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A convex polytope is a bounded intersection of finitely many closed half-spaces, equivalently the convex hull of finitely many points.

The line segment joining vectors $x$ and $y$ is

A closed half-space is a set of the form $\{x:a^{T}x\le b\}$ for a nonzero vector $a$; it is a convex set and is closed.

This is the non-strict version of a half-space; replacing the inequality by a strict one gives the open version.

A convex combination is a finite weighted average of points in a real vector space. For $n\ge 1$, it has the formThe nonnegative coefficients forbid subtracting a component, and their unit sum keeps the result on the same affine scale as the inputs. With two points it parametrizes their line segment; equal coefficients give their centroid. A convex set contains every finite convex combination of its points. Conversely, the two-point case already gives the defining property of a convex set.^[ref]

The distinction from a linear combination is concrete. Combining $(0,0)$ and $(2,0)$ with coefficients $-1$ and $2$ gives $(4,0)$, outside their line segment, despite the unit coefficient sum. Using two coefficients both equal to one gives $(2,0)$ here, but their sum is two, so it is not a convex combination with those coefficients. A numerical output may happen to coincide with a point admitting a different convex representation; the definition applies to the specified coefficients.

The nonnegative orthant consists of vectors whose coordinates are all nonnegative. Its interior is the positive orthant, where every coordinate is positive.

It is the closed orthant obtained by choosing the nonnegative sign for every coordinate.

For a convex set $C$ and a differentiable convex objective $F$, projected gradient descent uses

This is the proximal gradient method when the nonsmooth term is the indicator of the feasible convex set, whose proximal operator is the metric projection.

For a real matrix $P$, exactly one of the following holds: there is $\varphi$ with $P\varphi \ge 0$ and $P\varphi \ne 0$, or there is $q>0$ with $P^{T}q=0$. Separating $\mathrm{Im}P$ from the standard simplex proves the result; $q$ may then be normalized to sum to one.

This is an alternative related to Farkas' lemma. It is not the strict-primal Gordan alternative: for the column matrix $(1,0{)}^T$, a nonnegative nonzero primal vector exists, while a componentwise strictly positive one does not.

For a convex function $f$, the subdifferential at $x$ isThe point $x$ minimizes $f$ exactly when $0\in \partial f(x)$.

The swap test on pure states $|a⟩,|b⟩$ returns ancilla zero with probability

The Hadamard test estimates the real part of $⟨\psi |U|\psi ⟩$ from one ancilla measurement.

The controlled-NOT gate maps $|a,b⟩$ to $|a,b\oplus a⟩$.

Every set lying between a connected set and its closure is connected.

The product of any family of connected spaces is connected. For two factors, fix $y_0$ and use the connected setsThey cover $X\times Y$ and all contain the connected horizontal slice $X\times \{y_0\}$.

If connected subsets cover a space and every two of them intersect, then their union is connected. Equivalently, any continuous integer-valued function is constant on each member, and the intersections force all those constants to agree.

A topological space is connected exactly when every continuous map from it to the discrete space $\mathbb{Z}$ is constant. A disconnection produces a nonconstant indicator map, while the continuous image of a connected space in $\mathbb{Z}$ must be a singleton.

If $f:X\to Y$ is a continuous map and $X$ is a connected space, then its image $f(X)$, with the subspace topology from $Y$, is a connected subset of $Y$. The whole codomain $Y$ need not be connected; the assertion concerns the values actually attained. No injectivity is required.^[ref]

For example, the continuous function $f(x)=x^2$ maps the connected closed interval $[-2,2]$ onto the connected closed interval $[0,4]$, even though opposite inputs have the same output. The same continuous function maps the disconnected union $[-2,-1]\cup [1,2]$ onto $[1,4]$. Thus a connected image does not imply a connected source: a continuous map can merge different connected components.

For a continuous function $r:[0,1]\to \mathbb{R}$ representing a continuous reading, $r(0)=0$ and $r(1)=1$ force some input with $r(t)=1/2$. Its image is a connected subset of the real line, hence a real interval containing both endpoint readings. For the synthetic model $r(t)=t^2$, the missing-value test is explicit: $r(1/\sqrt{2})=1/2$. A step reading equal to $0$ for $t<1/2$ and $1$ for $t\ge 1/2$ skips $1/2$, but fails continuity at $t=1/2$.

Proof outline: a separation of a topological space of $f(X)$ into nonempty open subsets pulls back under $f$ to nonempty open subsets of $X$. Their intersection is empty and their union is $X$, contradicting connectedness. This also shows why a continuous map sends each connected component of its source into a single connected component of its target.

A connected component of a topological space $X$ is a nonempty connected subset that is maximal under subset inclusion. If $C$ is a connected component and $C\subseteq D\subseteq X$ with $D$ a connected subset, then $D=C$. Maximality means that no larger connected subset contains $C$; it does not mean largest cardinality among all connected subsets.

For example, the connected components of $[-2,2]\setminus \{0\}$ in the usual subspace topology of the real line are $[-2,0)$ and $(0,2]$. Each is a real interval, and no connected subset of the real line can meet both without containing the missing point $0$. For the finite set $\{-2,0,2\}$ in the same subspace topology, the three connected components are its singleton sets.

Every point $x\in X$ lies in exactly one connected component, constructed by taking the unionAll members contain $x$, so their union is connected by connected union with a connected hub. The union includes every connected subset through $x$, so it is maximal. Equivalently, defineThis is an equivalence relation: singleton sets give reflexivity, symmetry is immediate, and two connected subsets sharing the middle point have a connected union, giving transitivity. The equivalence classes are precisely the connected components. They form a partition of a set: they cover $X$ and any two are either equal or disjoint.^[ref]

Every nonempty connected subset lies within one connected component. Moreover, every connected component is closed in $X$: closure of a connected set makes its closure connected, and maximality forces that closure to equal the original connected component. A connected component need not be open. For example, in $S=\{0\}\cup \{1/n:n=1,2,\dots \}$ with its usual subspace topology from the real line, all connected components are singleton sets, but $\{0\}$ is not open: every neighbourhood of $0$ in $S$ contains some $1/n$.^[ref]

The map $z\mapsto \cosh (\pi z)$ is a conformal bijection fromonto the upper half-plane. Its three boundary intervals map respectively to $(1,\infty )$, $(-1,1)$ and $(-\infty ,-1)$.

If a holomorphic function $f$ satisfies $f^{\prime }(z_0)\ne 0$, then it has a holomorphic local inverse near $z_0$ and is conformal there. Its possible failures of local conformality are therefore its critical points.

Two plane domains are conformally equivalent when there is a bijective holomorphic map between them whose inverse is holomorphic.