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I've been thinking about open balls in $\mathbb{R}^n$ and whether the density of $\mathbb{Q}$ in $\mathbb{R}$ means that we can find open balls in $\mathbb{Q}^n$ to 'nest inside open balls in $\mathbb{R}^n$, if we take the 'centre' of the ball to be distinct. I've tried to formalise my question below, in an arbitrary dimension (the one-dimensional case seems more simple):

If we think of $\mathbb{R}^n$ endowed with the $\|\|_{\infty}$ norm, can we show that for any open ball around an arbitrary point $x \in \mathbb{R}^n$, $B(x,\epsilon)$ with $\epsilon>0$ given, that there is some $q \in \mathbb{Q}^n$, $\delta>0$ such that

$$x \in B(q,\delta) \subset B(x,\epsilon)$$

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up vote 3 down vote accepted

Yes. Choose $q \in \mathbb{Q}^n$ such that $\| q - x \|_{\infty} < \epsilon/2$, and take $\delta = \epsilon/2$. Then $B(q,\delta) \subseteq B(x,\epsilon)$ because if $y \in B(q,\delta)$, then $$\| y - x \| \leq \| y - q \| + \|q - x\| < \delta + \frac{\epsilon}{2} = \epsilon.$$ Note that which norm you choose on $\mathbb{R}^n$ is irrelevant to the truth of this statement.

Edit: To make the choice of $q$, suppose that $x = (x_1, x_2, \ldots, x_n)$. For each $x_i$ with $1 \leq i \leq n$, we can (using density of $\mathbb{Q}$ in $\mathbb{R}$), choose $q_{i}$ such that $|q_{i} - x_{i}| < \epsilon/2$. Then, if $q = (q_1,q_2, \ldots, q_n)$, we have $$\| q - x \|_{\infty} = \max_{1 \leq i \leq n} \{|q_i - x_i|\} < \frac{\epsilon}{2}$$

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Thanks for this, but how can we chose such a $q$ to satisfy that condition; does this follow instantly in n-dimensions from the density of $\mathbb{Q}$ in $\mathbb{R}$? – Mathmo Feb 13 '12 at 11:31
@MiamiMath: Yes, more or less; I've edited the answer. – Martin Wanvik Feb 13 '12 at 11:40
Many thanks for your edit, much clearer now - Answer accepted. – Mathmo Feb 13 '12 at 11:41

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