# Check my answer: Prove that every open set in $\Bbb R^n$ is a countable union of open intervals.

I have a question. I have solved this but please can you check my solution? Thank you.

If there are any mistakes or something is missing and so on, please tell me.

This is important to me. Is this proof enough to get a successful grade on an exam?

Btw, I underlined the question with pink a pencil.

• Also, what theorem of Lindelöf's are you referring to? Commented Jul 3, 2013 at 20:02
• I object to the sentence (and idea) that $A$ is countable implies $A = \mathbb{N}$. There is a bijection with $\mathbb{N}$ you can use. Commented Jul 3, 2013 at 20:09
• I’m working on it right now. Commented Jul 3, 2013 at 20:24
• @B11b Be very happy, not everyone is lucky enough to get an answer from Brian M. Scott. Commented Jul 3, 2013 at 20:28
• Really? But I got answer from Brian M. Scott previous time. Scott is the best to teach well and solve understandable @LordSoth
– 1190
Commented Jul 3, 2013 at 20:30

There are two problems with your argument. The first is when you have

$$V\subseteq\bigcup_{x\in A_0}B_{\epsilon_x}(x)\;,\tag{1}$$

where $A_0$ is a countable subset of $X$, and say that $A_0=\Bbb N$. $A_0$ is not $\Bbb N$: it’s a subset of $X$. If it’s a countably infinite subset of $X$, then there is a bijection from $\Bbb N$ to $A_0$, and you can enumerate $A_0=\{x_j:j\in\Bbb N\}$, let $B_j(x_j)=B_{\epsilon_j}(x_j)$ for each $j\in\Bbb N$, and say (as you did) that

$$V\subseteq\bigcup_{j\in\Bbb N}B_j(x_j)\;,\tag{2}$$

but you have to say that that is what you’re doing. However, $A_0$ might be finite, in which case there is no bijection from $\Bbb N$ to $A_0$.

There’s no need to do all of this, however: $(2)$ is an unnecessary rewriting of $(1)$ even when $(2)$ is correct. If $A_0$ is countable, then the union is $(1)$ is a countable union, and that’s all you need. However, I would strengthen $(1)$ and say that

$$V=\bigcup_{x\in A_0}B_{\epsilon_x}(x)\;:\tag{3}$$

you’ve actually proved this, and it’s what you need: $V$ is a countable union of open balls, not just a subset of a countable union of open balls.

Now we come to the larger problem. Your last step does not work at all: a ball in $\Bbb R^n$ is not an interval. To complete your proof, you need to show that an open ball in $\Bbb R^n$ is a countable union of intervals. Then you can argue like this:

For each $x\in A_0$ there is a countable family $\mathscr{I}_x$ of intervals in $\Bbb R^n$ such that $B_{\epsilon_x}(x)=\bigcup\mathscr{I}_x$. Let $\mathscr{I}=\bigcup_{x\in A_0}\mathscr{I}_x$; then $\mathscr{I}$, being the countable union of countable sets, is countable, and $$B=\bigcup_{x\in A_0}\bigcup\mathscr{I}_x=\bigcup\mathscr{I}$$ is a countable union of intervals.

To do this, though, you have to prove that if $B_r(x)$ is any open ball in $\Bbb R^n$, then $B_r(x)$ is the union of countably many sets of the form

$$[a_1,b_1)\times[a_2,b_2)\times\ldots\times[a_n,b_n)\;.$$

I suggest the following approach.

1. First prove that $B_r(x)$ is the union of countably many sets of the form $$(a_1,b_1)\times(a_2,b_2)\times\ldots\times(a_n,b_n)\;.$$ You can do this by showing that if $y\in B_r(x)$, there are rational numbers $p_1,\dots,p_n,q_1,\dots,q_n$ such that $$y\in(p_1,q_1)\times(p_2,q_2)\times\ldots\times(p_n,q_n)\;.\tag{4}$$ There are only countably many rational numbers, so there are only countably many open boxes like $(4)$.

2. Show that each open box like $(4)$ is the union of countably many intervals. HINT: Use the fact that in $\Bbb R$, $$[a,b)=\bigcup_{n\in\Bbb Z^+}\left(a+\frac1n,b\right)\;.$$

• You are amazing! Wonderful! Thank you so much! I read it and understand your solution. Thank you so much thank you:)
– 1190
Commented Jul 3, 2013 at 20:47
• @B11b: You’re very welcome! I’m glad that it helped. Commented Jul 3, 2013 at 20:51
• @Brian - That kind of feedback is priceless! ;-) Commented Jul 3, 2013 at 20:59
• Where did the products of half-open intervals come from? Commented Jul 3, 2013 at 21:00
• @dfeuer: That’s what is meant by interval in $\Bbb R^n$. I’ve seen the definition before and got an explicit confirmation that it was intended here; see the long string of comments under the question. Commented Jul 3, 2013 at 21:01