Answer:
inf(A) does not exist.
Explanation:
As per the question:
We need to prove that A is closed under multiplication,
If for every
![X_(1), X_(2)\in X](https://img.qammunity.org/2020/formulas/mathematics/college/qj5ekzhpy9cv2dddm2xp3o75xqexh8decm.png)
![X_(1)X_(2)\in X](https://img.qammunity.org/2020/formulas/mathematics/college/bdmkws0uixwybdv4xen79gng2qwnpjnwk0.png)
Proof:
Suppose, x, y
![\in A](https://img.qammunity.org/2020/formulas/mathematics/college/dwpqzncm69kb5azp6zqqsvv1op0x5j1e11.png)
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,
![B\subset A](https://img.qammunity.org/2020/formulas/mathematics/college/lts821bdf0b1bm4rd98h39ax1emlhzyosv.png)
⇒ xy
![\in A](https://img.qammunity.org/2020/formulas/mathematics/college/dwpqzncm69kb5azp6zqqsvv1op0x5j1e11.png)
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) =
![\beta](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ifxhqjas18460gou6t6tvzd0b2mntp5172.png)
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,
![n(A)\geq \beta](https://img.qammunity.org/2020/formulas/mathematics/college/bkwvva6re1woiqocccais6p6r02d1o9b87.png)
But
is an element of A and
![\beta - 1 \leq \beta](https://img.qammunity.org/2020/formulas/mathematics/college/df0pjk1pdxjnuqdm4j6jns256ofgrswnsh.png)
This is contradictory, thus inf(A) does not exist.
Hence, proved.