Question

Let A, B be sets
f:A-->B
let A1,A2,....An be subsets of A
Give an example show \(f(\cap_i A_i)< \cap_i f( A_i)\)
Please help

Answer 1

@myininaya

Answer 2

what does that subscript i mean next to the intersection
i understand the subscript on the A

Answer 3

that just show A_i is arbitrary

Answer 4

\[f(A_1 \cap A_2 \cap \cdots \cap A_n)=A_1 \cap A_2 \cdots \cap A_n \]
this is what we want to give an example for ?

Answer 5

and f is a function

Answer 6

f:A->B be defined by f(x)=x should work , right?

Answer 7

nope, f (intersection subsets) < intersection of f ( subsets)
strictly lesser than

Answer 8

omg i didn't see the inequality

Answer 9

do u mean
order of f(union ) < order (union )

Answer 10

typo ,intersection :P

Answer 11

for example
A1 = [0,1,2]
A2=[0,1]
then \(A_1\cap A_2= [0,1]\)
let define f(x) =x^2 , then \(f(A_1\cap A_2)\) = [0,1]
\(f(A_1) =[0,1,4]\)
\(f(A_2)= [0,1]
\)
and \(f(A_1)\cap f(A_2)=[0,1]\)
I always see they are = , not <

Answer 12

I mean \(f(A_1\cap A_2)=f(A_1)\cap f(A_2)=[0,1]\)
but my question is : is there any function which makes the left hand side strictly < the right hand side?

Answer 13

in that last part you didn't mean to say f(A1) instersect f(A2) right?
I don't see any f's on the right hand side or our inequality.

Answer 14

hmm i was looking for my note , i thought somewhere this is written as a theorem .
\(f(A_i\cap A_j)=f(A_i)\cap f(A_j) \)

thats why i asked what does th < notations even mean :O
like order or comparing sets hmmm

Answer 15

of* not or

Answer 16

and see no f in the right side , however
this is theorem
\(f(\cap_i A_i)=\cap_i (f A_i )\)

and what ur trying to show is different :3

Answer 17

https://www.youtube.com/watch?v=jZXHzpq-vmM&list=PLbMVogVj5nJSxFihV-ec4A3z_FOGPRCo-&index=2
at 7:02

Answer 18

oh ok :O
A1 = [0,2,4]
A2=[2,1,3]
then A1∩A2=[2]
let define f(x) =x! , then f(A1∩A2) = [2]
f(A1)=[1,2,24]
f(A2)=[2,1,6]
f(A1)∩f(A2)=[2,1]

does this mean
f(A1∩A2)< f(A1)∩f(A2) ?

Answer 19

Perfect!! thanks so much.

Answer 20

but f is not function?

Answer 21

since f(0) = f(1), right?

Answer 22

:O
yeah discreat xD

Answer 23

it is a discrete function
(loser , are you think about injective because this isn't an injective function)

Answer 24

Yes, It is good enough to show the case. Since the prof didn't ask for a special function, we can choose any of them to show.

Answer 25

ok then it make sense now
if f in injective then
\(f(\cap_i A_i)=\cap_i (f A_i )\)

if u take ur example f=x^2
u could only change sets to make an example for
ur question :) ( assuming its not asking for 1-1 or onto )

Answer 26

yes, my bad. hehehe.

Answer 27

hehe <3

Answer 28

I need watch a lot of video tapes to master the concept. I can't be failed. hehehe

Answer 29

whats ur class ?

Answer 30

real analysis

Answer 31

cool ^_^
ok then i would like to see all of ur post
even if i dint know answers i would like to review it , ok ?
i have took it 3 years ago , so i dont wanna forget it :O

Answer 32

good student.!!