Let us use the axiom of choice to well-order the set of open intervals of R, and suppose this well-ordering has order type gamma. For any ordinal alpha less than gamma, let I_alpha be the alpha'th interval according to the well-ordering.(adsbygoogle = window.adsbygoogle || []).push({});

Now let us use transfinite recursion to define a set of points in R. Let a_1 be any point in I_1. For any ordinal alpha less than gamma, define a_alpha as follows: if any of the a_beta's for beta less than alpha are in I_alpha, let a_alpha be the least such a_beta. Otherwise, let a_alpha just be any element of I_alpha.

Now if we take the set A of all a_alpha for all alpha less than gamma, then clearly A is dense in R, since it intersects every open interval by construction. But my question is, is it possible for A to have a nonempty interior, i.e. is it possible for A to contain any open intervals? I'd like to show that it's impossible, and I'd like to generalize this to arbitrary metric spaces.

Any help would be greatly appreciated.

Thank You in Advance.

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# Does this transfinitely defined set have empty interior?

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