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Regard the space of continuous functions vanishing at infinity $C_0({\mathbb{R}}^n)$ as a subspace of $L^{\infty }({\mathbb{R}}^n)$. Point evaluation at the origin,is a bounded linear functional of norm one on this subspace because a continuous function's supremum and essential supremum agree. By the Hahn-Banach theorem, ${\ell }_0$ extends to a bounded linear functional

Suppose that some $g\in L^1({\mathbb{R}}^n)$ represented this extension:Choose $\phi \in C_c({\mathbb{R}}^n)$ with $0\le \phi \le 1$, $\phi (0)=1$, and support in the unit ball, and set ${\phi }_k(x)=\phi (kx)$. Then $\ell ({\phi }_k)=1$ for every $k$. On the other hand, ${\phi }_k(x)\to 0$ for almost every $x$ and $|{\phi }_{k}g|\le |g|$. The dominated convergence theorem givesa contradiction. Thus $\ell$ is a singular functional on L infinity and has no $L^1$ density.

Choose a nonzero $\varphi \in C_c^{\infty }({\mathbb{R}}^n)$ supported in a ball of radius $1$, and letTranslation preserves both the $L^2$ norm and the weak-derivative norms, sofor every $k$. Thus $({\varphi }_k)$ is bounded in the Sobolev space $H^1({\mathbb{R}}^n)$.

Distinct translates have disjoint supports. Hence, for $j\ne k$,No subsequence is Cauchy in $L^2$, so no subsequence converges there. The embedding $H^1({\mathbb{R}}^n)\hookrightarrow L^2({\mathbb{R}}^n)$ is therefore not a compact operator. This is the noncompactness of a Sobolev embedding by translation caused by loss of compactness at infinity.

No. Take the Hilbert space $H={\ell }^2$ and its standard orthonormal sequence $(e_n)$. Letthe unit sphere. This set is norm closed but is not convex. By the Bessel inequality, for every $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$; equivalently, this is a weakly null orthonormal sequence. Every $e_n$ lies in $C$, but the weak limit $0$ does not. Hence convexity is essential.

Apply part (b). Since every $x_n$ belongs to the convex set $C$, every convex combination ${\tilde{x}}_k$ also belongs to $C$. Since $C$ is norm closed and ${\tilde{x}}_k\to x$ in norm, it follows that $x\in C$. Thus a norm-closed convex set is weakly closed.

Put $z_n=x_n-x$, so $z_n\rightharpoonup 0$. We show that the norm closure of the convex hull of every tailcontains zero. If it did not, the Hahn-Banach separation theorem would give a bounded linear functional $\varphi$, a real number $c>0$, and, after multiplying $\varphi$ by a complex scalar if necessary,for every $z$ in that closed convex hull. In particular $\mathrm{Re}\varphi (z_n)\ge c$ for all $n\ge k$. By the Riesz representation theorem, $\varphi (z)=⟨z,y⟩$ for some $y\in H$, contradicting $z_n\rightharpoonup 0$.

Consequently, for every $k$ there is a finite convex combinationwith $\Vert w_k{\Vert }_H<1/k$. DefineThen every ${\tilde{x}}_k$ is a convex combination of terms of the original sequence andThis is Mazur lemma in the present Hilbert-space setting.

For each $n$, define the bounded linear functionalThe hypothesis is precisely that $x_n$ converges weakly to $x$. In particular, for each fixed $y\in H$, the scalar sequence $T_n(y)$ converges and is therefore bounded. Thus the family $\{T_n\}$ is pointwise bounded.

The Uniform boundedness principle givesBy the Riesz representation theorem, $\Vert T_n\Vert =\Vert x_n{\Vert }_H$. Henceso the sequence is bounded.

Assume that $\zeta$ is a primitive fifth root of unity. The cyclotomic field $K=\mathbb{Q}(\zeta )$ has degree four and signature $(r_1,r_2)=(0,2)$. The Dirichlet unit theorem therefore gives free rankThe roots of unity in a rational cyclotomic field are the ten elements $\pm {\zeta }^j$, so

For explicit units, $-\zeta$ has order ten, whileHence $1+\zeta$ is a unit. It has infinite order because in the standard complex embeddingwhereas every root of unity has modulus one. Its image in the free $\mathbb{Z}$ factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequentlyhas finite index in ${\mathcal{O}}_K^{\times }$, as described by the unit group of the fifth cyclotomic field.

