Recently, I found your channel, and I must say it is one of the most amazing channels I've come across on these topics. The explanations of each topic are great, delivered with extraordinary enthusiasm
set A is a bounded set because elements are forward to wards or going close to 0, so if it is bounded set then they have upper and lower bounded Upper bounded= [1 to ∞) Lower bounded= (-∞ to 0]
in assinment 1st : A={1,1/2,1/3,.......} upper bound should (1,0) according to question at a point there will be 1/infinity which will be equal to 0. Am i right ??
Ye bonded hy set hy q k ye zero ko approach Kar rahay hen .ye zero or 1 k dermeyan rahy ga . Es ka upper bond bhi hy or lower bond bhi hy .(0,1] Sup=1 Inf= 0
Good method but sir jinko bilkul ku6 nhi aata unko UE math waali language smj ni ayegi...plzz thora Aur simple btaya kren ..I mean mathematical k saath apni wording men b likh diya kren
Jazak Allah sir.
Assignment
Upper bound=[1,infinity)
Lower bound=(−infinity,0]
As the upper and lower bound exist so this is bounded set
I'm totally satisfied with your method
This video saved my life. Finally a clear concept.
Recently, I found your channel, and I must say it is one of the most amazing channels I've come across on these topics. The explanations of each topic are great, delivered with extraordinary enthusiasm
Thanks for appreciation
Jazakillah Sir Thank you Soo much hma PHD doctor n uni m ya concept smjaya pr Zara b smj Nhi lgi pr ap k sb smj lg gai
I am totally satisfied with u
Finally understood this concept 🙏
Upper bound=[1,infinity)
Lower bound=(-infinity,0]
So it is bounded set
Really
Very good method.
I got it
Very well explained!
Simple teaching method that's helpful for me also for others💌
Just trying to be helpful ♥️
Thnks alot sir your lcture is really helpful for me
Excellent method sir
Thank you so much sir god bless you
Good way of teaching really impressed 👍
MashAllah sir great
Sir ap calculus mn talpur ka 4th chaptr upload kr dain because your method of teaching is very good
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Assignment:
Its upper bound is 1 and lower bound is zero so its bounded set
How lower bond is zero?
@@msbajwi3505 qk values km ho rahi hain. 1 divide ho raha h or divide hoty hoty infinite tk jae ga. 1/infinite = 0.
Good
Great method sir jii
Thanku sir
ASSIGNMENT
Upper bound [0,i.finity]
Lower bound[-infinity,1]
It's wrong
love you Sir 💟
Weldone sir
Upper bound [1, ♾️)
Lower bound (-♾️,0]
Yes it is bounded
Nice videos uploded by you
Thanks you so much sir
Your method of teaching very good sir
thanku very well explained
Good explanation ✌️✌️✌️
Great sr
set A is a bounded set because elements are forward to wards or going close to 0, so if it is bounded set then they have upper and lower bounded
Upper bounded= [1 to ∞)
Lower bounded= (-∞ to 0]
Gud
@@m.k.f.amathematicsknowledg1668 Thank you sir
Lower bond kese aya
Upper bound =[1, +infinity)
Lower bound =(-infinity ,0]
Is it correct sir
Right
Upper bound:1 to infinity
Lower bound:0 to -ve infinity
I have also noticed.
Uper bound =[ 1,infinity)
and
Lower bound =[0,-infinity)
Upper bound " 0 to infinity
Lower bound " -ive infinity to 1
Upper bound:{ 1,+infinity)
Lower bound:( _infinity, 0}
in assinment 1st : A={1,1/2,1/3,.......}
upper bound should (1,0)
according to question at a point there will be 1/infinity which will be equal to 0.
Am i right ??
Assignment
Upper bound =[1,infinity)
Lower bound not exist so it not bounded set
Love you sir
(-infinity,1] upper bound and it has no lower bound sir,so it is not bounded set
Aslamoalikum sir is topic ky related exam m question kis terha sy pochy jaty hai . Discrete mathematics ma please zaroor btaiy ga please🙏🙏🙏
upper bounds of A (0 , +infinity)
lower bounds of A (_infinity,1)
A is bounded set
Ye bonded hy set hy q k ye zero ko approach Kar rahay hen .ye zero or 1 k dermeyan rahy ga .
Es ka upper bond bhi hy or lower bond bhi hy .(0,1]
Sup=1
Inf= 0
Nice method..sir
Assignment wala set bounded set ho ga ya nhi????
Here lower bound is 1 and and upper bound is 0 bcs 1,1/2,1/3... Approaches to 0. And is bounded set.
Upper bound 1,infinity
Upper bound=[1, ♾️)
Lower bound=(- ♾️,0]
So,it is bounded.
Is upper bound bigger value and lower bound smaller value. Please tell.
Upper bound=[1,0)
Lower bound=o .it approach to o so it is bounded set. Sir yh dkhn correct hai kia?
Upper bound[0,infinity)
Lower bound(-infinity,1]
This is bounded set
Sir is this correct
upper bound 1 to positive infinity while no lower bound
It's not a bounded set because we don't know about the upper bound
[-infinity , 1]
[⅓ , infinity]
Prove that set of whole num is not bounded above??? Plz sir
Upper bound 1infinate
Here uper bound are 1 and lower bound 0 b.c it approch to zero so it s bounded set
Correct answer
Nice sir
1 uppor ifinty lgana sa 0 lower bounded
Great method sir
Good method but sir jinko bilkul ku6 nhi aata unko UE math waali language smj ni ayegi...plzz thora Aur simple btaya kren ..I mean mathematical k saath apni wording men b likh diya kren
Sir upper bound will be [1, ♾️)
And lower bound is 1/♾️ = 0 so lower bound is ( ♾️ ,0 ]
Lower bound infinity, 0
Aoa sir...may i th k a is also belong to R....but you said a belong to A.
Upper bound 1
Lower bound 0
Upper bound 1 to infinity
And lower bound 0 to _infinity
Plz sir, tell me my answer is correct or not
Sir it approaches to zero but not equal to zero
How we find lower bound??
Plzz answer me
Bounded above and below ni bataya sir
UB=1
LB=0
Upper bound is 1 and lower bound is 0
Please dear response our desirable result ,to solve chapter 1 book naame real analysis writer name wader roder
How [1,1/2.......) Are bounded??????
Because icki limit exist krti h jo k 0 h
upper bound 1 lower bound 0 .it is bounded
Correct answer
Upper bownd 1 to infinity and lower bound not exist
Upper bound doesn't exist
Lower bound is negative infinity to 1
This is not bounded se
Ye bounded b ni h or na ye upar h
Here uper bound are 1 and lower bound 0 b.c it approch to zero so it s bounded set