Question

Give reasons

Answer 1

0 is an element in A so we can write \[0 \in A\]

Answer 2

If thats what you asked for

Answer 3

okay
and what about second

Answer 4

what means P{P(P(A))} ?

Answer 5

we have to find the no. of elements,

Answer 6

power of A

Answer 7

power of A ? i cant caught it

Answer 8

given , A={1)
Power of set A=P(A) = {Φ, {1}}

Answer 9

acc. to me ,, answer should be 2^3=8

Answer 10

looks like subset ?

Answer 11

yes

Answer 12

but i think my answer is wrong

Answer 13

your question doesnt seem to be asking for the cardinality of it, but rather the elements of it

Answer 14

had to zoom out on it, number of elements is cardinality, ok

Answer 15

this should be simple enough:

A = {1}
P{A} = { { } , {1} }
P(P{A}) = { { } , { } , {1}, { { } , { 1 }} }
= { { } , {1}, { { } , { 1 }} }
= {a,b,c} , for simplicity

P(P(P{A})) = { { }, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c} }

= { { }, { }, {1}, {{ { } , { 1 }}}, { { } , [1]}, { { }, {{ },{1}}}, {{1},{{ },{1}}}, { } , {1}, { { } , { 1 }}}

ow my eyes!!

Answer 16

P(P(P{A})) = { {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c} } = 7 distinct elements

Answer 17

the problem is, that with each new power, you reintroduce a null set that is already an element.

Answer 18

k

Answer 19

actually my teacher solved this, and last step , i added myself

Answer 20

spose we renotate this as null = n, and A = {a}

P(A) = { n , a }

P(A(A)) = {n, n, a, {n,a}} = {n, a, {n,a}}

P(A(A(A))) = {n, n, a, {n,a}, {n,a}, {n,{n,a}}, {a,{n,a}}, {n, a, {n,a}} } so we actually intrduced 2 like elements that time

Answer 21

okay

Answer 22

P(A(A(A))) = {n, a, {n,a}, {n,{n,a}}, {a,{n,a}}, {n, a, {n,a}} } = 6 elements

Answer 23

what do we call a subset that is fully contained inside of another set?

Answer 24

@amistre64
thank you