The full question : Let $X = Y = \{ x \in \mathbb{R} : 0 <x<1 \}$. Define $h:X \times Y \to \mathbb{R}\;$ by $h(x,y) := 2x+y.$
(a) For each $x \in X$, find $f(x):= \sup\{ h(x,y) : y \in Y \}$; then find $\inf \{f(x) : x \in X \}$
(b) For each $y \in Y$, find $g(y):= \inf\{ h(x,y) : x \in X \}$; then find $\sup \{g(y) : y \in Y \}$. Compare with the result found in part (a).
I know that the answer is: $$\begin{aligned}
&(a) :f(x) = 2x+1,\; \inf \{ f(x) \} =1,\\
&(b) :\; g(y)=y,\; \sup\{ g(y) \} = 1,\; \text{ and } \sup\{ g(y)\} \leq \inf\{ f(x) \} \; (\sup\{g(y)\} = \inf\{ f(x)\})
\end{aligned}
$$
I need to understand the technicality behind it, for example here is my attempt at part (a) :
Notice : $$
0 < x < 1 ,\; 0<y<1 \;\Rightarrow\; 2x<2x+y<2x+1 $$
Let $u=2x+1$, Clearly
$$ \forall s \in \{ h(x,y) : y \in Y \}, \;s<2x+1=u$$ Then $u=2x+1$ is an upper bound for $\{ h(x,y) : y \in Y \}$
(Now show that $u$ is the supremum) :
Here I use the lemma that states : $u = \sup\{S\} \iff \forall \varepsilon >0, \exists s_{\varepsilon} \in S : u- \varepsilon < s_{\varepsilon} $
(Finding $y_{\varepsilon}$) Start by setting $$
\begin{aligned}
&u - \varepsilon < s_{\varepsilon} \\
&2x+1 - \varepsilon < 2x+y_{\varepsilon} \\
&1- \varepsilon < y_{\varepsilon}
\end{aligned}
$$
So in the formal answer : Let $\varepsilon >0$ be given, choose $y_{\varepsilon} > 1- \varepsilon \Rightarrow (\cdots) \Rightarrow 2x+1 - \varepsilon < s_{\varepsilon} = 2x+y_{\varepsilon}$
Thus $f(x)=2x+1$ is the supremum for $\{ h(x,y) : y \in Y \}$
I'm worried that my solution is incomplete, especially because I saw someone else's solution and it involved "let $ y_{\varepsilon} = \max\left\{1-\dfrac{\varepsilon}{2},\dfrac{1}{2}\right\}$" but then continues to do exactly as I did afterward.
I apologize in advance in case I was missing something obvious.
max is necessary to make $y_{\epsilon} \in \left ] 0,1 \right [$ since $\epsilon$ is assumed to be positive , so it can get too big making $1-\frac{\epsilon}{2}$ negative (doesn't belong to $Y$).of course in your proof u need to start like this:choose $1>y_{\epsilon}>\max (1-\epsilon,0)$