2

Am I doing this right?

  1. S={$(-1)^n$|n$\in$natural numbers}
    I got that
    S =\begin{cases} -1, & \text{if $n$ is odd} \\ 1, & \text{if $n$ is even} \end{cases}

  2. S={$(-1)^n$n|n$\in$natural numbers}
    I got that
    S =\begin{cases} -\infty, & \text{if $n$ is odd} \\ \infty, & \text{if $n$ is new} \end{cases}

ematth7
  • 719

1 Answers1

1

The supremum and infimum of a set are numbers, when they exist; you can tell from this a priori that your answers can't be correct. ($\sup S$ is never a case statement, as you've given -- it's either a number, or else it doesn't exist).

For 1), note that $x \leq 1$ for all $x \in S$. Further, $1 \in S$. These together imply that $\sup S = 1$. The same argument shows that $\inf S = -1.$

For 2), the important point is that for every $N \in \mathbb{N}$, there exists $x \in S$ with $|x| > N$. This implies that $\sup S$ does not exist, and similarly for the infimum.

  • 1
    And then if $x_n$=$(1.5)^n$ for all n$\in$ natural numbers, would inf(S) be 1.5 and sup(S) not exist since it just keeps increasing? – ematth7 Sep 17 '15 at 23:18
  • @ematth7 That's completely right. – Eric Tressler Sep 17 '15 at 23:20
  • and so I wouldn't be able to find the $lim supx_n$ – ematth7 Sep 17 '15 at 23:23
  • @ematth7 Not quite. Knowing that $\lim \sup S$ is equal to $\infty$ or $-\infty$ is not quite the same as not being able to find it; you have understood something about the behavior of $S$ if you can say something like that, and some authors will allow $\sup$ and $\inf$ to take extended real values (i.e. $\mathbb{R} \cup {\infty,-\infty}$). This stands to reason, since making the statement that "$\sup S = \infty$" is saying a lot more than "I don't know $\sup S$". Whether the infimum or supremum "exists" in these cases is left up to semantics, and it doesn't really matter how you say it. – Eric Tressler Sep 17 '15 at 23:28
  • @ematth7 to be complete, I'll add that in the case you described, $\lim \sup x_n$ and $\lim \inf x_n$ are both $\infty$, which is fine, and better than saying that we can't find them. – Eric Tressler Sep 17 '15 at 23:45