r_x$$ $a_t \in B_x \subseteq G$. $B_t$ is the largest open ball centred at $a_t$ entirely contained in $G$.
Consider the open ball centred at $a_t$, $K = \{ (l_1, l_2,.. , l_p) \ | \ \left| {l_i - t_i} \right| \lt \frac {r_x}{2\sqrt p}\}$. Since $\left| {l_i - x_i} \right| = \left| {(l_i - t_i) + (t_i - x_i)} \right| \le \left| {l_i - t_i} \right| + \left| {t_i - x_i} \right| \le \frac {5r_x}{6\sqrt p}$, $K \subseteq B_x \subseteq G \implies K \subseteq B_t$.
But clearly $x = (x_1, x_2, ... , x_p) \in K \implies x \in B_t$ for some $t \in \Bbb N \implies G \subseteq \bigcup_{n \in \Bbb N} B_n$
Q.E.D.
In any case it is correct. The author may have had a different characterisation of open sets in mind, but probably not, so most likely it is not only correct but also fine.
You did well. There are only a few nitpicks, nothing serious.
Nitpick 1: "Let $B_n$ be the largest of these open balls."
If $G = \mathbb{R}^p$, then no largest ball contained in $G$ exists, unless you consider the entire space a ball with infinite radius. Treating the case $G = \mathbb{R}^p$ when allowing only balls with finite radius is however easy.
Nitpick 2 (Well, that may be just a typo): You write
$$\left| {t - x} \right| = \sqrt { \sum_{i=1}^p (t_i - x_i)^2} \lt \sqrt { \sum_{i=1}^p \frac {(r_x)^2}{ 9p}} \lt r_x,$$
but $t$ is the index, the point is $a_t$.
It might have been better to use $k,m$ or $n$ for the index, in any case it is a bit confusing to denote the components of the point $a_t$ with $t_i$.