Question

Let x_n is in R and suppose that there is an M in R such that | x_n | less than or equal to M for n in N. Prove that s_n=sup{x_n, x_(n+1), ...} defines a real number for each n in N and that s_1>s_2>....

Answer 1

Solution

Let $x_n$ is in $R$ and suppose that there is an $M$ in $R$ such that $|x_n|$ less than or equal to $M$ for $n$ in $N$. Prove that $s_n = \sup\{x_n, x_n + 1, \ldots\}$ defines a real number for each $n$ in $N$ and that $s_{1} \geq s_{2} \geq m$:

$$\exists M, \forall n \in N: |x_n| \leq M \text{ (comment } M \geq 0 \text{)} \Rightarrow$$

By definition $s = \sup \{x_n \mid n \in N\}$ follows: $-M \leq \sup \{x_n\} \leq M \Rightarrow s \in [-M; M] \Rightarrow s \in R$

By condition: $s_r = \sup \{x_n \mid n \in N \setminus \{1, 2, 3, \ldots, r-1\}\}$

But $\forall r \in N: s_r = \sup \{x_n \mid n \in N \setminus \{1, 2, 3, \ldots, r-1\}\} = s_r = \sup \{x_n \mid n \in (N \setminus \{1, 2, 3, \ldots, r\}) \cup \{r\}\} = \sup \{s_{r+1}; x_r\} \geq \forall n \in N: s_{r+1} \Rightarrow s_1 \geq s_2 \geq s_3 \geq \ldots s_n \geq s_{n+1}$