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Basic analysis: Proof by induction on sets

  • Thread starter K29
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  • #1
K29
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Homework Statement


Prove by induction that the set [tex] [a_{n} | n_{0}\leq n \leq n_{1}][/tex] is bounded.
[tex]a_{n}[/tex] are the elements of the sequence [tex](a_{n})[/tex]
[tex]n \in N[/tex]

Homework Equations


Definition of set bounded above:
[tex] \forall x \in S, \exists M \in R[/tex] such that [tex]x \leq M[/tex]


The Attempt at a Solution


Just proving its bounded above here...

Base step: [tex][a_{1}][/tex] The set has only 1 element and
[tex]a_{1} \leq a_{1} +1[/tex]

Now assume true for [tex]a_{n}[/tex]
[tex] [a_{n}|n_{0}\leq n \leq n_{1}][/tex]
and [tex] \exists M \in R[/tex] such that [tex]a_{n} \leq M,
forall a_{n}[/tex]

For [tex] [a_{n+1}|n_{0} \leq n+1 \leq n_{1} ][/tex]
..... well I'm not really sure what to do here. Normally you use the assumption to prove it true for n+1, but I'm not sure how to incorporate the assumption here.
Please help
 
Last edited:

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
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Now assume true for [tex]a_{n}[/tex]
[tex] [a_{n}|n_{0}\leq n \leq n_{1}][/tex]
and [tex] \exists M \in R[/tex] such that [tex]a_{n} \leq M [/tex],
for all [tex] a_{n}[/tex]

For [tex] [a_{n+1}|n_{0} \leq n+1 \leq n_{1} ][/tex]
..... well I'm not really sure what to do here
To do induction, you should consider the set [tex] \{ a_n | n_0 \leq n \leq n_1+1 } [/tex]
Say its bound is max{a_{n+1}, M}.

"Bounded" might mean bounded below and above. If it means that in the problem, you should also do the workd for "bounded below".
 
  • #3
K29
108
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Great stuff. solved. Thanks!
 

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