Find a nonempty subset of Q (rational numbers) that is bounded above but has no least upper bound in Q. Justify your claim. Thanks for the help!
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Originally Posted by jstarks44444 Find a nonempty subset of Q (rational numbers) that is bounded above but has no least upper bound in Q. Justify your claim. Thanks for the help! What about the set $\displaystyle \{x| (x \in \mathbb{Q}) \cap (0 \le x^2 < 2) \}$
Originally Posted by jstarks44444 Find a nonempty subset of Q (rational numbers) that is bounded above but has no least upper bound in Q. Justify your claim. Thanks for the help! This is the quintessential example: $\displaystyle \{q\in\mathbb{Q}:q^2<2\}$ Now the details of the proof are yours to do.
in general, any set Q ∩ (a,b) (where (a,b) is an open interval of the reals) where b is irrational will do. why is this so?
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