Question

question set 2

Answer 1

question f ) show that every monotonic decreasing sequence which is bounded below ,is convergent

Answer 2

@BSwan

Answer 3

ok i found my old precious txt book of real analysis xD
and it happend that this is a thm
wanna meto type the proof ?

Answer 4

take snap n post ne?

Answer 5

my web cam is not working :O
something wrong with its sitting
ill type it

Answer 6

oh okay

Answer 7

its called
MONOTONE CONVERGENCE THEOREM
wait i might have a pdf ...

Answer 8

okay

Answer 9

lolz in book it only proved increasing xD
decreasing =exercise xD

Answer 10

oi you jumped to last question? :P see the image

Answer 11

https://proofwiki.org/wiki/Monotone_Convergence_Theorem_%28Real_Analysis%29#Proof_for_Decreasing_Sequence

Answer 12

i thought u only wanted f

Answer 13

no it wasnt clear,so i wrote it again

Answer 14

sweet
thanks

Answer 15

ok "a" is sandwich thm for limit of a function ,
state the thm then apply it

Answer 16

got it
next/

Answer 17

b-
\(\lim_{n \rightarrow \infty} \text{inf } x_n\le \lim_{n \rightarrow \infty} \text{sup } x_n\)
this need a proof :P
do u know wats the sup and inf ?

Answer 18

they would be two cases for \(x_n\)
1_ increasing
2_ decreasing

Answer 19

supremum n infimum,well not much..not as in i'd use them as data for solution

Answer 20

u should know the def of both :O
why jugdar why u dnt know them :'(

Answer 21

oh
its not like i dont know?
partialy ordered set's subset's greatest element is infimum?

Answer 22

ok ill give u example :-
let T be a set (5,8)
then we would say \(S_1 \) is the least element of T \(S_1 \subset \) T
so we would know S_1 would have elements like 5.0001 , 5.00001...ect
then the inf =5 ( the least of the least )
which is the smallest elemnt of all \(S_1\)

Answer 23

sorry imade a typo
lim inf =5
cuz 5 not in T

Answer 24

inf can be any number close to that

Answer 25

oh the least one

Answer 26

so inf of {1,2,3} would be 1,like that

Answer 27

and sup would be 3

Answer 28

yep, but u gave a distent set ( which is not familior to be given ) lolz

Answer 29

try with seq like 1/x or stuff

Answer 30

ok -.-
so how do we solve this one

Answer 31

by contradiction and epsilon delta def XD

Answer 32

...

Answer 33

okay so again u'd be trying to prove inverse but it contradicts so the given must be true?

Answer 34

i hv to go out n buy some stuff -.-
i will cya in 2 hours maybe :)
thanks for helping me <3

Answer 35

ok ill type the proof ^_^
for any \(X_n\)
sup \( x_n \le\text{ inf } x_n\)

so for all x\(\in\) upper bound
inf < =x
then limit inf <= x
limit inf <= sup


so for all x\(\in\) lower bound
sup < =x
then limit sup >= x
limit sup >= inf>=limit inf
limit sup >=limit inf

Answer 36

oh these all are so easy answers
wish i was tought these again,wouldnt even take lots of time to teach some basics

Answer 37

how to prove c

Answer 38

ok could u type the cauchy deff for me ?

Answer 39

if <Xn> is sequence in R then it is cauchy sequence if
∀ϵ∈R:ϵ>0:∃N:∀m,n∈N:m,n≥N:|xn−xm|<ϵ

Answer 40

if ϵ>0 is given we choose a natural number H = H(ϵ) such that H>2/ϵ , then
if m,n >= H we have \(\large \frac{1}{n^2}\le 1/H < \frac{ϵ}{2}\) and similary
\(\large \frac{1}{m^2} < \frac{ϵ}{2}\) therefore it follows that if m,n \(\geq H\) then :-

\(\large |\frac{1}{n^2}-\frac{1}{m^2}| \le \frac{1}{n^2}+\frac{1}{m^2} <\frac{ϵ}{2} +\frac{ϵ}{2} =ϵ\)
since ϵ > 0 is arbitrary , then 1/n^2 is a cauchy sequence

Answer 41

wew !

Answer 42

thats easy :O

Answer 43

ikr :)

Answer 44

will keep this proof in ur mind and only change the seqence lol
but the rest of the proof would be the same each time :P

Answer 45

yeah :D

Answer 46

ok rest i knw

Answer 47

thanks ^^

Answer 48

cool :)
np lol