In any ordered field F, we can define absolute value in the usual way: |x| = x if x ≥ 0 and |x| = −x if x<0. Using this we can define neighborhoods, and from neighborhoods obtain the other topological concepts of open sets, closed sets, accumulation points, and so on. Our proof that a closed, bounded set is compact used the completeness of ℝ in a crucial way. Show that this result does not necessarily hold in an ordered field that is not complete. For an ordered field F as given below, find a subset of F that is bounded and closed in F, but not compact.
(a) Let F be the ordered field of rational numbers ℚ.
(b) Let F be the ordered field of rational functions
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