Basic analysis: Proof by induction on sets

[K29]

Homework Statement

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

Homework Equations

Definition of set bounded above:
$$\forall x \in S, \exists M \in R$$ such that $$x \leq M$$

The Attempt at a Solution

Just proving its bounded above here...

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

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

For $$[a_{n+1}|n_{0} \leq n+1 \leq n_{1} ]$$
... 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

 

[Stephen Tashi]

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

For $$[a_{n+1}|n_{0} \leq n+1 \leq n_{1} ]$$
... well I'm not really sure what to do here

To do induction, you should consider the set $$\{ a_n | n_0 \leq n \leq n_1+1 }$$
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".

 

[K29]

Great stuff. solved. Thanks!