The construction is based on a well ordering of $R^3$ into the least ordinal of cardinality continuum. Let $\phi$ be that ordinal and let $R^3={p_\alpha:\alpha<\phi}$ R^3=\{p_\alpha:\alpha<\phi\}$ be an enumeration of the points of space. We define a unit circle $C_\alpha$ containing $p_\alpha$ by transfinite recursion on $\alpha$, for some $\alpha$ we do nothing. Here is the recursion step. Assume we have reached step $\alpha$ and some circles ${C_\beta:\beta<\alpha}$ \{C_\beta:\beta<\alpha\}$ have been determined. If some of them contains (=covers) $p_\alpha$, p_\alpha$, we do nothing. Otherwise, we choose a unit circle containing $p_\alpha$ that misses all the earlier circles. For that, we first choose a plane throu $p_\alpha$ that is distinct from the planes of the earlier circles. This is possible, as there are continuum many planes throu $p_\alpha$ and less than continuum many planes which are the planes of those earlier circles. Let $K$ be the plane chosen. The earlier circles intersect $K$ in less than continuum many points, so it suffices to find, in $K$, a unit circle going throu $p_\alpha$ which misses certain less than continuum many points. That is easy: there are continuum many unit circles in $K$ taht pass throu $p_\alpha$ and each of the bad points disqualifies only 2 of them.
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The construction is based on a well ordering of $R^3$ into the least ordinal of cardinality continuum. Let $\phi$ be that ordinal and let $R^3={p_\alpha:\alpha<\phi}$ be an enumeration of the points of space. We define a unit circle $C_\alpha$ containing $p_\alpha$ by transfinite recursion on $\alpha$, for some $\alpha$ we do nothing. Here is the recursion step. Assume we have reached step $\alpha$ and some circles ${C_\beta:\beta<\alpha}$ have been determined. If some of them contains (=covers) $p_\alpha$, we do nothing. Otherwise, we choose a unit circle containing $p_\alpha$ that misses all the earlier circles. For that, we first choose a plane throu $p_\alpha$ that is distinct from the planes of the earlier circles. This is possible, as there are continuum many planes throu $p_\alpha$ and less than continuum many planes which are the planes of those earlier circles. Let $K$ be the plane chosen. The earlier circles intersect $K$ in less than continuum many points, so it suffices to find, in $K$, a unit circle going throu $p_\alpha$ which misses certain less than continuum many points. That is easy: there are continuum many unit circles in $K$ taht pass throu $p_\alpha$ and each of the bad points disqualifies only 2 of them. |
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