Proving Properties of Lim Sup: Sequences and Limits

[Bashyboy]

Homework Statement

The limit superior of a sequence $(a_n)$ is defined as $\overline{\lim}_{n \to \infty} a_n = \lim_{n \to \infty} \sup \{a_k ~|~ k \ge n \}$. Letting $L = \overline{\lim}_{n \to \infty} a_n$, I am asked to prove the following:

(i) For each $N$ and for each $\epsilon > 0$, there exists some $k > N$ such that $a_k \ge L - \epsilon$.

(ii) For each $\epsilon > 0$, there is some $N$ such that $a_k \le L + \epsilon$ for all $k > N$.

(iii) $\overline{\lim}_{n \to \infty} ca_n = c \overline{\lim}_{n \to \infty} a_n$

Homework Equations

The Attempt at a Solution

(iii) is rather easy to prove: it just follows from the fact that $\sup(cA) = c \sup (A)$ and $\lim_{n \to \infty} c x_n = c \lim_{n \to \infty} x_n$ for $c \ge 0$, so I will move on to the first and second part

Let $\epsilon > 0$ and $N \in \Bbb{N}$. Since $L = \overline{\lim}_{n \to \infty} a_n$, then there exists a $K \in \Bbb{N}$ such that $| \sup \{a_k ~|~ k \ge n \} - L | < \epsilon$ for every $n \ge K$ or $L - \epsilon < \sup \{a_k ~|~ k \ge n \} < \epsilon + L$ for every $n \ge K$.. On the one hand, we get $a_k \le \sup \{a_k ~|~ k \ge n \} < \epsilon + L$ for every $k \ge K$, which proves part (ii). However, I don't see how to prove part (i). Also, as you may have noticed, I can only get strict inequality in part (ii).

 

[andrewkirk]

To prove part (i), assume it is false, so that there is some $\epsilon>0$ and positive integer $N$ such that $\forall k>N:\ a_k<L-\epsilon$.

Can you get an upper bound for lim sup of the sequence from that, that is less than $L$? If so, that's the contradiction we need.