A metric space is complete when every Cauchy sequence in it converges to a point of that space. A sequence is Cauchy when, for each $ϵ>0$, there is $N$ such that $d(x_m,x_n)<ϵ$ whenever $m,n\ge N$. The requirement that the limit lie in the space distinguishes completeness from having limits only in a larger ambient space.

If $A$ is closed and $a_n\in A$ converges to $x\in X$, then $x$ cannot lie in the open complement: an open ball about such an $x$ would eventually contain all $a_n$ while missing $A$. Thus $x\in A$.

Conversely, suppose $A$ contains every limit of a sequence in $A$. If $x\in \overline{A}$, every ball $B(x,1/n)$ meets $A$; choose $a_n$ in that intersection. Then $a_n\to x$, so $x\in A$. Hence $\overline{A}=A$, proving that a subset of a metric space is a closed set exactly when it is sequentially closed.

No such sequence exists. By Fatou's lemma on $(0,1)$,But the proposed nonnegative measurable integrand is positive at every point of $(0,1)$, so its integral cannot be zero. Explicitly, the sets where it is at least $1/k$ cover $(0,1)$; at least one has positive measure, forcing a positive integral. This contradicts the displayed bound.

For each integer $m\ge 1$, list the $m^2$ functionsin increasing $j$, and concatenate the blocks in increasing $m$. Each function has $L^1$ norm $1/m$, so the resulting sequence has norm tending to zero. Every $x\in (0,1)$ lies in exactly one interval of each block, where the function takes value $m$. ThereforeThis unbounded typewriter sequence with vanishing L1 norm is a moving-spike construction: the support shrinks fast enough to beat the increasing height.

Let $N$ be the vector subspace of integrable functions that vanish almost everywhere. The Lebesgue space is the quotient vector space $L^1(\mathbb{R})={\mathcal{L}}^1(\mathbb{R})/N$: its elements are equivalence classes of functions equal almost everywhere. Addition and scalar multiplication are $[f]+[g]=[f+g]$ and $c[f]=[cf]$.

Set $\Vert [f]{\Vert }_1=\int |f|$. This is well-defined because replacing $f$ by an equal-almost-everywhere representative does not change the integral. It is nonnegative, homogeneous, and obeys the triangle inequality by $|f+g|\le |f|+|g|$. Part (i) says its only zero element is the zero class. Hence it is a norm on $L^1$.

Exactly those integrable functions that vanish almost everywhere:Thus this expression is a seminorm on actual functions, becoming a norm after the quotient in part (ii).

For each $n$, let $N_n$ be the null set where $f_n\ne g_n$. Their union $N$ is null. At any $x\notin N$, the two sequences agree term by term, so their pointwise limits are equal: $f(x)=g(x)$. Therefore $f\sim g$. Taking the countable union first is essential: it produces one exceptional set outside which all equalities hold simultaneously.

The relation is reflexive because the exceptional set for $f=f$ is empty, and symmetric because the exceptional sets for $f=g$ and $g=f$ agree. If $f=g$ off a null set $N_1$ and $g=h$ off a null set $N_2$, then $f=h$ off $N_1\cup N_2$, which is null by part (b). Hence it is transitive as well, so equality almost everywhere is an equivalence relation.

For each $n\ge 1$, choose a countable interval cover of $E_n$ of total length below $ϵ/2^n$. The union of these covers is countable and covers ${\bigcup }_n{E}_n$, with total length less than ${\sum }_nϵ/2^n=ϵ$. As $ϵ$ is arbitrary, a countable union of null sets is null.

Given $ϵ>0$, cover $E_1$ by open intervals of total length less than $ϵ/2$, and cover $E_2$ similarly. Combining the two countable covers gives a cover of their union of total length less than $ϵ$. Thus $E_1\cup E_2$ is a null set.

A property holds almost everywhere when its exceptional set, where the property fails, is a null set. The exceptional set need not be empty or countable.

A null set $E\subseteq \mathbb{R}$ has Lebesgue outer measure zero. Equivalently, for every $ϵ>0$ it can be covered by countably many open intervals whose total length is less than $ϵ$. This definition also applies before assuming $E$ measurable; every such set is Lebesgue measurable with measure zero.

For distinct odd primes $p,q$, the quadratic reciprocity law isThus the two symbols have opposite signs exactly when both primes are three modulo four; otherwise they agree. Distinctness excludes zero symbols.

Multiplicativity of the Legendre symbol givesThis includes the case where $p$ divides a factor, when both sides are zero.

The second supplementary law for quadratic reciprocity isThe exponent is an integer because every odd square is one modulo eight.

The first supplementary law for quadratic reciprocity isIt also follows immediately from Euler's criterion applied to $-1$.

For an odd prime $p$, the Legendre symbol is zero if $p\mid a$, one if $a$ is a nonzero quadratic residue modulo $p$, and minus one if it is a quadratic nonresidue. ThusIt depends only on the congruence class of $a$ modulo $p$.

The symmetric group $S_3$ has subgroups of orders $1,2,3,6$. There is one trivial subgroup, three conjugate subgroups generated by transpositions, one order-three subgroup $A_3$, and the whole group. These are all subgroups by Lagrange's theorem. Thus again there is one conjugacy class for each index $6,3,2,1$.

The coset action classification in part (c) supplies exactly one transitive orbit type of each of these sizes. The same orbit-multiplicity equation as in part (i) applies, givingThe equality of the two counts comes from their identical lists of transitive orbit types, rather than from the groups being isomorphic.

A cyclic group of order six has exactly one subgroup of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of subgroups. Thus there is one transitive group action of each size $1,2,3,6$.

An action on six points is a disjoint union of its orbits. Its equivalence type is therefore determined by nonnegative integers satisfyingIf $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. HenceThis is counting finite group actions by transitive orbit types. Equivalently, a generator acts as a permutation whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.

The finite sum of truncated cubes is $C^2$ by part (b) and is cubic on each knot interval. For $x\le 0$, every term vanishes. For $x\ge 4h$, every truncated power equals its ordinary polynomial, so part (a) applied to $P(x)=x^3$ givesIt is therefore zero on both exterior intervals. Its $C^2$ regularity makes the endpoint second derivatives zero too, proving naturality.

Evaluation at the five knots givesFor example the three interior values in units of $h^3$ are $1$, $8-4=4$, and $27-4\cdot 8+6=1$. At $4h$ the cancellation is $64-4\cdot 27+6\cdot 8-4=0$.

For $h=1$ an explicit graph description isIt is symmetric about $x=2$, increases from zero to its maximum four at the centre, and decreases symmetrically back to zero. It joins the zero tails with zero first and second derivatives.