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A solvable group has a finite subnormal series from the identity to the group with abelian group factors. An equivalent definition uses the derived series: set $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$; the series must reach the identity. For completeness, an abelian-factor series forces successive derived subgroups into its successive terms, while a terminating derived series itself supplies such a series.

If $H\le G$, induction gives $H^{(j)}\le G^{(j)}$: every commutator of elements of $H^{(j)}$ is also a commutator in $G^{(j)}$. Thus termination for $G$ implies termination for $H$, proving that every subgroup of a solvable group is solvable.

In $S_4$, useThe Klein four-group is invariant under conjugation in $S_4$; the other normality assertions follow from the alternating group being the kernel of the sign map. The quotient group factors are $V_4,C_3,C_2$, all abelian. Hence $S_4$ is solvable.

For the last example take $A_5\times C_2$. Its subgroup $\{1\}\times C_2$ is nontrivial, proper and normal, so the product is not a simple group. Projection onto $A_5$ is surjective, and every quotient of a solvable group is solvable, because a homomorphism carries each derived subgroup onto the derived subgroup of its image. But the nonabelian simple group $A_5$ has $A_5^{\prime }=A_5$: its derived subgroup is nontrivial and normal. Thus the product is not solvable either. One standard justification of the simplicity used here is that the nonidentity conjugacy class sizes of $A_5$ are $15,20,12,12$; no proper nontrivial union with the identity has size dividing $60$, as a normal subgroup would require by Lagrange's theorem for finite groups.

For any topological target $Y$, $f(K)$ is compact. Pull an open cover of $f(K)$ back by $f$, take a finite subcover of $K$, and map the covering statement forward. Without a separation assumption on $Y$, the image need not be closed, as a constant map into a nonclosed singleton of an indiscrete space shows.

If $Y$ is a metric space, $f(K)$ is also closed in $Y$, bounded, totally bounded, complete with its inherited metric, and sequentially compact. Closedness uses the Hausdorff theorem from part (iii); total boundedness follows by taking finite subcovers of the covers by balls of any fixed radius. Compact metric spaces are sequentially compact, and a Cauchy sequence with a convergent subsequence converges to the same limit, proving completeness. These properties are asserted for the image subset, without requiring the whole ambient metric space to be complete.

A best approximation to $x$ from $U_n$ is a vector $u_{\ast }\in U_n$ such thatThe definition requires attainment of the infimum. This finite-dimensional subspace does have such an approximation, as the explicit projection below shows; completeness of the whole inner-product space is unnecessary.

Separation of variables replaces a compatible partial differential equation by ordinary differential eigenvalue equations. For example, with a thermally diffusive rod, $u_t=D{u}_{xx}$, try $u=X(x)T(t)$. Wherever division is valid,The two sides depend on independent variables, so each must be constant. Homogeneous end temperatures $u(0,t)=u(L,t)=0$ give $X_n=\sin (n\pi x/L)$ and ${\lambda }_n=(n\pi /L{)}^2$. The general initial-temperature profile is represented by superposing these separated modes:Each spatial mode decays at its own rate. For a fixed-end wave equation the same spatial eigenfunctions instead have oscillatory time factors $A_n\cos (cn\pi t/L)+B_n\sin (cn\pi t/L)$, with coefficients fixed by initial displacement and velocity. Similar constructions separate Laplace's equation in suitable coordinate systems.

A single product is usually only one mode, not the most general solution. Linear superposition and completeness of the spatial eigenfunctions are what extend the method to general allowed data. Nonhomogeneous boundary values can first be removed by a known stationary or particular solution. Separation is effective when the geometry, coefficients and boundary conditions admit a common separated eigenproblem; an arbitrary nonlinear equation or inseparable geometry need not do so. Self-adjoint Sturm-Liouville problems explain the weighted orthogonality and completeness in many physical applications.

Orthogonality means zero inner product, not merely that two objects look different. For complex vectors, $⟨u,v⟩={\sum }_i{u}_i^{\ast }{v}_i$; for functions a common choice is $⟨u,v⟩=\int w(x)u(x{)}^{\ast }v(x)dx$ with positive weight $w$. An orthonormal set has $⟨u_m,u_n⟩={\delta }_{mn}$.

