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Every bounded sequence of real numbers contains a subsequence with a finite limit of a sequence. If $|x_n|\le M$ for every positive integer $n$, there are integers $n_1<n_2<\cdots$ and a real number $\ell$ such thatThe increasing indices matter: a subsequence retains the original order and selects terms arbitrarily far along the original sequence. The theorem guarantees the existence of at least one such selection; it does not claim that the entire sequence converges or that the subsequential limit is unique.^[ref]

For example, the bounded sequence $x_n=(-1{)}^n+1/(n+1)$ has two distinct subsequential limits:The even terms form a convergent sequence, and so do the odd terms. The full sequence cannot have a limit of a sequence because its two subsequences have different limits. Its limit superior is $1$ and its limit inferior is $-1$.

This is useful when a construction produces many candidate points but does not immediately produce a convergent sequence. For a continuous function on a nonempty closed bounded feasible region in a finite-dimensional Euclidean space, take a sequence whose objective values approach the infimum. A convergent subsequence supplies a candidate point; a closed subset retains that point, and continuity passes the objective values to the limit of a sequence. This is the mechanism behind attainment in the extreme value theorem. The diagonal subsequence argument extends the real-sequence theorem to finite-dimensional Euclidean space.

The theorem is named for Bernard Bolzano and Karl Weierstrass.^[ref]

Consider a stipulated continuous-allocation problem. Amounts $x,y$ of two activities use $x+2y$ units of a common resource, with budget six. The feasible regionis a triangle with extreme points $(0,0)$, $(6,0)$ and $(0,3)$. It is the intersection of three closed half-spaces, hence a closed convex set. The general intersection argument is short: a line segment between two points in every constraint set belongs to every constraint set, so it belongs to their intersection.^[ref]

For an original numerical check, take $A=(1,2)$ and $B=(5,1/4)$. Their resource totals are $5$ and $11/2$. Every convex combination $Z(t)=(1-t)A+tB$ is nonnegative and has resource totalIn particular the midpoint is $(3,9/8)$, using $21/4$ units. This verifies the full line segment, rather than just a plotted sample of it.

This is useful whenever divisible quantities share linear limits: the feasible choices remain compatible under averaging. A new linear equation also preserves this property, because averaging two identical constraint values gives the same value. The affine subspace and closed half-space viewpoints explain this without requiring a separate argument for every budget.

The convex hull of a set consists of every finite convex combination of its points. It collects all choices obtainable by mixing the original choices with nonnegative weights summing to one. For $S$ in a real vector space,Equivalently this is the smallest convex set containing $S$. Mixing two already mixed points produces another finite convex combination, so the displayed set is convex; every convex set containing $S$ must contain all these mixtures.^[ref]

In an original planar example, take four points $A=(0,0)$, $B=(4,0)$, $C=(3,3)$, $D=(0,2)$ and four additional samples $(1,1)$, $(2,1.7)$, $(3,0.5)$, $(1,0.5)$. Their convex hull is the filled convex polygon $ABCD$, including its boundary. None of the samples is $p=(2,1)$, butThe coefficients are nonnegative and sum to one, providing an explicit membership certificate. Geometrically, first form $U=(3/2,0)=\frac{5}{8}A+\frac{3}{8}B$ on the line segment $AB$, then mix $p=\frac{2}{3}U+\frac{1}{3}C$. The additional interior samples do not enlarge the convex hull.

For a finite nonempty $S\subset {\mathbb{R}}^d$, the convex hull is the image of a probability simplex of weights under the linear map $\lambda \mapsto {\sum }_i{\lambda }_i{s}_i$. The probability simplex is a compact set, and the map is continuous, so this convex hull is compact and closed. A one-point input stays one point; a collinear input stays in a line segment. Filling the convex hull does not necessarily produce a full-dimensional region.

The construction is useful whenever mixtures preserve the intended meaning of a choice. Mixing resource-allocation vectors, for example, interpolates their coordinate requirements. The convex hull describes the resulting feasible region. A coordinate-wise box can admit combinations that no mixture produces, as the outside-point certificate in linear optimization on a convex hull shows.

A convex set permits every intermediate mixture of two of its points. If two choices meet its constraints, continuously averaging them also meets those constraints. This geometric property makes many feasible regions easier to explore: there is no forbidden gap along a straight move between feasible choices.

Formally, a subset $C$ of a real vector space is convex whenThe parameter $t$ traverses the entire line segment, including both endpoints. The definition does not require a closed subset, a bounded set, or a nonempty interior. The empty set, the whole vector space, and every affine subspace are convex; an open unit disk and a closed unit disk are both convex.^[ref]

The triangle inequality verifies the disk example directly: if $\Vert x\Vert ,\Vert y\Vert \le 1$, then $\Vert (1-t)x+ty\Vert \le (1-t)\Vert x\Vert +t\Vert y\Vert \le 1$. For the open disk, the last inequality is strict. The unit circle fails the same test: $(1,0)$ and $(-1,0)$ belong to it, but their midpoint $(0,0)$ does not. A curved route along the unit circle connects the endpoints, so being path-connected is weaker than being a convex set.

