Real Analysis Proof

[Mr Davis 97]

Homework Statement

Suppose that $( s_n )$ and $(t_n)$ are bounded sequences. Given that $A_k$ is an upper bound for $\{s_n : n \ge k \}$ and $B_k$ is an upper bound for $\{t_n : n \ge k \}$ and that $A_k + B_k$ is an upper bound for $\{s_n + t_n : n \ge k \}$, show that $\sup \{s_n + t_n : n \ge k \} \le \sup \{s_n : n \ge k \} + \sup \{t_n : n \ge k \}$

Homework Equations

The Attempt at a Solution

Here is what I know. Since $A_k$, $B_k$ and $A_k + B_k$ are upper bounds, we can conclude that $\sup \{s_n + t_n : n \ge k \} \le A_k + B_k$, that $\sup \{s_n : n \ge k \} \le A_k$ and that $\sup \{t_n : n \ge k \} \le B_k$. I think this might be a start, but I am not sure where to go from here...

 

[fresh_42]

Homework Statement

Suppose that $( s_n )$ and $(t_n)$ are bounded sequences. Given that $A_k$ is an upper bound for $\{s_n : n \ge k \}$ and $B_k$ is an upper bound for $\{t_n : n \ge k \}$ and that $A_k + B_k$ is an upper bound for $\{s_n + t_n : n \ge k \}$, show that $\sup \{s_n + t_n : n \ge k \} \le \sup \{s_n : n \ge k \} + \sup \{t_n : n \ge k \}$

Homework Equations

The Attempt at a Solution

Here is what I know. Since $A_k$, $B_k$ and $A_k + B_k$ are upper bounds, we can conclude that $\sup \{s_n + t_n : n \ge k \} \le A_k + B_k$, that $\sup \{s_n : n \ge k \} \le A_k$ and that $\sup \{t_n : n \ge k \} \le B_k$. I think this might be a start, but I am not sure where to go from here...

Say $S=\sup \{s_n : n \ge k \}$ and $T=\sup \{t_n : n \ge k \}$. Then $S+T \le A_k+B_k$. Now is $S+T$ an upper bound for $\{s_n + t_n : n \ge k \}\,?$

 

[Mr Davis 97]

Say $S=\sup \{s_n : n \ge k \}$ and $T=\sup \{t_n : n \ge k \}$. Then $S+T \le A_k+B_k$. Now is $S+T$ an upper bound for $\{s_n + t_n : n \ge k \}\,?$

For $S+T$ to be an upper bound, wouldn't $S+T \ge A_k + B_k$ have to be the case?

 

[fresh_42]

For $S+T$ to be an upper bound, wouldn't $S+T \ge A_k + B_k$ have to be the case?

No, they are both only upper bounds. Which one is smaller cannot be concluded. The trick is, that $(s_n+t_n)$ is a construction with only one index, $(s_n) + (t_n)=(s_n)+(t_m)$ has two indexes, will say the single sets aren't paired, so the supremums could "happen" at two different places. However, $S+T \leq A_k + T \leq A_k+B_k$, so what's left is why $S+T$ is an upper bound for the paired set.

 

[Mr Davis 97]

No, they are both only upper bounds. Which one is smaller cannot be concluded. The trick is, that $(s_n+t_n)$ is a construction with only one index, $(s_n) + (t_n)=(s_n)+(t_m)$ has two indexes, will say the single sets aren't paired, so the supremums could "happen" at two different places. However, $S+T \leq A_k + T \leq A_k+B_k$, so what's left is why $S+T$ is an upper bound for the paired set.

Why can't I just say that $S = \sup \{s_n : n \ge k \} \ge s_n$, $T = \sup \{t_n : n \ge k \} \ge t_n$, and thus $S + T \ge s_n + t_n$, and hence $S+T$ is an upper bound of $\{s_n+t_n : n \ge k \}$, so hence $S+T \ge \sup \{s_n+t_n : n \ge k \}$?

 

[fresh_42]

Why can't I just say that $S = \sup \{s_n : n \ge k \} \ge s_n$, $T = \sup \{t_n : n \ge k \} \ge t_n$, and thus $S + T \ge s_n + t_n$, and hence $S+T$ is an upper bound of $\{s_n+t_n : n \ge k \}$, so hence $S+T \ge \sup \{s_n+t_n : n \ge k \}$?

I don't know. I think you can. I would write it the same way as the one above, but that's only another way to say the same: $\{s_n+t_n : n \ge k \} \leq S + \{t_n : n \ge k \} \leq S+T$

 

[WWGD]

I think this is a nice case to consider: $s_n =(-1)^n ; t_n=(-1)^{n+1}$. EDIT: It may be a nice exercise to determine conditions when the Sup of the sum is the sum of the Sups.