DSE 2012 Option A

[MI]

QUESTION 18. A and B are two non empty sets.
A-B = {x belongs to A | x does not belong to B} and

A + B = (A - B) U (B- A)

Consider the following statements is true or false
Statement 1: A + B = B implies A IS SUBSET OF B

The doubt regarding this problem is for statement 1 to be true A has to be a empty set. Which contradicts the given condition.

Please help regarding this.

[srawan]

getting same problem

[duck]

Hi.. :)

Given, A+B = B
⇒ A+B ⊆ B and B ⊆ A+B
⇒ (A-B) ∪ (B-A) ⊆ B   [From A+B ⊆ B]
⇒ (A-B) ⊆ B
⇒ A ⊆ B
Hence, proved!

[MI]

Hi Duck :)

Th problem I am facing is as follows

A + B = (A-B) U (B-A)

This can also be written as (using venn diagram) as

A + B = (A U B) - (A ∩ B) ......(I)

Lets assume

A⊆B

=> (A U B) = B   ALSO  (A ∩ B) = A    ...........(II)

Using (I) AND (II)

A+B=A-B

WHICH IS TRUE ONLY WHEN B= ∅

Doesn't this contradict with the initial assumption " A and B are two non-empty sets"?

Please let me know if there are any logical fallacy in my approach.

Thanks :)

[Sinistral]

How did u infer  A ⊆ B from (A-B) ⊆ B (in general, given A and B being non empty) ?

[Sinistral]

duck wrote

Hi.. :)

Given, A+B = B
⇒ A+B ⊆ B and B ⊆ A+B
⇒ (A-B) ∪ (B-A) ⊆ B   [From A+B ⊆ B]
⇒ (A-B) ⊆ B
⇒ A ⊆ B
Hence, proved!

focussing at the bold part of your text:

A-B means all those elements of A which are not present in B. ie if we remove all the elements of A which are in B and if still this set is a subset of B implies it still contains some elements (at least 0 elements) of B. But this contradicts our definition of A-B since A-B cant have any element which is present in B. So A has to be an empty set...
what am I missing?

[duck]

Hey Thanks Sinistral for pointing out that.. :)
Sorry for the earlier proof. Please ignore that.

Please see the below proof:
Given: A+B = B
To show : A⊆B
Proof: Let, x∈A . If we could show that x∈B then, we'll be done!
Lets prove it by contradiction. Supose, x∉B
⇒ x∈ (A-B)
⇒ x∈ A+B
⇒x∈ B which is a contradiction.

Therefore, x∈B.
Hence, A⊆B

[duck]

Hey MI.. :)

What you're showing is "If A⊆B then A+B=B".
And the above statement doesnot hold as pointed by you.
Because, if A⊆B then, A-B will be an empty set.
⇒A+B = B-A ≠ B ( as A is a non empty set)

[Sinistral]

ok mathematically ur proof is just fine. but somehow its not convincing me. (read below quoted message)

duck wrote

Hey Thanks Sinistral for pointing out that.. :)
Sorry for the earlier proof. Please ignore that.

Please see the below proof:
Given: A+B = B
To show : A⊆B
Proof: Let, x∈A . If we could show that x∈B then, we'll be done!
Lets prove it by contradiction. Supose, x∉B
⇒ x∈ (A-B)
⇒ x∈ A+B
⇒x∈ B which is a contradiction.

Therefore, x∈B.
Hence, A⊆B

now suppose this proof is good. and  (given) A+B = B ⇒ (to prove) A⊆B
lets verify this. I'll go in sync with ur proof.
now since its given that A⊆B and u said it will be all fine if for x∈A if we can show that x∈B. And it has been proved also.
so now lets verify it:
    x∈B ⇒ x∉ A-B
    x∈A ⇒ x∉ B-A
⇒ x∉ A+B
⇒ x ∉  B
contradiction. why???

or putting it more simply can you give me example of such sets A and B which satisfy this: A+B = B ⇒ A⊆B

[duck]

First of all, i didnt assume A⊆B. It was what i proved.
Secondly, i couldn't understand.
"so now lets verify it:
    x∈B ⇒ x∉ A-B
    x∈A ⇒ x∉ B-A
⇒ x∉ A+B
⇒ x ∉  B
contradiction. why??? "
Its not at all sync with the proof which i gave. I think you haven't read the proof carefully Or you're misinterpreting!

Thirdly, A+B = B can never hold for any non empty sets A and B. So, the antecedant is always false. Hence, the statement A+B=B ⇒ A⊆B will hold trivially.

[Sinistral]

I went ahead with all these proofs and verification because I dint want to use the line A+B=B ⇒ A⊆B will hold trivially at the very first instance. Now this will hold trivially; doesn't it violate the initial assumption given in the question A and B are two non empty sets??

ofcourse statements 1,2,3 are true but only if we assume A and B to be any sets (including non empty sets). whereas statement 2 & 3 are valid for even non trivial cases.

[duck]

If the antecendant is false then, P⇒Q holds. Thats what "trivially hold" mean.
Now, A+B = B ⇒ A⊆B holds even for non empty sets A and B. As the antecendant (A+B = B) is always false for non empty sets A and B.

[Sinistral]

thank you so much. :)
In your last reply you said "so the antecedent is always true". this got me confused.

I think now its clear. thanx once again :)

[maahi]

i didn't get d last step . when x belongs to a +b why does it imply that it belongs to b ? doesn't it mean that it can be in a or b ?

[Sinistral]

because its given that A+B=B

[maahi]

ooh ya thanks :-)

[MI]

Hey duck...,

I lot happened when I was away.... :P Just trying to catch up

I got my mistake. Thanks for pointing that out.

I didn't get the following step in the revised proof you gave.

-------------------------------
Given: A+B = B
To show : A⊆B
Proof: Let, x∈A . If we could show that x∈B then, we'll be done!
Lets prove it by contradiction. Supose, x∉B
⇒ x∈ (A-B)
⇒ x∈ A+B
⇒x∈ B which is a contradiction.

Therefore, x∈B.
Hence, A⊆B

--------------------------------------

How  x∈ A+B can imply x∈ B??

Thanks in advance :)

[MI]

Yeah got it....Missed that in a hurry.

Thanks :D