Extremums of $\sin(\frac{n\pi}{6}): n \in \mathbb{Z}$, and $\frac{1}{n} + \sin(\frac{n\pi}{2}) : n \in \mathbb{N}$

Question

I can't formalize an answer, but I can visualize it as functions. Any help here?

Supremum, infimum, maximum and minimum of the sets:

• $\sin(\frac{n\pi}{6}): n \in \mathbb{Z}$, and

• $\frac{1}{n} + \sin(\frac{n\pi}{2}) : n \in \mathbb{N}$.

PS: (My Answer)

For the first one: Supremum is 1 and infimum is −1. The maximum is 1 and minimum is −1. The second, supremum is 2, infimum is −1 and maximum is 2 and minimum is −1. Am I right?

Answer 1

I'm gonna help you with the supremum cases for the first set, in a more rigorous way, and with that I think you can resolve the others by yourself.

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Supremum for $S = \{x \in \Re : x = sen(\frac{n\pi}{6}), n \in Z\}$:

• First, you need to show that the set is superiorly limited, i.e., $\exists$ $c$ $\in \Re $ such that $\forall x \in S, x \leq c$. How do we do that? In this case, with a guess. Choose $4$ and prove that all elements of $S$ are less or equal to $4$. Now we know that $c$ is, at least, a upper quota. We can do it for $1$, I've only choose $4$ for clarity.
• To find out our lower upper quota, let's say $sup\{S\} = s$, there's several approach, but in the end, we must show that $\forall \varepsilon > 0, \exists x' \in S: s - \varepsilon < x' \leq s$. It can be confuse, but we are saying that, if we move a little bit to the left in the real line, we are already jumping an element of $S$.

Now, let's jump to the analysis.

1. $\forall x \in S, x \leq 1$. Using the Fundamental Theorem of Trigonometry, we can prove that by contradiction. Suppose there exists such $x > 1$, so $x^2 + cos^2(\frac{n\pi}{6}) = 1 \Leftrightarrow cos^2(\frac{n\pi}{6}) = 1 - x^2 < 0$. Absurdo.
2. $\forall \varepsilon > 0, \exists x' \in S: 1 - \varepsilon < x' \leq 1$. Plugging our function $\sin(\frac{n\pi}{6}) > 1 - \varepsilon \Leftrightarrow \varepsilon > 1 - \sin(\frac{n\pi}{6})$. Choosing $n = 3$, we get $\sin(\frac{\pi}{2}) = 1$, so $\varepsilon > 0$.

Ok, $1 = sup\{S\}$

Detail: since the exercise is so simple, in the second step we could just say that, $1 \in S$, so $1 = max\{S\}$, thus $1$ is $sup\{S\}$, since in every subset of the Real numbers, if this subset has a maximum, the maximum is the supremum of this set because the real number is a complete ordered field.