# A subset of $\Bbb R^p$ is open iff it is the union of a countable collection of open balls

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.

• possible duplicate of Union of a countable collection of open balls – Luiz Cordeiro Feb 14 '14 at 18:12
• 'Clearly an open ball is an open set' -- if you are on a beginner level and need to show the fact from your heading this is not all clear. – Thomas Feb 14 '14 at 18:13
• @Luiz Cordeiro: I did read through that question before posting this. All the answers contain terms that I am not familiar with. They are solved in metric space.. – Ishfaaq Feb 14 '14 at 18:14
• @Thomas You're right. I plugged it in. – Ishfaaq Feb 14 '14 at 18:37
• Well lemme just say that marking this as a duplicate is a little unfair.. – Ishfaaq Feb 14 '14 at 18:39

I have also chosen to interpret an open set in $\mathbb{R}^p$ 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??

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.

Took me a good couple of days but did do it entirely on my own.

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 $t_i$.

• Thanks for the answer.. Yeah Nitpick 2 was just a typo. But you're absolutely right - should have used another index letter. Now Nitpick 1: How would I go about proving the case $G = \Bbb R^p$? By defining an arbitrary open ball centred at each point in $\Bbb Q^n$ and claiming their union should form the whole set?? – Ishfaaq Feb 14 '14 at 20:00
• Just choose the balls with radius $n$ and centre $0$. Or the balls with rational centres and radius $1$. Or - there are a lot of ways. – Daniel Fischer Feb 14 '14 at 20:04
• Rational centres and radius $1$. That was my initial idea. Will try both. Thanks loads for the advice.. – Ishfaaq Feb 14 '14 at 20:06