Minkowski's lemma states that if $\Lambda$ is a full lattice in ${\mathbb{R}}^n$ and $C\subset {\mathbb{R}}^n$ is measurable, convex, and centrally symmetric, thenimplies that $C$ contains a nonzero point of $\Lambda$.

Let ${\sigma }_1,{\sigma }_2:K\to \mathbb{R}$ be the two real embeddings. The Minkowski embedding of a real quadratic field identifies ${\mathcal{O}}_K$ with a latticeof covolume $\sqrt{|D_K|}$. Choose a constant $c$ satisfyingFor $R>0$, consider the closed diamondIt is convex and centrally symmetric, and its area isMinkowski's lemma supplies a nonzero ${\alpha }_R\in {\mathcal{O}}_K$ whose two embeddings lie in $C_R$.

For $(x,y)\in C_R$, the arithmetic-geometric mean inequality givesSince the coordinate product is the field norm,Finally, $|{\sigma }_2({\alpha }_R)|\le c/R\to 0$ as $R\to \infty$. No fixed nonzero algebraic integer can occur for arbitrarily large $R$, because its second embedding is nonzero. Consequently the elements ${\alpha }_R$ obtained along an unbounded sequence of $R$ contain infinitely many distinct values. Here $N(\alpha )$ denotes the absolute norm, as usual in this inequality.

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.

Letbe the terminal set, and let $C$ be the vertex set of the given complete graph $K_{2k}$. Any vertex set $S$ meeting every $T$-$C$ path has size at least $2k$: if $|S|<2k$, choose $t\in T\setminus S$ and $c\in C\setminus S$. Since $G$ is $2k$-connected, $G-S$ is connected, so it contains a $t$-$c$ path avoiding $S$, a contradiction.

The Set version of Menger theorem therefore gives $2k$ pairwise vertex-disjoint $T$-$C$ paths. Truncate them at their first and last vertices in $T\cup C$. Since both sets have $2k$ vertices, every terminal is the endpoint of one path and the other endpoints are distinct vertices of $C$. Denote the path from a terminal $z$ to its clique endpoint by $R_z$, and call that endpoint $c_z$.

For each $i$, the clique contains the edge $c_{x_i}{c}_{y_i}$. Concatenatinggives an $x_i$-$y_i$ path in a graph. The linkage paths are mutually vertex-disjoint, their clique endpoints are all distinct, and the joining clique edges pair those endpoints without introducing a new vertex. The resulting paths $P_1,\dots ,P_k$ are therefore mutually vertex-disjoint. In particular, under the stated clique hypothesis, the $2k$-connected graph is $k$-linked for these terminals.

Construct a bipartite graph with left vertices $A_1,\dots ,A_k$, right vertices $B_1,\dots ,B_k$, and an edge $A_i{B}_j$ exactly when $A_i\cap B_j\ne \varnothing$. For $S\subseteq [k]$, every point of ${\bigcup }_{i\in S}{A}_i$ lies in some $B_j$ adjacent to one of those $A_i$. ThusUsing finite additivity of Lebesgue measure and the equal-volume hypotheses,Hence $|N(S)|\ge |S|$ for every $S$. The Hall marriage theorem supplies a perfect matching, which has the form $A_i{B}_{\sigma (i)}$ for a permutation $\sigma$ of $[k]$. Every matched edge means precisely that $A_i\cap B_{\sigma (i)}\ne \varnothing$.

The Hall marriage theorem says that a bipartite graph with classes $A,B$ has a matching saturating $A$ if and only ifwhere $N(S)$ is the graph neighbourhood of $S$. Necessity follows because the matching sends the vertices of $S$ to distinct vertices of $N(S)$.

For sufficiency, add vertices $x,y$, join $x$ to every vertex of $A$, and join every vertex of $B$ to $y$. Let $W$ be an $x$-$y$ vertex separator, and write $W_A=W\cap A$ and $W_B=W\cap B$. If an edge joined a vertex of $A\setminus W_A$ to a vertex of $B\setminus W_B$, it would give an $x$-$y$ path avoiding $W$. HenceHall's condition givesand therefore $|W|\ge |A|$. By Menger theorem, there are $|A|$ internally vertex-disjoint $x$-$y$ paths. Each has the form $xaby$ with $a\in A$ and $b\in B$; disjointness makes all the $a$ and all the $b$ distinct. Their middle edges form a matching saturating $A$.

