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The Arzelà-Ascoli theorem for $C([0,1])$ states that a family has compact closure in the uniform norm exactly when it is uniformly bounded and equicontinuous. To prove sufficiency, choose a countable dense set. From any sequence of functions, successive subsequence extraction at each dense point and a diagonal selection give convergence at all those points. Given $\epsilon >0$, equicontinuity gives a common $\delta$; cover the interval by finitely many $\delta$-balls centered at dense points. On a sufficiently late tail the functions differ by less than $\epsilon /3$ at those finitely many centres, and equicontinuity makes each of the other two differences less than $\epsilon /3$. Thus the selected subsequence is uniformly Cauchy. Completeness of $C([0,1])$ gives a continuous uniform limit. In a metric space this proves compactness of the closure. Conversely, a compact family has finite uniform $\epsilon$-nets. The finitely many net functions are bounded and uniformly continuous; approximation by them gives a common bound and equicontinuity for the family. This proves both directions.

For $f\in S_N$, a positive interior maximum would have $f^{\prime }=0$ and $f^{\prime \prime }\le 0$, contradicting $f^{\prime \prime }=f>0$. A negative interior minimum likewise contradicts $f^{\prime \prime }=f<0$. The endpoint bounds therefore give $|f|\le N$ everywhere.

Now put $w=f^{\prime }$. Differentiating the equation, using the stipulated third derivative, gives $w^{\prime \prime }=w+2w{w}^{\prime }$. At an interior extremum of $w$, $w^{\prime }=0$, so $w^{\prime \prime }=w$. The same maximum/minimum argument and the derivative endpoint bounds imply $|f^{\prime }|=|w|\le N$. Consequently $|f(x)-f(y)|\le N|x-y|$ by the mean value theorem. Thus $S_N$ is uniformly bounded and equicontinuous, and the theorem makes it relatively compact, hence totally bounded in $C([0,1])$. Its closure need not be asserted to remain thrice differentiable.

An action assigns each $g\in G$ a homeomorphism of $X$, with $e\cdot x=x$ and $g\cdot (h\cdot x)=(gh)\cdot x$. For the covering assertion use a properly discontinuous free action: every $x$ has an open neighbourhood $U$ such that $gU\cap U=\varnothing$ for every $g\ne e$. Then the translates $gU$ are pairwise disjoint. The quotient map $p$ is open because $p^{-1}(p(U))={\bigcup }_{g}gU$ is open. Each restriction $p:gU\to p(U)$ is a continuous open bijection and hence a homeomorphism. Thus $p(U)$ is evenly covered and $p:X\to X/G$ is a covering map.

For the surjective homomorphism one additionally needs $X$ path connected; choose $x_0$ and its image $y_0$. Lift a based loop at $y_0$ from $x_0$. Its endpoint has a unique form $g{x}_0$. Homotopy lifting makes this assignment well-defined on the fundamental group. With concatenation meaning first the first loop and then the second, the second lift from $g{x}_0$ is $g$ times the lift from $x_0$, so the endpoint is $gh{x}_0$. This proves the homomorphism property. For each $g$, a path in $X$ from $x_0$ to $g{x}_0$ projects to a loop realizing $g$. HencePointwise freeness alone is insufficient for a covering: an irrational rotation action of $\mathbb{Z}$ on the circle is free but its orbits are dense. Path connectedness is also essential for the final surjectivity claim as written: the discrete space $X=G$ with left translation is a free properly discontinuous action whose quotient is a point, and its trivial fundamental group cannot surject onto nontrivial $G$. The proof supplies the standard connected covering-space interpretation rather than hiding these hypotheses.

