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Real Analysis Proof

  • #1
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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...
 

Answers and Replies

  • #2
12,336
8,725

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 \}\,?##
 
  • #3
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44
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?
 
  • #4
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8,725
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.
 
  • #5
1,456
44
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 \}##?
 
  • #6
12,336
8,725
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##
 
  • #7
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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.
 

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