Let X and Y be non-empty sets, and let h : X × Y → R be a bounded function.(i) Prove thatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y) = supx∈Xsupy∈Yh(x, y).Hint) Interchanging the roles of X and Y if necessary, it is enough to prove thatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y)
Question
Let X and Y be non-empty sets, and let h : X × Y → R be a bounded function.(i) Prove thatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y) = supx∈Xsupy∈Yh(x, y).Hint) Interchanging the roles of X and Y if necessary, it is enough to prove thatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y)
Solution
To prove the equality, we will show that both sides of the equation are less than or equal to each other.
First, let's show that supy∈Y supx∈X h(x, y) ≤ sup(x,y)∈X×Y h(x, y):
For any y in Y, supx∈X h(x, y) is less than or equal to sup(x,y)∈X×Y h(x, y) because the supremum over a larger set is greater than or equal to the supremum over any of its subsets. Since this is true for any y in Y, it follows that supy∈Y supx∈X h(x, y) is also less than or equal to sup(x,y)∈X×Y h(x, y).
Next, let's show that sup(x,y)∈X×Y h(x, y) ≤ supy∈Y supx∈X h(x, y):
For any (x, y) in X×Y, h(x, y) is less than or equal to supx∈X h(x, y) because the supremum over a set is greater than or equal to any element in the set. Since this is true for any x in X, it follows that h(x, y) is also less than or equal to supy∈Y supx∈X h(x, y). Since this is true for any (x, y) in X×Y, it follows that sup(x,y)∈X×Y h(x, y) is also less than or equal to supy∈Y supx∈X h(x, y).
Therefore, we have shown that supy∈Y supx∈X h(x, y) = sup(x,y)∈X×Y h(x, y).
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