# Is an infinite linearly independent subset of $\Bbb R$ dense?

Suppose $(a_n)$ is a real sequence and $A:=\{a_n \mid n\in \Bbb N \}$ has an infinite linearly independent subset (with respect to field $\Bbb Q$). Is $A$ dense in $\Bbb R?$

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One can certainly rescale any linearly independent set (by individual scalars) so that they each lie in $[0,1]$... –  Erick Wong Feb 28 '13 at 14:09

Not necessarily. We can use recursion to construct a linearly independent set $A = \{ a_n : n \in \mathbb{N} \}$ so that $a_n \in ( n , n+1 )$ for all $n$. Such a set is clearly not dense, but even more it is a closed discrete subset of $\mathbb{R}$.
Even a Hamel basis $B$ for $\mathbb{R}$ may not be dense in $\mathbb{R}$ (just ensure that all elements of $B$ avoid some open interval), however unlike the simple infinite linearly independent case by the Baire Category Theorem $B$ cannot be nowhere dense (since we have $\mathbb{R} = \bigcup_{q \in \mathbb{Q}} ( q + B )$).
If $A$ is a linearly independent subset of $\mathbb R$, for each $a\in A$ there is a positive integer $n(a)$ such that $n(a)>|a|$. The set $\left\{\dfrac{a}{n(a)}:a\in A\right\}$ is a linearly independent set with the same cardinality and span as $A$, but it is a subset of $(-1,1)$.