Approximating $\inf\sup$

Question

Let $f\colon A \times B \rightarrow \mathbb{R}$. I am reading a paper in which some quantity of the form $$ \inf_{a\in A}\sup_{b\in B}f(a,b) $$ is equal to $$ \inf_{a\in A^{\prime}}\sup_{b\in B^{\prime}}f(a,b) $$ where $A^{\prime}\subset A$ and $B^{\prime}\subset B$.

To establish equality, the author claims it is sufficient to show that for all $\epsilon>0$, $a\in A\setminus A^{\prime}$, and $b\in B\setminus B^{\prime}$ there exists $a^\prime\in A^{\prime}$ and $b^\prime\in B^{\prime}$ such that $$ \left|f(a,b)-f(a^{\prime},b^{\prime})\right|\leq\epsilon. $$ However, I am not sure I follow this argument. Is this true? Why?

Answer 1

Let $A' = A$ be some subset of $\mathbb R$, $B' = [0,1]$, $B = [0,2]$ and $f(a,b) = b$.

Then $$\inf_{a\in A} \sup_{b\in B} f(a,b) = \inf_{a\in A} \sup_{b\in [0,2]} b = 2$$ and $$\inf_{a\in A'} \sup_{b\in B'} f(a,b) = \inf_{a\in A'} \sup_{b\in [0,1]} b = 1.$$

These quantities are not equal, but $A\setminus A' = \emptyset$, so the stated condition is true.

Answer 2

There is also a simple counterexample if the set inclusions are strict:

Let $c\in [0,1)$ be some constant, $A = [c,\infty)$ with subset $A' = [1,\infty)$, and $B = [0,2]$ with subset $B' = [0,1]$. Define $$f\colon A\times B\to \mathbb R,\quad (a,b) \mapsto ab.$$

Then the $\varepsilon$-condition holds, but

$$\inf_{a\in A}\sup_{b\in B} f(a,b) = \inf_{a\in [c,\infty)} \sup_{b\in[0,2]} ab = \inf_{a\in[c,\infty)}2a = 2c$$ and $$\inf_{a\in A'}\sup_{b\in B'} f(a,b) = \inf_{a\in [1,\infty)} \sup_{b\in[0,1]} ab = \inf_{a\in[1,\infty)}a = 1$$

and we have $2c > 1$, $2c = 1$ or $2c < 1$ if $c > 1/2$, $c= 1/2$ or $2 < 1/2$, respectively.