= Empty and open feasible regions
Adding an <optimization constraint> removes possibilities: if it permits a <set> $C$, the new <feasible region> is $F\cap C\subseteq F$. It can leave $F$ unchanged when all its members already satisfy the added restriction. For a fixed <objective function>, restricting the <feasible set> cannot lower a <minimization problem>'s <infimum>. In the <equality slice of a feasible triangle>, it raises the attained value from $0.5$ to $1$.
Instead add $x+y\geq4$ to the <feasible triangle with a squared-distance objective>. No point can have both $x+y\leq3$ and $x+y\geq4$. The resulting <feasible set> is the <empty set>, so there is no <feasible point> and no optimizing decision. Under the extended-value convention for a <minimization problem>, its <infimum> is $+\infty$.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[§4.1.1, printed p. 127: infeasibility and the empty-set infimum convention]{ref}
A nonempty <feasible set> can also lack an optimizing decision. With domain $D=\mathbb R$, minimize $f(x)=x$ subject to the strict <inequality> $x>0$. The <feasible region> $(0,\infty)$ is an <open set>. Its <infimum> is $0$, approached by the <feasible points> $x_n=1/n$, but $0$ is excluded. Every <feasible point> $x$ is beaten by $x/2$, so no <minimum> is attained. Replacing $x>0$ by $x\geq0$ includes the missing optimum. Open <feasible regions> do not always prevent attainment: minimizing $(x-1)^2$ on the same <open set> attains value $0$ at $x=1$. A useful sufficient condition is instead the <extreme value theorem>: a <continuous> <objective function> on a nonempty <compact set> attains its <minimum>.
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