We show that for a reductive algebraic group \(G\) there exists an integer \(r(G)\), such that any finite set of elements in \(G\) of size more than \(r(G)\) that generates a Zariski-dense subgroup must be redundant i.e. we can remove some elements and still generate a Zariski-dense subgroup. We use this to deduce the analogous result for compact Lie groups. Thus, for example, if you have \(1000\) rotations that generate a dense subgroup of \({\rm SO}(3)\), some of them must be redundant. For non-compact Lie groups (e.g \({\rm SL}_2(\mathbb{C})\)) this fails: there are arbitrarily large irredundant topologically generating sets. The proof is mostly arithmetic: we ensure generators live in a number field in order to reduce the problem to finite groups via strong approximation and other results of this sort.