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For a complex number $z=x+iy$, its real part is $\mathrm{Re}z=x$.

If $G$ is the principal complex inverse sine, all values reached by analytic continuation areThis follows from the monodromy reflections about the branch values $-\pi /2$ and $\pi /2$.

A theory is $\kappa$-categorical when all of its models of cardinality $\kappa$ are isomorphic.

For the adjacency matrix $A$ of $K_n$,Hence $I,A,A^2$ are linearly dependent for every $n\ge 2$.

Every compactly supported distribution is a finite sum of distributional derivatives of bounded continuous functions. One proof convolves it with a sufficiently high-order Bessel potential and then applies a power of $1-\Delta$.

A space is locally compact when every point has a base of neighbourhoods containing compact neighbourhoods. A compact Hausdorff space is locally compact because regular separation produces $x\in V\subseteq \overline{V}\subseteq U$.

Every compact Hausdorff space is normal: disjoint closed subsets have disjoint open neighbourhoods. Pointwise Hausdorff separation becomes uniform first over one compact closed set and then over the other by taking finite subcovers.

Every open cover of a compact metric space has a number $\delta >0$ such that every subset of diameter less than $\delta$ lies in one member of the cover.

Every quotient of a compact space is compact because the quotient projection is continuous and surjective.

Let $X,Y$ be topological spaces, let $X$ be compact, and let $f:X\to Y$ be a continuous map. Then its image $f(X)$ is compact with the subspace topology inherited from $Y$. More generally, a continuous map sends every compact subset $K\subseteq X$ to a compact subset $f(K)\subseteq Y$, by applying the same statement to its restriction to $K$. Neither $X$ nor $Y$ needs to be Hausdorff.^[ref]

For example, the polynomial function $f(x)=x^2$ sends the compact closed interval $[0,1]$ onto $[0,1]$. The open intervals $V_1=(-1/10,16/25)$ and $V_2=(9/25,11/10)$ give an open cover of the image. Their inverse images in the domain areBoth are open sets in the subspace topology of $[0,1]$, even though neither is an open set in the ambient real line. They cover the domain, so the corresponding two target open sets cover the image.

Matching colors identify the target open sets and their inverse images. Filled endpoints are included and hollow endpoints excluded. This particular open cover illustrates the mechanism; compactness requires a finite subcover for every open cover.

For the general proof, write an arbitrary open cover of $f(X)$ as $\{f(X)\cap V_i:i\in I\}$ with each $V_i$ an open set in $Y$. By continuity, the inverse images $f^{-1}(V_i)$ form an open cover of $X$. Its compactness provides indices $i_1,\dots ,i_m$ such thatThis gives the required finite subcover of $f(X)$. The argument uses inverse images because continuous maps need not send open sets to open sets. For the empty set, the empty selection suffices.

The image of a compact Hausdorff space under a continuous map need not itself be Hausdorff. Take $D=\{0,1\}$ with the discrete topology and $S=\{0,1\}$ with the Sierpiński space topology $\{\varnothing ,\{1\},S\}$. The identity map $D\to S$ is a continuous bijection, since the inverse image of every target open set is an open set in $D$. Its image is compact, but the two points of $S$ have no disjoint open neighborhoods. If the target $Y$ is Hausdorff, its subspace topology does make $f(X)$ Hausdorff, and the compact subset of a Hausdorff space theorem also makes $f(X)$ closed in $Y$.

The practical payoff is existence: the extreme value theorem uses the compact image of a nonempty compact space under a real-valued continuous function to obtain an attained minimum and maximum. When $f$ is also a bijection onto a Hausdorff target, the compact-to-Hausdorff continuous bijection theorem supplies continuity of its inverse without a separate formula.

A closed densely defined operator has compact resolvent when $(A-zI{)}^{-1}$ is compact at one, equivalently every, resolvent point. Its spectrum consists only of isolated eigenvalues of finite multiplicity, with possible accumulation only at infinity.

