Answer to Question #123529 in Real Analysis for Tau

Question #123529
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.
1
Expert's answer
2020-06-23T20:18:56-0400

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\n\u200b" , 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


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