For an orthogonal set, the component of $f$ along $u_n$ has coefficientA complete orthonormal set gives $f={\sum }_n{c}_n{u}_n$ in the appropriate norm and Parseval's identity $\Vert f{\Vert }^2={\sum }_n|c_n{|}^2$. Without completeness there can be a nonzero residual orthogonal to the entire chosen set. Orthogonal nonzero vectors are linearly independent, but independence alone does not imply orthogonality.

Fourier coefficients are projections becauseLikewise, for a self-adjoint operator, $A{u}_n={\lambda }_n{u}_n$ gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n⟩=0$, so distinct eigenvalues have orthogonal eigenvectors. Degeneracy does not itself force a chosen pair to be orthogonal, but an orthonormal basis can be selected within each eigenspace. The differential version uses the correct Sturm-Liouville weight and boundary conditions.

For mechanical normal modes with unequal masses the relevant inner product is $⟨u,v{⟩}_{\mathsf{M}}=u^{†}\mathsf{M}v$. This makes the modes in D14 orthogonal even when an ordinary unweighted dot product does not. Expanding in such coordinates diagonalizes the kinetic and potential quadratic forms, explaining why modal motions evolve independently and their energies add.

Act by left multiplication on the five cosets of $H$. The kernel is the core ${\bigcap }_{g\in G}gH{g}^{-1}$, a normal subgroup contained in $H$, so the hypothesis makes it trivial. Thus $G$ embeds in $S_5$, and its image has order $60$ and index two. An index-two subgroup is normal, so the quotient is of order two. Every three-cycle maps trivially to that quotient, since its order divides three. Three-cycles generate $A_5$, so the image contains $A_5$; their equal order gives $\boxed{G\cong A_5}$.

For completeness, prove simplicity rather than merely invoking it. The nonidentity even cycle types in $S_5$ are three-cycles, products of two disjoint transpositions and five-cycles. In $A_5$, a three-cycle has centralizer of order three, so its conjugacy class has size $20$. A double transposition has centralizer of order four in $A_5$, giving class size $15$. A five-cycle has centralizer its cyclic subgroup of order five, giving class size $12$; the $24$ such elements therefore split into two classes of size $12$. These account for all $60$ elements with the identity.

A normal subgroup is a union of whole conjugacy classes, containing the identity, and its order must divide $60$. No nonempty proper sum of the class sizes $15,20,12,12$, plus one, is a divisor of $60$ greater than one: the possible proper nontrivial sums are $13,16,21,25,28,33,36,40,45,48$. Only the full union has order $60$. Hence $\boxed{A_5\text{ is simple}}$, and so is $G$.

Yes. A Cauchy sequence of continuous functions in the supremum norm is Cauchy in the given complete space of bounded functions, so it has a uniform limit there. A uniform limit of continuous functions is continuous: at a point, use two uniform errors smaller than $\epsilon /3$ and continuity of one sufficiently late term for the remaining error. Thus the limit is in $C([0,1])$, proving completeness. Equivalently this is a closed subspace of the given complete space.

Set $u={\sum }_{k=1}^n⟨x,e_k⟩e_k$, with the convention that the inner product is linear in its first argument. Orthonormality gives $⟨x-u,e_j⟩=0$, hence $x-u\perp U_n$. For any $v\in U_n$ the Pythagorean identity yieldsThus this $u$ is the best approximation, and uniqueness gives $\boxed{P_{n}x=\sum _{k=1}^n⟨x,e_k⟩e_k}$. No completeness assumption on $H$ is necessary for this finite-dimensional construction.

The Caratheodory measurability criterion yields a complete sigma-algebra ${\mathfrak{M}}_{\mu }^{\ast }$, and $\mu$ restricted to it is a countably additive measure. Completeness means that every subset of a measurable null set is measurable.

For the requested algebra proof, $X$ and $\varnothing$ plainly satisfy the criterion. Replacing $A$ by $X\setminus A$ merely exchanges the two terms, proving closure under complements. If $A,B$ are measurable, apply their splitting identities successively to any $E$:Subadditivity applied to the first two pieces impliesThe reverse inequality is outer subadditivity. Hence $A\cup B$ is measurable, proving closure under finite unions and therefore the algebra of sets property. The argument never subtracts infinite values.

