Question

TRUE or FALSE

Answer 1

1. If A\[If A \cup B = A, then B =A\]

Answer 2

2. Given 3 non-empty sets A, B & C, if \[A \neq B\] and \[B \neq C\], then \[A \neq C\]

Answer 3

3. For any two sets X and Y, if x \[\notin \] Y and y \[\notin \] X, then X and Y are disjoint

Answer 4

4. All sets are disjoint with the empty set

Answer 5

5. Let A \[\subseteq\] B. If n(B) = 5 and n(A intersection B) = 3, then n(B-A) = 2

Answer 6

1) \(\mathbb{R} \ne \mathbb{Z}\) but \(\mathbb{R}\cup \mathbb{Z}=\mathbb{R}\)

Do you see this is a counter example?

Answer 7

yes :)

Answer 8

so its false for no. 1? :)

Answer 9

correct

Answer 10

is no.2 true?

Answer 11

can you think of a set A where A = C but both A, C are different from some set B?

Answer 12

There is no real math here

replace set, with number

so a,b, and c are numbers
if a is different than b
and b is different than c
does this imply a is different than c?

Answer 13

yes?

Answer 14

from what I know, equal means they all have the same elements ..

Answer 15

this question is strange...

there is no assumption that A, B, and C are different (the question would not make sense if it did). I would answer it with any two sets that are not the same.

e.g.

Let \(A=C= \mathbb{Z}\) and let \(B=\mathbb{R}\)
then \(A\ne B\) and \(B\ne C\) but \(A=C\)

Answer 16

uhm, yes. I get your point but based on the definition they gave us

A and B are equal, written as A=B, if and only if they have PRECISELY THE SAME ELEMENTS.

???

Answer 17

correct. do you see how I gave a counter example?

Answer 18

is this the first day or so of class?

Answer 19

not exactly, I'm reviewing for a test on Monday :'(

Answer 20

yes yes, I get the example :) thankyou

but the answer for this question is .. ? XD so sorry

Answer 21

if you understood my example you would know the answer...

The question is asking the following.

if two sets A and B are not equal
and B is not equal to some other set C

does this imply that A and C are different?

I gave a counter example.

Answer 22

You need to focus on what the question is asking. If you understood the question, you would see that the answer is very easy to see,

simply replace the word "set" by "number" and see if you understand what is asking, then try and understand the question as its asked.

Answer 23

yep true

Answer 24

like so

if a,b,c are numberes
and \(a\ne b\) and \(b\ne c\)

Does this imply \(a\ne c\)?

OF COURSE NOT
let a = 2, b = 3, c = 2

Answer 25

now similarly

let \(A=\mathbb{R}\) (the set of real numbers)
let \(B=\mathbb{Z}\) (the set of integers)
let \(C=\mathbb{R}\) (again the set of real numbers)

then \(A\ne B\) and \(B\ne C\) but \(A=C\)

Thus the statement is true/false ?

Answer 26

3 is false

consider {1,2,3} and {3,4,5}

Answer 27

they each have an element that is not in the other set, but they are not disjoint because they both contain 3

Answer 28

sorry

its false

I used

A={2,3,4}
B={2,3,5}
C={3,4,6}

Answer 29

for 2?

Answer 30

that does not show its false... we want A and C to be the same. then its a counter example to the claim...

Answer 31

yeah

Answer 32

yeah, its false, didn't think of the other 'claims', sorry

Answer 33

I really don't get #3 :(

Answer 34

what does it mean for two sets to be disjoint?

Answer 35

what part do you not understand?

Answer 36

A and B are said to be disjoint if A intersection B = null set

Answer 37

uhm, x & y are elements right? and x is not an element of Y & y is not an element of X.

so what's this have to do with X & Y being disjoint? :O

Answer 38

ok so can you think of two sets, where the sets are different, but not disjoint

Answer 39

like {1,2,3} and {3,4,5}

Answer 40

yeah

Answer 41

so X={ 1, 2, 3}
Y={3,4,5}

Answer 42

1 is in {1,2,3} but 1 is NOT in {3,4,5}
4 is in {3,4,5} but not in {1,2,3}
They are NOT disjoint because 3 is in both of them.

