Answer to Question #123529 in Real Analysis for Tau

Question

Prove this
Let {xm} be a sequence in Kn, say xm = (x1m,...,xnm). Then
lim m infinity
xm = (x1,..., xn)
with respect to || ||2 if and only if
lim
m infinity
xim = xi
for i = 1,..., n.

Answer 1

let us start with, $\lim_{m \to \infty} {x_m} = (x_1,x_2,x_3, . . . . . . . . . , x_m)$ = {x}

for $\epsilon > 0,$ $n_0 \isin \N$ such that

$|| {x_n} - {x} || < \epsilon$ for $\forall$ $n \geq n_0$

so $n \geq n_0 ​$ , we can say that

$||(x_{1m},x_{2m}, x_{3m}, . . . . . . . . . ,x_{nm}) - ( x_1,x_2,x_3,......,x_n )|| < \epsilon$

$|| (x_{1m}-x_1),(x_{2m}-x_2),.....(x_{nm}-x_n)|| < \epsilon$

$\sqrt { (x_{1m}-x_1)^2+(x_{2m}-x_2)^2+.....(x_{nm}-x_n)^2} < \epsilon$

$\sqrt{{(x_{im} - x_i)}^2} < \epsilon$

$||{(x_{im} - x_i)} ||< \epsilon$

$\lim_{x \to \infty} {x_{im}} = x_i$

Conversely, assume that $\lim_{x \to \infty} {x_{im}} = x_i$

since, for $\epsilon > 0,$ $\exists$ $n_{0i} \isin \N$ such that

$|| x_{im} - x_i|| < \epsilon$ for every value of $n > n_{0i}$ (1)

for each $i=0,1,2,3,4,. .......,n$

Consider the maximum of $(x_{01},x_{02},........,x_{0n})$

then (1) will hold true simultaneously for each i.

$|| x_{im} - x_i || < \epsilon$ $\forall$ $n\geq n_{0i}|_{max}$

$(x_{im}-x_i)^2 < \epsilon^2$ for each i, $n\geq n_{0i}|_{max}$

$(x_{1m}-x_1)^2+(x_{2m}-x_2)^2+......+(x_{nm}-x_n)^2 < n\epsilon^2$

$|| x_m - x || < \sqrt{n}\epsilon$ for each values of i, $n\geq n_{0i}|_{max}$

$\lim_{m \to \infty} {x_m} =$ {x}

Hence proved