The distinction between a convex set and a convex function matters. The former concerns where one may choose a point; the latter concerns how a value changes along line segments. Convex optimization combines both: averaging preserves feasibility, and a convex function obeys an upper bound by the corresponding average of endpoint values. A convex set alone does not make an arbitrary objective a convex function or eliminate its inferior local minima.

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 $x+y\ge 4$ to the feasible triangle with a squared-distance objective. No point can have both $x+y\le 3$ and $x+y\ge 4$. 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 $+\infty$.^[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 $x\ge 0$ 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.

Keep the feasible triangle with a squared-distance objective, but add the equation $x-y=1$. An equality requires an exact relation, whereas an inequality allows a range. Here the equality selects a line, and intersecting that line with the original feasible region leaves just the line segment from $(1,0)$ to $(2,1)$:The lower bound comes from $y\ge 0$; the upper bound comes from $x+y=2t+1\le 3$. This is a nonempty closed set even though it has empty interior in the plane. An equality can reduce the dimension of the allowed decisions without making the feasible region empty.

The old optimum $(1.5,1.5)$ violates the added equality. On the surviving line segment, the objective function isThe derivative shows that the unique global minimum is at $t=1$, giving $(x,y)=(2,1)$ and value $1$. The nearest point on the entire equality line would be $(2.5,1.5)$, at $t=1.5$ with value $0.5$, but it violates $x+y\le 3$. Both the equality and the original inequalities must hold simultaneously.

Use $g=x+y-3$ and $h=x-y-1$ to fix the multiplier sign convention in the optimization Lagrangian $L=f+\lambda g+\nu h$; the Lagrange multipliers of $-x$ and $-y$ are again zero. At $(2,1)$, Karush-Kuhn-Tucker conditions hold with $\lambda =1$ and $\nu =-1$:The Lagrange multiplier of an equality has no nonnegativity restriction. For an independent global certificate, completing the square gives $f+g-h=(x-2{)}^2+(y-1{)}^2+1$. On the feasible set, $h=0$ and $g\le 0$, so $f\ge 1$, with equality exactly at $(2,1)$.

The set of all feasible points of an optimization problem. It depends on the optimization constraints and the decision domain; the objective function ranks its members.

Think of three successive questions. Which decisions can be represented in the model? Which satisfy every optimization constraint? Which of those receive the best objective function value? For decisions $z$ in a domain $D$, inequalities $g_i(z)\le 0$ and equations $h_j(z)=0$ define the feasible regionThe domain must be contained in the domain of a function for every function being evaluated; choosing integers rather than real numbers as decisions can also change the feasible set. Constraint satisfaction determines membership of $F$. Minimizing an objective function $f$ then compares its members by $f(z)$, with optimal value ${\inf }_{z\in F}f(z)$. Changing the scores while keeping $D$ and the optimization constraints fixed leaves $F$ unchanged, provided the new objective function is defined throughout $D$. These distinctions separate an allowed decision from a best decision.^[ref]

Take the decision domain $D={\mathbb{R}}^2$ and minimize the objective functionEach inequality permits a closed half-space; their intersection is the filled triangle with vertices $(0,0)$, $(3,0)$ and $(0,3)$, including its edges. The feasible region is a compact set and a convex set. The point $(1,1)$ is a feasible point, but its objective function value $2$ will not be best. Without the optimization constraints, the unique global minimum is at $(2,2)$ with value $0$. That point is infeasible because $2+2>3$.

The objective function is the squared Euclidean distance from $(2,2)$. Its positive level sets are circles centered there. Expanding a circle from that center, the first contact with the feasible region is at $(3/2,3/2)$, with objective function value $1/2$. Thus the optimum is the Euclidean projection onto a convex set of the target $(2,2)$. Squaring a nonnegative Euclidean distance preserves which point is nearest.^[ref]

Here is a direct certificate, without relying on the picture. Put $s=x+y$ and $d=x-y$. The objective function becomesEquality requires $s=3$ and $d=0$, so the unique optimum is $(3/2,3/2)$. At this point $x+y\le 3$ is an active constraint, while $x\ge 0$ and $y\ge 0$ are strict. An active constraint identifies a tight restriction at a specified point; it alone does not establish optimality. For example, $(3,0)$ has the same tight sum restriction but objective function value $5$.

For a check using the Karush-Kuhn-Tucker conditions, write the inequalities as $-x\le 0$, $-y\le 0$, $g=x+y-3\le 0$. Their Lagrange multipliers at the optimum are $(0,0,1)$. The gradient equation isFeasibility, nonnegative Lagrange multipliers and complementary slackness also hold. With this strictly convex function and affine optimization constraints, the Karush-Kuhn-Tucker conditions certify the unique global minimum.^[ref]

If the amounts in the budget example must be integers, the actual feasible region is $C\cap {\mathbb{Z}}^2$. Both $(0,0)$ and $(1,0)$ are feasible, but their midpoint $(1/2,0)$ is not. Thus the continuous feasible region is a convex set while the integer-restricted choices are not. Averaging two manufacturing plans may specify half an indivisible item. Integer programming must respect that restriction; using the surrounding convex set is a relaxation, not a proof that its averages are admissible integer plans.

