Supremum of Union: Proving Sup(A)=SupiEI(Sup(A))

[CarmineCortez]

Homework Statement

let A_i be a subset of the reals and i is element of I = (1,...,n)

now let A = UNION _{i is element of I} A_i

show that sup(A) = sup_iEI(sup(A))

Homework Equations

The Attempt at a Solution

My idea of solving was taking the limits of both sides but I'm not sure what the limit of the union of A_i is

 

[mighty2000]

. Another approach could be using the definition of supremum and proving that both sides are equal to the least upper bound of the set A.

To start, let's define the supremum of a set A as the least upper bound of the set, denoted as sup(A). This means that sup(A) is the smallest number that is greater than or equal to all elements in A.

Now, let's start with the left side of the equation: sup(A). By definition, this is the smallest number that is greater than or equal to all elements in A. Since A is the union of all sets A_i, this means that sup(A) must also be greater than or equal to all elements in each A_i. In other words, for every element x in A_i, we have sup(A) ≥ x. This is true for all i, since i ranges from 1 to n. Therefore, we can say that sup(A) ≥ sup(A_i) for all i.

Now, let's look at the right side of the equation: sup i∈I (sup(A_i)). Here, we are taking the supremum of each individual set A_i, and then taking the supremum of those values. This means that for each A_i, we have a value that is greater than or equal to all elements in A_i. Therefore, sup(A_i) is an upper bound for A_i, and since i ranges from 1 to n, sup i∈I (sup(A_i)) is an upper bound for all A_i.

Since sup(A) is the least upper bound of A, and sup i∈I (sup(A_i)) is an upper bound for A, we can conclude that sup(A) ≤ sup i∈I (sup(A_i)). This is true for all i, so we can also say that sup(A) ≤ sup(A_i) for all i. Therefore, we have shown that sup(A) is both greater than or equal to all sup(A_i) and less than or equal to all sup(A_i), which means that sup(A) = sup i∈I (sup(A_i)).