Question

For non-empty sets A and B, if A ⊂ B then (A × B) ∩ (B × A) is equal to 
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Answer 1

Answer:

see ur answer!!!

Step-by-step explanation:

If A & B are two non- empty subsets and we have to prove AxB=BxA iff A=B

Proof:

We will prove this in two parts. In first part we will prove that if A=B then AxB=BxA and in the second part we will prove that if AxB=BxA then A=B.

i) For the first part let us assume that A=B

Then in AxB we can first replace first ‘A’ with ‘B’ (As by assumption A=B) so that it becomes BxB. Now we have AxB=BxB. In the next step we replace ‘B’ with ‘A’ so that BxB can be written as BxA.

Thus we have AxB=BxB=BxA

ii)For the second part we assume that AxB=BxA and then we will prove that A=B. We will prove this by double containment. We will prove that A is subset of B and then B is subset of A .

Let x∈ A and y∈B

Now (x, y) ∈ AxB

But by our assumption AxB and BxA are equal

So (x,y) ∈ BxA implying that x ∈B , sine x is an arbitrarily chosen element, so A is subset of B.

Now let x∈B and y∈A

So (x,y) ∈ BxA

But again since BxA=AxB; therefore x∈A. Thus implying as before that Bis subset of A

Thus from double containment viz. A being subset of B and B being subset of A; we get A=B.

Thus (i) and (ii) complete the proof.

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Answer 2

Answer:

For non-empty sets A and B, if A ⊂ B then (A x B) ∩ (B x A) is equal to :

A = B.

Step-by-step explanation:

If A and B are two non-empty subsets, and if A=B, then we must demonstrate that AxB=BxA.

Proof – There are two elements to our argument. We shall demonstrate in the first section that if A=B, then AxB=BxA, and in the second section, that if AxB=BxA, then A=B.

Considering that A=B, let's go on to part one.

Then, in AxB, we can first swap out the initial "A" with "B" (on the assumption that A=B), resulting in BxB. Since then, AxB=BxB. In the following step, "B" is changed to "A" so that BxB can be written as BxA.

This gives us AxB=BxB=BxA.

In order to prove that A=B in the second section, we shall make the assumption that AxB=BxA. We'll use double confinement to demonstrate this. We will demonstrate that A is a subset of B, after which B is a subset of A.

Suppose xe A and ye B

Here, (x, y) E AXB

However, based on our presumption, AxB and BxA are equal.

Given that x is a randomly chosen element and that (x,y) E BxA implies that x EB, A is a subset of B.

Let EB and vEA now. 

So (x,y) E BxA

However, as BxA=AxB once more, EA follows. As before, this implies that Bis is a subset of A. 

A=B is the result of double containment, where A is a subset of B and B is a subset of A.

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