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The Baire category theorem states that a complete metric space is not a countable union of closed sets with empty interior. To prove it, let $G_1,G_2,\dots$ be open dense sets and start with any nonempty open set $V$. Inductively choose closed ballswith the first ball inside $V\cap G_1$. The centres are Cauchy. Completeness supplies a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense, which is the equivalent form of the theorem.

Now each $nS$ is closed and ${\bigcup }_{n\ge 1}nS=X$. Baire givesfor some $x,r,n$. Symmetry gives $B(-x,r)\subset nS$. If $\Vert z\Vert <r$, then $x+z$ and $-x+z$ lie in these two balls; convexity makes their midpoint $z$ belong to $nS$. Hencewhich proves the closed convex absorbing set has an origin neighbourhood result.

Convexity cannot be dropped. In $X=\mathbb{R}$, setThis is closed and symmetric but has gaps arbitrarily close to zero. For any $x\ne 0$, choose $k$ so that the intervalhas length at least one, and choose an integer $n$ in it. Then $x/n\in S$, so ${\bigcup }_{n}nS=\mathbb{R}$, while $S$ is not a neighbourhood of zero. This is a closed symmetric absorbing set without an origin neighbourhood.

The Seifert-van Kampen theorem says that if $X=U\cup V$, where $U,V,U\cap V$ are path connected open sets containing $x_0$, then

The cell attachment is the quotientwith the image of $x_0=f(\ast )$ as base point. For $n=2$, take one open set that deformation retracts onto $X$ together with a boundary collar, and another that consists of the interior of the disc with a collar and is contractible. Their intersection deformation retracts onto $S^1$. Its generator maps to $[f]$ in the first set and to the identity in the second. Van Kampen therefore proves the fundamental group after attaching a 2-cell formula

Applying van Kampen to the two circles givesAttach three discs along loops representing $a^2$, $b^3$, and $(ab{)}^2$. The resulting presentation complex for the symmetric group on three letters has groupSending $a$ to $(12)$ and $b$ to $(123)$ gives a surjection onto $S_3$. The relations imply $aba=b^{-1}$, so every word reduces to one ofThe presented group has at most six elements and surjects onto the six-element group $S_3$, so this map is an isomorphism.

The Dirichlet unit theorem states that for a number field of signature $(r_1,r_2)$,

For $K=\mathbb{Q}(\sqrt{5})$,The field is real quadratic, so the unit rank is one and its only roots of unity are $\pm 1$. To see that $\varphi$ is fundamental, suppose a positive unit satisfies $1<u<\varphi$. If its norm is $1$, its integral trace $u+u^{-1}$ lies strictly between $2$ and $\sqrt{5}$; if its norm is $-1$, its trace $u-u^{-1}$ lies strictly between $0$ and $1$. Neither interval contains an integer. Reducing any positive unit by a suitable power of $\varphi$ now proves the units of the quadratic field Q square root of five formula

If ${\mathcal{O}}_L^{\times }/{\mathcal{O}}_K^{\times }$ is finite, the two unit groups have the same rank, namely one. Put $d=[L:K]$ and let $(r_1,r_2)$ be the signature of $L$. ThenTherefore $r_1=4-2d$ and $r_2=2d-2$. For a proper extension, nonnegativity forceswith signature $(0,2)$.

This degree occurs: take $L=K(i)$. It is a totally imaginary quadratic extension of $K$, so the unit ranks agree and the quotient is finite. It is nontrivial because $i\in {\mathcal{O}}_L^{\times }$ but $i\notin {\mathcal{O}}_K^{\times }$ (and its coset has order two). This is the finite relative unit quotient over a real quadratic field example.

A reduced positive definite binary quadratic form $[a,b,c]$ satisfieswith $b\ge 0$ when either $|b|=a$ or $a=c$. For $d\equiv 0$ or $3(\mathrm{mod}4)$, the class number of a negative discriminant $h(-d)$ is the number of proper equivalence classes of primitive positive definite integral forms of discriminant $-d$, equivalently the number of their reduced representatives.

WritePartitioning the $k$ complete prime-power factors between $r$ and $s$ gives $2^k$ ordered factorizations $d=rs$ with $\gcd (r,s)=1$. After identifying $(r,s)$ with $(s,r)$, there are $2^{k-1}$ choices with $r\le s$. Each gives the primitive reduced formwhose discriminant is $-4rs=-4d$. The uniqueness of reduced representatives makes these classes distinct, proving the coprime-factorization lower bound for a quadratic-form class number

The inequality can be strict. For $d=5$, there is one distinct prime factor, while the reduced primitive forms of discriminant $-20$ areThus $h(-20)=2>1=2^{1-1}$.