Let $P_N$ project ${\ell }^2$ onto its first $N$ coordinates. Although $P_N\to I$ only strongly on the whole unit ball, convergence is uniform on every compact subset. Hence compact $T$ satisfiesand each $P_{N}T$ has finite rank.

A compact Hermitian operator has real nonzero eigenvalues of finite multiplicity, with zero as their only possible accumulation point, and an orthonormal eigenbasis after a basis of its kernel is included.

This is the discrete-eigenbasis case of the spectral theorem for a compact operator that is self-adjoint.

A finite-rank operator has finite-dimensional image and is compact.

A bounded linear operator on a Hilbert space is Hilbert-Schmidt when ${\sum }_n\Vert T{e}_n{\Vert }^2<\infty$ for one, equivalently every, orthonormal basis. Every Hilbert-Schmidt operator is a compact operator.

Peculiar velocity is motion relative to the homogeneous Hubble flow. In comoving coordinates,where $\mathbf{u}$ is physical velocity and $\mathbf{v}$ is peculiar velocity.

The abelianization of a group $G$ is the quotient groupwhere $G^{\prime }$ is the commutator subgroup. Its elements are cosets, and its operation is $(g{G}^{\prime })(h{G}^{\prime })=gh{G}^{\prime }$. The result is an abelian group: the two products $gh$ and $hg$ differ by a group commutator, which disappears in the quotient group.

For the symmetric group $S_3$, the sign of a permutation realizes $S_3^{\mathrm{ab}}\cong C_2$, as shown in commutator subgroup of the symmetric group on three letters. For the dihedral group of a square, abelianization of the square symmetry group gives $C_2\times C_2$: two independent yes-or-no pieces of information survive. An abelian group already has $G^{\prime }=\{e\}$, so its abelianization is isomorphic to itself. At the other extreme, a perfect group has trivial abelianization.

Calling this the largest abelian quotient means that every group homomorphism into any abelian group factors uniquely through $q$. It does not require $G$ to be finite, or compare the sizes of unrelated abelian groups. The universal property of abelianization makes this precise. It lets us solve all such mapping problems in one simpler group: any invariant that respects multiplication and takes values in an abelian group sees only $G^{\mathrm{ab}}$.

For every finite alphabet $X$ and every positive integer $k$, there is $n$ such that every $k$-coloring of $X^n$ contains a monochromatic combinatorial line.

The impulse of a force over a time interval is its vector integral:Its International System of Units unit is $\mathrm{N}\mathrm{s}=\mathrm{kg}m/s$. The impulse-momentum theorem equates the net impulse to the change in momentum. For one fixed-mass point mass, integrate its net force; for a closed collection obeying Newton's third law, integrate the sum of external forces. In one dimension, the signed area under the force versus time graph gives the impulse; negative force contributes negative area. ^[ref]

Three stipulated rightward net force pulses illustrate the same $J=6\mathrm{N}\mathrm{s}$: constant $12\mathrm{N}$ for $0.5\mathrm{s}$, a triangle rising from zero to $24\mathrm{N}$ at $0.25\mathrm{s}$ and falling to zero at $0.5\mathrm{s}$, and constant $3\mathrm{N}$ for $2\mathrm{s}$. Their areas areApplied separately to the same fixed mass $m=2\mathrm{kg}$, each gives $\Delta v=J/m=3m/s$. Their peak forces differ: $12$, $24$, and $3\mathrm{N}$. Thus impulse fixes the final momentum change, while the force history determines how abruptly it occurs.

The mean net force over the chosen interval is $\overline{\mathbf{F}}=\mathbf{J}/(t_f-t_i)$. Extending the interval for a fixed impulse reduces this mean force. A lower peak force additionally depends on the pulse shape: the two short pulses above have the same mean $12\mathrm{N}$ but different peaks. This explains why controlling the duration and shape of a stopping impulse matters in cushioning a collision.

In a frictionless oblique collision, impulse acts along the common normal and tangential velocity components remain unchanged.