A feasible point $x$ is an extreme point solution of a linear program if it cannot be written as a nontrivial convex combination of two distinct feasible points. Explicitly, ifthen $y=z=x$. Equivalently, $x$ lies in no open line segment joining two distinct feasible points. Being an extreme point concerns the feasible region; it does not by itself mean that $x$ optimizes the objective function.

Only $z,w$ have printed nonnegativity restrictions; $x,y$ are unrestricted. In four-dimensional space, a vertex of a linear polyhedron needs four independent active constraints. There are exactly four available constraints, so all must be active. Solving them givesand therefore the only possible vertex of a linear polyhedron is $(x,y,z,w)=(2,3,0,0)$. It is feasible, and its four active normals are independent, so it is indeed a vertex of a linear polyhedron.

One can also see uniqueness geometrically from the slack variables $s_1=1-2x+y-5w$ and $s_2=3-y-z-w$. The feasible region is in affine bijection with the nonnegative orthant in $(s_1,s_2,z,w)$:That orthant has only the origin as an extreme point, giving the same unique vertex of a linear polyhedron.

After removing the two redundant inequalities, the feasible region isIts lower boundary consists of three line segments or rays: $y_2=6-3y_1$ for $0\le y_1\le 4/3$, then $y_2=3-3y_1/4$ for $4/3\le y_1\le 16/7$, then $y_2=y_1-1$ for $y_1\ge 16/7$. The feasible region is everything on or above this boundary in the first quadrant.

The boundary extreme points areAlong its three successive pieces, the objective function $2y_1+y_2$ has respective slopes $-1$, $5/4$ and $3$ as $y_1$ increases. Thus it decreases to the middle extreme point and increases thereafter. For fixed $y_1$, moving above the boundary only increases the objective function. HenceAs an algebraic check of the graphical answer, the active inequalities givewith equality at the stated extreme point.

The two primal constraints are equalities, so let the two dual variables $u,v$ be free. Primal feasibility restricts $x$ to the nonnegative orthant, givingThus the feasible region lies above all four lines:Its lower boundary follows the third line up to $(5/9,7/9)$, the fourth line from there to $(10/3,-1/3)$, and then the first line. The second line never determines the envelope. Along these three pieces $u+v$ respectively has slopes $-3$, $3/5$, and $3/4$, so its unique minimum is at the first corner:

At this dual point the first two inequalities are strict, while the last two are equalities. complementary slackness therefore suggests $x_1=x_2=0$. The remaining primal equations become $4x_3+2x_4=300$, $x_3+5x_4=300$, giving $x_3=x_4=50$. This is nonnegative and satisfies both equations. Its objective function is $3\cdot 50+5\cdot 50=400$, equal to the feasible dual value. weak duality provesNo nonnegativity restriction was imposed on the dual variables; the negative-$v$ second corner is part of the feasible boundary, even though it is not optimal.

The solution set of a mathematical condition $P$ in a specified domain $D$ collects every allowed value that satisfies it:It describes the answer without committing to the steps used to find it. For the equation $x^2=1$ over the real numbers, the solution set is $\{-1,1\}$; restricting to positive real numbers leaves $\{1\}$. For $2x=1$, the solution set is $\{1/2\}$ over the rational numbers and the empty set over the integers. The domain is part of the question.

An equation can have one solution, several solutions, no solutions, or an entire domain of solutions. Over the real numbers, $2x+3=9$ gives the singleton set $\{3\}$, $x^2+1=0$ gives the empty set, and the mathematical identity $(x+1{)}^2=x^2+2x+1$ gives the whole real line. Changing that second equation's domain to the complex numbers gives $\{-i,i\}$. The elementary definition by values that satisfy an equation and the classification by solution sets appear in College Algebra 2e, §2.2: Solving Linear Equations in One Variable.

Requiring several conditions simultaneously takes the intersection of their solution sets. The squared equation $x=(x-2{)}^2$ has solution set $\{1,4\}$. Adding the restriction $x\ge 2$ giveswhich is also the solution set of $\sqrt{x}=x-2$. The extraneous solution example explains why the sign restriction is needed. Requiring at least one of several conditions instead takes their union. This turns logical combinations of constraints into reusable operations on sets.

For several variables, a solution is an ordered assignment, usually represented by a tuple. The real equation $x+y=1$ has solution set $\{(t,1-t):t\in \mathbb{R}\}$, a straight line in the Euclidean plane with coordinates $(x,y)$. Adding $x-y=0$ intersects that straight line with another, leaving the single tuple $(1/2,1/2)$. A system of linear inequalities has no solution exactly when the intersection of its individual solution sets is the empty set. For optimization constraints, this common solution set is the feasible region.

Because a solution set is a set, the order in which its members are listed is irrelevant. The multiple root of $(x-1{)}^2=0$ is still the one solution $1$; its multiplicity of a root is additional polynomial information, rather than a second occurrence inside the solution set. Equivalent equations preserve the solution set on the stated domain, even if their algebraic forms differ.