Use normalized Haar measure $dm=dt/(2\pi )$. A homogeneous Banach space on the circle is a Banach subspace $B\subseteq L^1(\mathbb{T})$ with $\Vert f{\Vert }_1\le \Vert f{\Vert }_B$, such that translations ${\tau }_{s}f(t)=f(t-s)$ preserve $B$, satisfy $\Vert {\tau }_{s}f{\Vert }_B=\Vert f{\Vert }_B$, and obey $\Vert {\tau }_{s}f-f{\Vert }_B\to 0$ as $s\to 0$.

For continuous $g$, the map $s\mapsto g(s){\tau }_{s}f$ is continuous with values in the complete space $B$. Define its Banach-valued Riemann integral, equivalently its Bochner integral, byThe triangle inequality for Riemann sums and passage to their norm limit give the canonical estimateThe continuous inclusion into $L^1$ carries the same Riemann sums to the ordinary convolution, so this definition agrees almost everywhere with $\int f(t-s)g(s)dm(s)$. Density of continuous functions in $L^1$, completeness of $B$, and the estimate extend the operation uniquely to all $g\in L^1$. The ordinary $L^1$ convolution estimate identifies that extension with the same integral formula.

For an orthonormal family in a Hilbert space, the squared norm of every finite partial sum is the sum of squared coefficients. Square summability therefore makes the net of finite partial sums Cauchy, proving unconditional convergence by completeness. If the orthonormal family is complete, Bessel's inequality makes the coefficients of each vector square summable, and the residual after summing is orthogonal to the whole family, hence zero. This recovers the orthonormal expansion and Parseval identity.

A family of seminorms separates points when every nonzero vector $z$ has $p(z)>0$ for some member $p$. Equivalently, its kernels of seminorms have intersection $\{0\}$, or any two distinct vectors $x,y$ have $p(x-y)>0$ for some $p$. Individual seminorms can have large kernels of seminorms; the whole family must have no common invisible direction.

The associated seminorm topology is Hausdorff if and only if the family separates points. Indeed, if $p(x-y)=a>0$, the neighbourhoods $\{u:p(u-x)<a/3\}$ and $\{v:p(v-y)<a/3\}$ are disjoint by the triangle inequality. Conversely, if every $p$ vanishes on a nonzero $z$, then zero and $z$ belong to exactly the same open sets and cannot be separated. This criterion keeps separation distinct from completeness: detecting all directions does not guarantee that all Cauchy sequences have limits.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

Let $X$ be a complete metric space and let $F_n$ be nonempty closed bounded subsets with $F_{n+1}\subseteq F_n$ and Hausdorff distance $d_H(F_n,F_{n+1})\le a_n$, where $a_n\ge 0$ and ${\sum }_n{a}_n<\infty$. Then $G={\bigcap }_n{F}_n$ is nonempty and closed, and $d_H(F_n,G)\le {\sum }_{k=n}^{\infty }{a}_k\to 0$.

To prove this, start with any $x_n\in F_n$ and choose $x_{k+1}\in F_{k+1}$ with $d(x_k,x_{k+1})<a_k+\eta 2^{n-k-1}$. The triangle inequality makes the resulting sequence a Cauchy sequence. Its limit exists by completeness and belongs to every $F_j$ by nesting and closedness. Its distance from the starting point is at most ${\sum }_{k=n}^{\infty }{a}_k+\eta$. Let $\eta \downarrow 0$ and take the supremum over starting points. The reverse directed distance is zero since $G\subseteq F_n$. Unlike the usual shrinking-diameter form of Cantor's intersection theorem, the sets need not have diameters tending to zero.

For a finite unramified extension of non-Archimedean local fields, the field norm on residue units is surjective: on a cyclic group of order $q^n-1$ it is exponentiation by $(q^n-1)/(q-1)$ and has image of order $q-1$. For a common uniformizer $\pi$ and $r\ge 1$, expansion of the conjugate product givesThe field trace of a finite finite field extension is onto: its polynomial $X+X^q+\cdots +X^{q^{n-1}}$ is nonzero of degree below $q^n$, so defines a nonzero linear map to the base field. Lift a residue norm and correct successively with elements of $1+{\pi }^r{\mathcal{O}}_L$. The infinite product converges by completeness, and its norm is the prescribed unit. This argument also works in positive characteristic and when $p$ divides $n$.