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Sep 9, 2011 at 13:52 comment added François G. Dorais I was analyzing some results in intuitionistic analysis. Some were perplexing: I could prove the existence of the finite set of all solutions but I couldn't always prove the existence of a single solution. It turns out the problem boils down to the existence or non-existence of continuous selectors for finite sets...
Sep 9, 2011 at 2:51 comment added Adam Bjorndahl That's a nice argument; I especially like the use of countability to ensure that $(x_{n}, y_{n})$ never lies on the boundary of any $B_{\varepsilon_{k}}(x_{k}, y_{k})$, $k < n$. I'd have to think about whether this process can ever correspond to a "linear" division of $\mathbb{Q}^{2}$ as in my construction. Your construction is clearly much more general, of course; both show, though, that $\mathbb{Q}^{2}$ admits very many distinct continuous selectors (in contrast to $\mathbb{R}$), which was not obvious to me before. Just curious -- how did this question originally come up?
Sep 7, 2011 at 21:33 history edited François G. Dorais CC BY-SA 3.0
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Sep 7, 2011 at 21:27 history answered François G. Dorais CC BY-SA 3.0