Answer to Question #184017 in Real Analysis for Kosma

Question

Let a be a sequence of real numbers and let c ∈ R be a cluster point of a.

Let π:N → N be defined by

π(1) = min{k∈N||ak −c|<1},

π(n+1) = min{k∈N|k>π(n), |ak −c|< 1 } foralln∈N. n+1

(i) Justify the definition of π. (i.e Show that π is well defined.) (ii) Show that π is strictly increasing.

(iii) Prove that the subsequence (aπ(n))N of a converges to c

Answer 1

Solution :-

Let a_n be the sequence of real numbers

And c $\epsilon$ R

$\Pi : N\rightarrow N$

$\Pi(1) = min{[K\epsilon N || ak-c I < 1]}$ ,

$\Pi$ (n+1) = min[k∈N|k>$\Pi$ (n), |ak −c|< 1 ] for

all n∈N. n+1

(i) as we can see $\Pi$ is defined $: N\rightarrow N$

and minimum value of function $\Pi$ comes in $\Pi(1) = min{[K\epsilon N || ak-c I < 1]}$ ,

and also we can see that $\Pi$ (n+1) = min[k∈N|k>$\Pi$ (n), |ak −c|< 1 ] for

all n∈N. n+1

we can say that $\Pi$ is well defined in the given interval

(ii) we can say that , when we do differentiation of the a_n and $\Pi$ is well defined in the given interval

interval is given for $\Pi$

$\Pi : N\rightarrow N$

$\Pi(1) = min{[K\epsilon N || ak-c I < 1]}$ ,

$\Pi$ (n+1) = min[k∈N|k>$\Pi$ (n), |ak −c|< 1 ] for

all n∈N. n+1

we can say it is differentiaval in the given interval , by differentiation function Π is strictly increasing function

(iii)

 (a$\Pi$ (n))N of a converges to c

as we have limits and we can see that

$\Pi : N\rightarrow N$

$\Pi(1) = min{[K\epsilon N || ak-c I < 1]}$ ,

$\Pi$ (n+1) = min[k∈N|k>$\Pi$ (n), |ak −c|< 1 ] for

all n∈N. n+1

by this we can see that every bounded function is convrgence function

{an}n∈N is convergent. Then by Theorem we know that the limit is unique and so we can write it as l

$lim_{n\rightarrow \infin}a_n=l$

or $a_n \rightarrow l$

as $n \rightarrow \infin$

by this we have proved that  the subsequence (a$\Pi$ (n))N of a converges to c , AS $\boxed{c \space\epsilon R}$