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There are questions on Math.SE that seem relevant, but I will explain why they do not answer my question:

1) $\mathbb R^n$ has countable basis of open balls? (Yes): this question doesn't prove that rational centre/radii balls form a base; it proves that $\mathbb{R}^n$ has such a base

2) Open ball with rational radii forms a basis.: makes a claim that I do not know how to show (it is not obvious to me):

Now let $B$ be an open ball with centre $a$, and irrational radius $\rho$. Then $B$ is the union of all the open balls with centre $a$ and rational radius $r\lt \rho$.

3) How to cover an open subset of $\mathbb{R}^n$ with balls?: the answer to this question gets me the closest, I think, but I am still having trouble with it. Let me break it down my issues:

HINT: Every open set in $\mathbb{R}^n$ is a union of open (or closed) balls whose centres have rational coordinates and whose radii are rational; how many such balls are there?

Alright, assuming that I can show that, I still don't know how to show its countable. My guess is I have to play around and find a bijective function that connects each ball with an integer. Is this the right approach?

In case the first statement isn’t obvious, suppose that $B$ is the ball of radius $r$ about a point $x\in\mathbb{R}^n$. If $x$ has all rational coordinates, there’s nothing to be done.

How is there nothing to be done? Don't we still have to show that $r$ (the radius) is rational?

Otherwise there is a point $y$ with all coordinates rational inside the ball of radius $r/2$ centred at $x$.

Why is this an obvious statement? I don't know how to begin to show it. Could a hint be provided here?

Let $d$ be the distance between $x$ and $y$, let $q$ be a rational number such that $d<q<r/2$, and let $B'$ be the ball of radius $q$ centred at $y$; then $x\in B'\subseteq B$.

This part I am totally good with.

Anyway, so if I can show that there a countable number of such balls, then by definition of second-countability, $\mathbb{R}^n$ is second countable.

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    $\begingroup$ What is a "second countable base"? Do you mean a countable base? $\endgroup$
    – bof
    Commented Sep 11, 2016 at 4:38
  • $\begingroup$ @bof Yes! That is an error/redundant wording on my part, now that I look at the definition with more care. Should be fixed now $\endgroup$
    – bzm3r
    Commented Sep 11, 2016 at 4:41
  • $\begingroup$ @user89 There are many questions given here, mainly in five yellow boxes. Could you point out which one you would like answered certainly, so that we can get down to that one immediately, and address the rest later? I am a little confused at which question to answer, though I know the questions themselves inside out. $\endgroup$ Commented Sep 11, 2016 at 4:52
  • $\begingroup$ @астонвіллаолофмэллбэрг There are 3 questions I would like answered -- they are not in the yellow boxes, but rather, are in bold text. Can we start the second of the three: "Don't we still have to show that $r$ (the radius) is rational?". Then I would like to get a hint for how to show that "Otherwise there is a point $y$ with all coordinates rational inside the ball of radius $r/2$ centred at $x$." Lastly, I would like confirmation on my quess that: "[to show countability]...my guess is I have to play around and find a bijective function that connects each ball with an integer..." $\endgroup$
    – bzm3r
    Commented Sep 11, 2016 at 5:10
  • $\begingroup$ The confirmation for the last one is received from my side. $\endgroup$ Commented Sep 11, 2016 at 5:30

2 Answers 2

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Let $(X,d)$ be a metric space and let $E$ be a dense subset of $X.$ Let $B=\{B_d(x,q): x\in E\land q\in \mathbb Q^+\}.$ Then $B$ is a base for $X.$

Proof: Let $p\in U$ where $U$ is open in $X.$ It suffices to show that $p\in b\subset U$ for some $b\in B.$ There exists $r>0$ such that $B_d(p,r)\subset U,$ and there exists $q\in (0,r/2)\cap \mathbb Q^+.$ And there exists $e\in E$ with $d(p,e)<q/2$ because $E$ is dense in $X.$ So $p\in B_d(e,q)\in B.$

And $y\in B_d(e,q)\implies d(y,p)$ $\leq d(y,e)+d(e,p)<q+q/2<2q<r$ $\implies y\in B_d(p,r)\subset U.$ So $p\in B_d(e,q)\subset U.$.... Q.E.D.

If $E$ is countable then $B$ is countable. When $X=\mathbb R^n$ with the usual topology, let $E$ be the set of points with rational co-ordinates. Then $E$ is dense and countable so $B$ is a countable base for $\mathbb R^n.$

EDIT: formatting second paragraph in proof

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  • $\begingroup$ Can someone tell me what I did to get this special effect with blue paint,etc., or how to undo it? $\endgroup$ Commented Sep 13, 2016 at 6:33
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"Don't we have to show that the radius is rational?"

I think the author has missed something. It is obvious only when the radius is rational, since then the ball is itself of the category "ball with rational radius and rational center".

So what we will do is, we will assume that $B$ has both rational center and rational radius in that step. Then it ok, right? (and obvious,that $B$ is a union of balls with rational radius and rational center).

So from now on, we will assume that both the center and radius are irrational.

There is a point $y$ with rational coordinates inside the ball of radius $0.5r$ centered at $x$.

To start,note that the hypercube (cube in $n$ dimensions) of side length $0.25r$ is contained inside the ball of radius $0.5r$ centered at $x$.

Now, let $x=(x_1,x_2,...,x_n)$, the coordinates. By what we know about the rationals, there is a rational number between (this means not equal to either side) $x_1$ and $x_1+0.25r$. Let's call it $y_1$.

Similarly, there is a rational number between $x_2$ and $x_2+0.25r$. Let's call it $y_2$. Going similarly, there is a rational number between $x_i$ and $x_i+0.25r$. Let's call it $y_i$.

Having done this for all the corrdinates, look at $y=(y_1,y_2,...,y_i)$. It is easy to see that $y$ lies inside the hypercube of length $0.25r$, and hence inside the ball of radius $0.5r$ centered at $x$, and $y$ has all rational coordinates.

My guess is I have to play around and find a bijective function that connects each ball with an integer.

Yes, you have to do that.

Please get back on any kind of doubts. I want to make sure you are clear with this.

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  • $\begingroup$ Thank you. I want one more confirmation: "By what we know about the rationals, there is a rational number between (this means not equal to either side) $x_1$ and $x_1+0.25r$. Let's call it $y_1$." -- this is because the rationals are "dense" in the reals? $\endgroup$
    – bzm3r
    Commented Sep 11, 2016 at 17:31
  • $\begingroup$ @user89 It is a consequence of the fact that the rationals are dense in the reals. If you do not know this, I will prove it as well. $\endgroup$ Commented Sep 12, 2016 at 5:54

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