The vertex form of Menger theorem states that if $x$ and $y$ are distinct nonadjacent vertices of a finite graph, then the maximum number of pairwise internally vertex-disjoint paths from $x$ to $y$ equals the minimum cardinality of an $x$-$y$ vertex separator.

Part (b) proves one implication. Conversely, assume in ZF thatfor every infinite cardinal $\kappa$. Let $X$ be any set and let $\gamma =h(X)$ be its Hartogs ordinal. Finite $X$ is already well-orderable, so suppose $X$ is infinite and form the disjoint unionBy hypothesis, $A\times A$ is bijective with $A$. Since $\gamma \times X$ injects into $A\times A$, we obtain

Apply part (c), with $K=\gamma$ and $L=X$. It gives either an injection $\gamma \to X$ or a surjection $s:\gamma \to X$. The first alternative contradicts Hartogs theorem. In the second, every fiber $s^{-1}(\{x\})$ is a nonempty set of ordinals and therefore has a least member. The mapis an injection of $X$ into the ordinal $\gamma$ and pulls its well-order back to $X$. Thus every set is well-orderable, so the well-ordering theorem gives the axiom of choice. This proves the Tarski cardinal-square theorem.

Letbe the injection witnessing $\kappa \lambda \le \kappa +\lambda$, and fix $(k_{\ast },l_{\ast })\in K\times L$. Extend the inverse of $i$ to a surjectionby settingThis construction uses only the fixed pair, not the axiom of choice.

Restrict $p$ to the $K$-summand and take its second coordinate:If $r$ is onto, it is the required surjective function from $K$ to $L$.

Otherwise choose one $l_0\in L\setminus r[K]$. For each $k\in K$, surjectivity of $p$ gives a preimage of $(k,l_0)$. No such preimage lies in the $K$-summand, by the choice of $l_0$, so it lies in the $L$-summand. Unless $(k,l_0)=(k_{\ast },l_{\ast })$, this preimage is exactly $i(k,l_0)$ and is unique: all points outside the range of $i$ map only to the default pair. Distinct $k$ give distinct preimages because $i$ is injective.

If $l_0\ne l_{\ast }$, these unique preimages therefore define an injective function $K\to L$. The only exceptional case is $l_0=l_{\ast }$, where they initially define an injectionIf $j$ is onto, extend it arbitrarily at $k_{\ast }$ to obtain a surjection $K\to L$. If it is not onto, choose one $l^{\prime }\in L\setminus j[K\setminus \{k_{\ast }\}]$ and set $j(k_{\ast })=l^{\prime }$; this gives an injection $K\to L$. These cases prove the product-sum comparison lemma entirely in ZF.

Under the axiom of choice, identify every infinite cardinal $\kappa$ with its initial ordinal. We prove ${\kappa }^2=\kappa$ by transfinite induction on infinite cardinals. The countable case follows from the usual diagonal enumeration of $\mathbb{N}\times \mathbb{N}$.

For the induction step, well-order the Cartesian product $\kappa \times \kappa$ by first comparingand then using lexicographic order among pairs with the same maximum. The predecessors of $(\alpha ,\beta )$ lie in $(\eta +1)\times (\eta +1)$, where $\eta =\max \{\alpha ,\beta \}<\kappa$. Put $\mu =|\eta +1|<\kappa$. If $\mu$ is infinite, the induction hypothesis makes this predecessor set have cardinal at most ${\mu }^2=\mu <\kappa$; if $\mu$ is finite, the same conclusion is immediate.

Let $\theta$ be the order type of this well-order. Every proper initial segment of $\theta$ therefore has cardinal less than $\kappa$. Hence $\theta <{\kappa }^{+}$, since otherwise its initial segment of order type $\kappa$ would have cardinal $\kappa$. Thus $|\kappa \times \kappa |=|\theta |\le \kappa$. The map $\alpha \mapsto (\alpha ,0)$ gives the reverse injection, and the Cantor-Schröder-Bernstein theorem yields