Let X and Y be non-empty sets, and let h : X × Y → R be a bounded function. Provethatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y) = supx∈Xsupy∈Yh(x, y).Hint) By symmetry, we can interchange the roles of X and Y if necessary. Therefore, it isenough to prove thatsupy∈Ysupx∈Xh(x, y) = sup(x,y)∈X×Yh(x, y)

To prove this, we will use the fact that for any two real numbers a and b, we have max(a, b) = max(b, a). This is a direct consequence of the definition of the maximum function.

Step 1: We first show that supy∈Y supx∈X h(x, y) ≤ sup(x,y)∈X×Y h(x, y).

Let ε > 0 be arbitrary. Since supy∈Y supx∈X h(x, y) is the least upper bound of the set {supx∈X h(x, y) : y ∈ Y}, there exists y0 ∈ Y such that supx∈X h(x, y0) > supy∈Y supx∈X h(x, y) - ε.

Similarly, since supx∈X h(x, y0) is the least upper bound of the set {h(x, y0) : x ∈ X}, there exists x0 ∈ X such that h(x0, y0) > supx∈X h(x, y0) - ε.

Therefore, we have h(x0, y0) > supy∈Y supx∈X h(x, y) - 2ε. Since ε > 0 was arbitrary, we conclude that supy∈Y supx∈X h(x, y) ≤ sup(x,y)∈X×Y h(x, y).

Step 2: We now show that sup(x,y)∈X×Y h(x, y) ≤ supy∈Y supx∈X h(x, y).

Let (x0, y0) be an arbitrary element of X × Y. Then we have h(x0, y0) ≤ supx∈X h(x, y0) ≤ supy∈Y supx∈X h(x, y).

Since (x0, y0) was arbitrary, we conclude that sup(x,y)∈X×Y h(x, y) ≤ supy∈Y supx∈X h(x, y).

Combining Steps 1 and 2, we obtain the desired equality.