Aset x is said to be closed under multiplication if for every x1, x2 e x we have x1x2 e x. let a be the union of all bounded subsets x cr that are closed under multiplication. does inf(a) exist? if it does, find it.

inf(A) does not exist.

Step-by-step explanation:

As per the question:

We need to prove that A is closed under multiplication,

If for every

Proof:

Suppose, x, y

Since, both x and y are real numbers thus xy is also a real number.

Now, consider another set B such that:

B = {xy} has only a single element 'xy' and thus [B] is bounded.

Since, [A] represents the union of all the bounded sets, therefore,

⇒ xy

Therefore, from x, y , we have xy .

Hence, set a is closed under multiplication.

Now, to prove whether inf(A) exist or not

Proof:

Let us assume that inf(A) exist and inf(A) =

Thus is also a real number.

Let C be another set such that

C = { - 1}

Now, we know that C is a bounded set thus { - 1} is also an element of A

Also, we know:

inf(A) =  

Therefore,

But

is an element of A and  

This is contradictory, thus inf(A) does not exist.

Hence, proved.

step-by-step explanation: step 3 should really go to step seven, but anyway never mind.

1 ∠oln ≅ ∠lno

i believe the answer is the polygon, heptagon.