Proposition 4. Suppose (an) is a bounded sequence and let L = lim sup on. Then the set {n : an 2 p} is nite for all p gt; L, and innite for all p...

Prove the following by working backwards from the proof I attached (this is the converse of the proof I attached).

Suppose {an : n ≥ n0} is a bounded sequence of real numbers and L ∈ R. Show if for every ρ>L there is an N so that for all n≥N ,an <ρ and for every ρ<L and for every N there is an n≥N such that an ≥ρ, then L=limsup(an). 

Proposition 4.29. Suppose (an) is a bounded sequence and let L = lim sup on. Thenthe set {n : an 2 p} is finite for all p &gt; L, and infinite for all p &lt; L. Conversely, if L’ is a real number such that {n : an 2 p} is finite for all p &gt; L’ , andinfinite for all p &lt; L’, then L’ = L. T Proof. Suppose p &gt; L. With notation as above7 with e = p — L, there is an N such thatan — L &lt; p — L for all n 2 N. Thus there is an N such that am &lt; p and thus, on &lt; p forall n 2 N. Now suppose p &lt; L. In this case, om 2 L &gt; p for all N. In particular, thereexists an element of the set {am : m 2 N} which exceeds p; i.e., there is an n 2 N suchthat an &gt; p.