Question

Let f(x,y): R^2-->R, and for each fixed y, let inf(x)f(x,y) denote the greatest lower bound of the set of numbers {f(x,y)|x element of R}. Show that

infx(inf(y)f(x,y)) = infy(inf(x)f(x,y)).

noting that inf(y)f(x,y) <= f(x,y) for all x and y.

Answer 1

We need to prove

$$\inf x \left(\inf (y) f (x, y)\right) = \inf y \left(\inf (x) f (x, y)\right)$$

Let's prove a lemma:

$$\inf _ {x} \left(\inf (y) f (x, y)\right) = \inf _ {x, y} f (x, y)$$

Suppose $\inf_{x,y} f(x,y) = \alpha$ (here $\alpha$ may be $-\infty$ ). Then there exists a sequence of pairs $(x_n, y_n)$ such that

$$\lim _ {n \to \infty} f (x _ {n}, y _ {n}) = \alpha$$

Such inequalities hold:

$$\inf (y) f (x _ {n}, y) \leq f (x _ {n}, y _ {n})$$

Thus

$$\inf _ {x} \inf (y) f (x, y) \leq \inf _ {n} \inf (y) f (x _ {n}, y) \leq \inf _ {n} f (x _ {n}, y _ {n}) = \alpha$$

On the other hand,

$$\forall x \in R \inf (y) f (x, y) \geq \inf _ {x, y} f (x, y) = \alpha$$

Thus,

$$\inf _ {x} \inf (y) f (x, y) \geq \alpha$$

Taking two obtained inequalities together we get:

$$\inf _ {x} \inf (y) f (x, y) = \alpha$$

and the lemma is proved.

Interchanging variables $x$ and $y$ we get such statement:

$$\inf _ {y} \inf (x) f (x, y) = \alpha$$

Taking all together we have:

$$\inf _ {x} \inf (y) f (x, y) = \inf _ {x, y} f (x, y) = \inf _ {y} \inf (x) f (x, y)$$

Thus

$$\inf _ {x} \inf (y) f (x, y) = \inf _ {y} \inf (x) f (x, y)$$