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I need to prove that $\sup A = \frac{1}{2}$ and $\inf A = -1$ for:

$$A = \{\frac{(-1)^n}{n}, n\in \mathbb{N}\}$$

Well, for $n\in 2\mathbb{N}$, we have:

$$A' = \{\frac{1}{n}, n\in 2\mathbb{N}\} = \{\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \cdots\} = \{\frac{1}{2n}, n\in \mathbb{N}\}$$

In this case, we have that $n\ge 1\implies 2n\ge 2 \implies \frac{1}{2n}\le \frac{1}{2}$. Well, $\frac{1}{2}$ is an upper bound of $A'$ and is in $A'$ therefore there can't be another upper bound $c$ that is both in the middle ($\frac{1}{2n}<c<\frac{1}{2}$) because is must be higher than $\frac{1}{2}$ too. So $\sup A = \frac{1}{2}$

And, for $n\in 2\mathbb{N}-1$, we have:

$$A' = \{\frac{-1}{n}, n\in 2\mathbb{N}-1\} = \{-1, \frac{-1}{3}, \frac{-1}{5}, \cdots\} = \{\frac{-1}{2n-1}, n\in \mathbb{N}\}$$

Well, $n\ge 1\implies -n \le -1\implies \frac{-1}{n}\ge -1$ but $\frac{-1}{2n-1}\ge\frac{-1}{2n}\ge\frac{-1}{n}\ge-1$, therefore, $-1$ is a lower bound AND is in the set $A$, therefore by the same argument as before, $\inf A = -1$

Well, since I split the set $A$ in $2$ sets, I must also prove that $\sup A\ge \sup A'$ and $\inf A\le \inf A'$, but this is a theorem, right?

Andrés E. Caicedo
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Guerlando OCs
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1 Answers1

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Observe that $\frac12\in A$ is an upper bound of $A$ and that $-1\in A$ is a lower bound of $A$. As you remarked yourself in your former question that allready justifies the conclustion that $\inf A=-1$ and $\sup A=\frac12$.

I tried to explain that in my answer to your former question.

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drhab
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  • yes but suppose I couldn't use the argument that if the element is in the set and it's a lower bound, it's the inf, is my reasoning correct? – Guerlando OCs Apr 27 '16 at 16:06
  • You proved for subset $A_{+}={a\in A\mid a>0}$ and $A_{-}={a\in A\mid a<0}$ that $\sup A_{+}=\frac12$ and $\inf A_{-}=-1$. From that you are indeed allowed to conclude that $\sup A\geq\sup A_{+}=\frac12$ and $\inf A\leq\inf A_{-}=-1$ by what you call a theorem but there is no profit in that. To come to an end you must show that $\sup A=\sup A_{+}$ and $\inf A=\inf A_{-}$. – drhab Apr 27 '16 at 19:16