The meaning of active and non-active constraints

Question

Consider the constrained minimization problem

min $f(x), x \in \mathbb{R^n}$

s.t $h_i(x)=0, i=1,2,...m$

$g_i(x) \leq 0 , i=1,2,..k$

Now the author states:

" For a feasible solution $x$, some of the inequality constraints can be satisfied at $x$ as strict in- equalities (i.e., $g_i (x) < 0$), and some – as equalities: $g_i (x) = 0$. The inequality constraints of this latter type are called active at $x∗$ , and those of the former type – nonactive. The reason for this terminology is clear: for a nonactive at $x$ inequality constraint we have $g_i(x) < 0$; from continuity of $g_i$ it follows that the constraint is satisfied in a neighbour- hood of $x$ as well; in other words, such an inequality locally does not participate in the problem: it makes no influence on feasibility/infeasibility of candidate solutions close to $x$ (of course, “far” from $x$ such an inequality can also come into the play). In contrast to this, an active at $x$ inequality cannot be neglected even in a small neighborhood of the point: normally, it influences feasibility/infeasibility of close to $x$ candidate solutions."

I'm really confused at a feasible point $x$ where $g_i(x) < 0$ and by the continuity of $g_i$ now we know that in some neighborhood $N_r(x)$ of $x$ the constraint $g_i(y) < 0$ is met $\forall y \in N_r(x)$ so it tells us that all the $ys$ in $N_r(x)$ are feasible also. Then why he states that "such an inequality locally does not participate in the problem:it makes no influence on feasibility/infeasibility of candidate solutions close to $x$" it has influence since it tells us the $ys$ are feasible!!

Answer 1

The inequality constraints essentially give you the set where you are allowed to search for your solutions (i.e. the domain of your problem). If an inequality constraint is not binding, this roughly means that you are in the interior of the set where you are allowed to search for solutions. In particular, since you are in the interior, you may fit a small ball between this "feasible point" and the "walls" enforced by your constraint.

Interpreting this small ball we may fit in between the candidate $x$ and the constraints as candidates close to $x$, one reobtains:

such an inequality locally does not participate in the problem:it makes no influence on feasibility/infeasibility of candidate solutions close to $x$.

The point here is that it does not locally change the behavior of your problem, although it for sure may impact the global properties (such as deciding the feasible set).

Answer 2

Let's look at a simple example, with two constraints. The domain is the real number line. The constraints are $$ x \le 2 \\ x \le 3. $$ Now look at the point $x = 2$. Both constraints are satisfied; the first (at this point) is an equality constraint, the second is a strict inequality, because $2 < 3$.

Near $x = 2$ (say, for $1.9 < x < 2.1$) the second constraint is always satisfied, so all points in that region are feasible for the second constraint. That does not make them feasible solutions to the set of inequalities however. So your assertion that " it has influence since it tells us the ys are feasible!!" is mistaken. The strict inequality tells us that the $y$s are feasible for that one constraint, not for the overall problem.