Answer 43

correct

Answer 44

but what about {5 6 7} and {4 9 8}

Answer 45

this is another claim right?

Answer 46

ohh, I kinda get it. If there are 'other' claims ...

Answer 47

okk so its false, I'll just have to consider other claims when answering these true/false questions right? :))

Answer 48

first, i'll find the 'claim' that makes that agrees with that statement

and i'll have to find 'other' claims that will make it wrong

so I can tell if it's true or false? O_O

Answer 49

everything that comes after "if" and before "then" is the assumptions they are allowing you to make, everything after the "then" is what you need to verify is true, with the given assumptions.

Answer 50

ohh okay

Answer 51

is #4 false?

Answer 52

when we want to show a if/then statement is true, we prove it.
when we want to show a /if/then statement is false, we show a counter example.

so far they have all been false so we are showing counter examples, so if we can think of anything that satisfies the given assumptions, but does NOT agree with the conclusion, then the statement is false. (<-- notice this last sentence was an if/then statement)

Answer 53

can you think of any set that contains an element that is in the empty set?
is there any element in the empty set that is in any other set?

Answer 54

\(\emptyset \cap A = \emptyset\) for any set \(A\)

Answer 55

im confused haha, wait. im trying to understand it

Answer 56

but null sets are subsets of all the sets right?

Answer 57

ohhkay, so its true

Answer 58

riight, we're talking bout the elements here, right?

Answer 59

the null set is a subset of any set, but this is just for ease of notation and other things, sort of like 1^0=1. It has no elements, so its sort of strange thing...

having said that, since it has no elements, it cant "share" any elements with any other set. and if two sets share no elements they are disjoint.

Answer 60

correct

Answer 61

ohh, yes. got it thanks

Answer 62

the last one is true, only because A is a subset of B, if this was not the case it would not be true in general.

Do you need to prove it?

Answer 63

noo, its fine. my answer's true too :D

Answer 64

let \(B = \{a,b,c,d,e\}\), Since \(A\subset B\) and \(|A\cap B|=3\) we may assume without loss of generality that \(A=\{a,b,c\}\). Then we see that \(|B-A| = |\{d,e\}|=2\)

Answer 65

ahh great.

Answer 66

another way to write \(n(A)\) is \(|A|\)

Answer 67

ohh okaay thanks!

Answer 68

this too please

Answer 69

be fast. its 2am

Answer 70

if A is finite set and B is infinite set, then A x B is a finite set. true or false?

Answer 71

if A = {x,y,\[\emptyset\]} and B = {m,\[\emptyset\],n}, then A intersection B = null set

Answer 72

the first one is true, A x B is all the possible pairs (a,b) where a is in A and b is in B

since there are infinite elements in B, surely there will be infinite pairs possible to make with A

Answer 73

the second one is true

these are sets of sets ( we some times call the collections of sets, to make it easier to think about)

so our "collections" have sets as elements.
they only share one element, and that element is the null set.

Answer 74

^^ this is confusing I know, but sets can be elements, it just depends on the context. We could also have collections of sets, where the sets elements are also sets, so a collection of collection of sets.

Answer 75

uhm, the first question says 'then AxB is a finite set' then your answer is 'there are infinite pairs .." O_O

Answer 76

then false... it will be infinite

Answer 77

yay, same answers :D

Answer 78

thankyou so much for your time & patience

Answer 79

all you need is one element in A
say that element is a
then {a} x B is infinite because B is infinite.
similarly \(\emptyset \times B\) is infinite when B is infinite.

Answer 80

ohh, okay. got it

Answer 81

np. This stuff is very confusing at first, just take your time and pm me anytime. Just tell me you are working on set theory stuff and I will help if I am here. I love this stuff:)

Answer 82

good night

Answer 83

good night :) thanks again!

Answer 84

will do :D