The character of $(\rho ,V)$ isThe stated action is a representation because successive action by $(g_2,h_2)$ and $(g_1,h_1)$ gives

Identifying ${\mathrm{Hom}}_{\mathbb{C}}(V,W)$ with $V^{\ast }\otimes W$ gives the character of a Hom representationHere the final equality follows after unitarizing the finite-group representation.

The permutation character of $\mathbb{C}G$ counts fixed points:Thus

On the other hand, the character of
${⨁}_i\mathrm{Hom}(V_i,V_i)$ isThe column form of character orthogonality says that this has exactly the same two values above. Complex representations of a finite group are semisimple, and semisimple representations with the same character are isomorphic. Hence the two-sided regular representation decomposition isas a $G\times G$ representation.

The Turan theorem states that every $n$-vertex graph with no $K_{r+1}$ has at most $e(T_r(n))$ edges, where $T_r(n)$ is the complete $r$-partite graph with part sizes differing by at most one.

Here is an induction on $n$ and $r$. If a $K_{r+1}$-free graph $G$ contains no $K_r$, induction on $r$ givesOtherwise choose a copy $S$ of $K_r$. Every vertex outside $S$ has at most $r-1$ neighbors in $S$, and induction on $n$ givesThe last identity is obtained by removing one vertex from every part of $T_r(n)$. This proves the theorem; the standard equality analysis forces the balanced complete $r$-partite graph.

Now suppose $G$ is rhombus-free. If it is triangle-free, the $r=2$ case just proved gives
$e(G)\le e(T_2(n))$. Otherwise remove the three vertices of a triangle. No remaining vertex can be adjacent to two vertices of that triangle, since those two triangle vertices and the outside vertex would form a second triangle sharing an edge with the first. Induction therefore givesThis proves the rhombus-free edge bound and hence the requested strict contrapositive.

For equality at $n=6$, take the triangular prism graph: two disjoint triangles joined by a matching. It has $6+3=9=e(T_2(6))$ edges. Every edge belongs to at most one triangle, so there is no rhombus, while the presence of triangles proves that it is not isomorphic to $T_2(6)=K_{3,3}$.

The Hartogs theorem says that for every set $X$ there is an ordinal that does not inject into $X$. Indeed, the well-orderings of subsets of $X$ form a set, since each is a relation in $\mathcal{P}(X\times X)$. By replacement, their order types form a set of ordinals. Its supremum plus one cannot be the order type of a well-ordered subset of $X$, and hence cannot inject into $X$. The least such ordinal is the Hartogs ordinal $h(X)$.

Ordinal exponentiation is defined byfor nonzero limit $\lambda$. Fix $\alpha ,\beta$ and induct on $\gamma$. The zero case is immediate. If the identity holds at $\gamma$, thenAt a limit $\lambda$, continuity of exponentiation and right continuity of ordinal multiplication give

Put $\rho ={\omega }^{\alpha }$. The setis nonempty and bounded, so let $\beta =\sup B$. Continuity of multiplication in its right argument gives $\rho \beta \le \gamma$. There is a unique tail $\delta$ with $\gamma =\rho \beta +\delta$. If $\delta \ge \rho$, then
$\rho (\beta +1)=\rho \beta +\rho \le \gamma$, contrary to maximality. Hence $\delta <\rho$. If
$\rho \beta +\delta =\rho {\beta }^{\prime }+{\delta }^{\prime }$ with both remainders below $\rho$ and $\beta <{\beta }^{\prime }$, thena contradiction. This proves existence and uniqueness in the division by an additively indecomposable ordinal.

Now let $X$ be nonempty and well ordered. Its least element is not a limit point, so $X^{\prime }\ne X$. Every derivative is itself a well-ordered subset. If $X^{(\alpha )}$ were nonempty for every $\alpha <h(X)$, then each differencewould be nonempty. Choosing its least member would inject $h(X)$ into $X$, since the nested successive differences are disjoint. This contradicts Hartogs' lemma. Thus some derivative is empty, as asserted by the derived-set iteration of a well-order.

For an ordinal $\xi$, a point $\gamma <\xi$ is a limit point exactly when it is a nonzero limit ordinal. Division by $\omega$ writes $\gamma =\omega \beta +n$ with $n<\omega$; this is a nonzero limit exactly when $n=0$ and $\beta >0$. HenceWithin this derived set, $\omega \beta$ is a limit point exactly when $\beta$ is a nonzero limit ordinal, equivalently $\beta =\omega \eta$ for some $\eta >0$. Therefore the derived sets of an ordinal satisfyFinally ${\omega }^{\prime }=\varnothing$, so the index of $\omega$ is $1$, while
$({\omega }^2{)}^{\prime }=\{\omega ,2\omega ,3\omega ,\dots \}$ and
$({\omega }^2{)}^{\prime \prime }=\varnothing$, so the index of ${\omega }^2$ is $2$.