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Let $X$ and $Y$ be nonempty sets and let $h : X\times Y \to R$ have bounded range in $\mathbb{R}$. Let $F: X \to\mathbb{R}$ and $G : y \to \mathbb{R}$ be defined by $F(x):=\sup\{h(x,y) : y\in Y\}$, and $G(y) := \sup\{h(x,y) : x\in X\}$. Establish the Principle of the Iterated Suprema: $\sup\{h(x,y) :x \in X, y \in Y\} = \sup\{F(x) : x \in X\} = \sup\{G(y) : y \in Y\}$.

This proof is quite long but im having trouble with it. Here is what I have so far.

Proof Since $h(x,y)$ is bounded we know that $\sup\{h(x,y) :x \in X, y \in Y\}$ exists. Let $T=\sup\{h(x,y) :x \in X, y \in Y\}$. By definition we know that $T \geq h(x_0,y_0)$ for arbitrary $x_0 \in X$ and $ y_0 \in Y$ Since $h(x_0,y_0) \in\{h(x,y) :x \in X, y \in Y\}$ and $h(x_0,y_0) \in \{h(x_0,y) : y \in Y\}$ Since $x_0$ and $y_0$ are arbitrary it follows that T is a upper bound for $F(x_0)= \sup\{h(x_0,y) : y\in Y\}$. $\sup\{F(x) : x \in X\}$ exists since $F(x_0)$ was arbitrary. Thus $T \geq \sup\{F(x) : x \in X\}$

This is where I stopped at since I do not know if what I have done so far is correct and I am stuck proving $T \leq \sup\{F(x) : x \in X\}$ in order to show $T=\sup\{F(x) : x \in X\}$.

Community
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user60887
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    Please not that one can write $X\times X\to R$; one need not write $X\times X->R$. Similarly, just writing \sup rather than sup cause the reader to see $\displaystyle a\sup_{b\in c} d$ instead of $a sup_{b\in c} d$. Thus the backslash results in (1) non-italicization; (2) proper spacing; (3) positioning of subscripts. ${}\qquad{}$ – Michael Hardy Jun 22 '13 at 06:10

2 Answers2

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I am omitting the $x \in X$, etc, for simplicity.

$\sup_{x',y'} h(x',y') \ge h(x,y)$ for all $x,y$. Hence $\sup_{x',y'} h(x',y') \ge G(y)=\sup_{x'} h(x',y)$ for all $y$, hence $\sup_{x',y'} h(x',y') \ge \sup_{y'}G(y')$. Similarly for $F$.

Let $\epsilon>0$, then we have $x_1,y_1$ such that $h(x_1,y_1) \ge \sup_{x',y'} h(x',y') - \epsilon$. Clearly $\sup_{y'}G(y') \ge G(y_1) \ge h(x_1,y_1) \ge \sup_{x',y'} h(x',y') - \epsilon$. Since $\epsilon$ was arbitrary, we have $\sup_{y'}G(y') \ge \sup_{x',y'} h(x',y')$. Similarly for $F$.

copper.hat
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To show $T \leq \sup_x F(x)$, note that $\sup_x F(x) \geq h(x,y)$. This implies that $\sup_x F(x) $ is an upper bound for $h(x,y)$. Therefore, $T \leq \sup_x F(x)$.

UberRose
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