-2
$\begingroup$

In an exercise we supposed that a number $b$ was the supremum of the set defined in the title.

Question: Is $\cos b < 1/4$ or =$1/4$ ?

Can anybody enlighten me please ?

  • 2
    Sure you are not after the supremum of $\{x\in[0,2\pi]\mid\cos x<\frac14\}$? – 2017-01-17
  • 0
    Yes you are right it makes more sense ! And would the cosinus of the supremum be equal or less than 1/4 ? – 2017-01-17
  • 0
    @Asaf You do not like that the tag (set-theory) is used for all and everything? But set theory **is** all and everything, right? :-) – 2017-01-17
  • 0
    @Did: I don't understand why questions involving sets of *numbers* get tagged with [set-theory] and not with [number-theory]... :| – 2017-01-17
  • 0
    @Asaf Because every number is a set (or a class of sets?) while not every set is a number? Or is it? – 2017-01-17

1 Answers 1

3

Your set has no supremum (it has no upper bound at all, much less a least one). Therefore the assumption that $b$ is such a supremum leads to a contradiction, and you can conclude that both $\cos b < \frac14$ and $\cos b = \frac14$ are true.

Perhaps what the exercise really intended was something like the supremum of $\{x\in[0,2\pi]\mid \cos x < \frac 14 \}$? In that case you would first solve the inequality to find that the set is equal to the open interval $(\arccos\frac14, 2\pi-\arccos\frac14)$, and the supremum of an open interval is its right endpoint, so $b=2\pi-\arccos\frac14$. Then you just need to take the cosine of this $b$, giving $\frac14$.

Note however, that this result depends critically on what you guess is supposed to be the range of $x$. For example, if we instead suppose that $x\in(-\pi,\pi]$, then we have $b=\pi$ and $\cos b = -1 < \frac14$.

  • 0
    True, but you might add the argument the author of the exercise expects, that, if $b$ existed, one would have $\cos b=\frac14$. – 2017-01-17