Apply the reciprocal-and-integer-part algorithm for the continued fraction. The successive complete quotients after the integer part two are $(\sqrt{7}+2)/3$, $(\sqrt{7}+1)/2$, $(\sqrt{7}+1)/3$, and $\sqrt{7}+2$. Their integer parts are $1,1,1,4$, and taking the next reciprocal repeats the first complete quotient. ThereforeThe first four continued fraction convergents are $2,3,5/2,8/3$. The last gives $8^2-7\cdot 3^2=1$. Squaring its unit $8+3\sqrt{7}$ gives $127+48\sqrt{7}$, whose norm is again one. Thus two positive solutions of the Pell equation areDirect verification gives $127^2-7\cdot 48^2=16129-16128=1$.

The permutation representation on the complex vector space with basis $X$ has permutation character ${\pi }_X(g)=|{\mathrm{Fix}}_X(g)|$. The multiplicity of the trivial character is $⟨{\pi }_X,1_G⟩=|G{|}^{-1}{\sum }_g|\mathrm{Fix}(g)|$. Count pairs $(g,x)$ with $gx=x$: each orbit contributes $|G|$, by the orbit-stabilizer theorem. Hence this multiplicity equals the number of orbits.

For a transitive action the average number of fixed points is one. The identity fixes $|X|>1$ points, so not every other element can fix at least one; otherwise the average would exceed one. Thus there is a derangement $g$ with ${\pi }_X(g)=0$.

If ${\pi }_X=1_G+m\chi$, evaluating at that derangement gives $\chi (g)=-1/m$. Character values are algebraic integers, since a finite-order representation matrix has roots of unity as eigenvalues, and the sum of algebraic integers is integral. A rational algebraic integer is an integer. Therefore $-1/m$ is an integer, forcing $m=1$.

The diagonal action on $X^2$ is $g(x,y)=(gx,gy)$. Its permutation character is ${\pi }_X(g{)}^2$, since both coordinates must be fixed. The orbit-count result and the real-valued character ${\pi }_X$ giveIf any $m_j\ge 2$ and another nontrivial constituent occurs, this is at least $1+4+1=6$. If just one nontrivial constituent occurs, the previous result forces its multiplicity to one. Thus $r\le 5$ implies every multiplicity is one, and the formula then gives $k=r$.

Hall marriage theorem says that a bipartite graph with equal finite vertex-class sizes has a perfect matching exactly when $|\Gamma (A)|\ge |A|$ for every $A\subseteq X$. Necessity follows because distinct matched partners of vertices of $A$ belong to $\Gamma (A)$.

For sufficiency, induct on $n$, the case $n=1$ being immediate. If a nonempty proper $A$ is tight, $|\Gamma (A)|=|A|$, the induced graph between $A$ and $\Gamma (A)$ satisfies Hall's condition and has a matching by induction. The remaining graph also satisfies it: for $B\subseteq X\setminus A$, the condition on $A\cup B$ gives $|\Gamma (B)\setminus \Gamma (A)|\ge |B|$. Induction gives the second matching, and their union is perfect. Otherwise every nonempty proper subset has strictly more neighbours. Choose an edge $xy$ and remove its ends. Every remaining nonempty subset loses at most one neighbour and originally had at least one spare neighbour, so Hall's condition survives. Induction matches the remaining graph, and restoring $xy$ completes the proof.

The last argument also proves flexibility under the strict condition: it works for every selected edge, so every vertex is flexible. The case $n=1$ is again immediate.

Now suppose only that a perfect matching exists. Choose a minimal nonempty tight set $A\subseteq X$; such a set exists since $X$ is tight. Every nonempty proper subset of $A$ satisfies the strict condition in the induced graph on $A,\Gamma (A)$. Hence every edge from a vertex of $A$ extends to a matching of that induced graph. There are no edges from $A$ to the complement of $\Gamma (A)$. Restrict the original perfect matching to the complementary vertex classes and append it to this new induced matching. This makes every vertex of $A$ flexible in the whole graph, proving that at least one flexible vertex exists.

For the requested counterexample, take $X=\{x_1,x_2,x_3,x_4\}$ and $Y=\{y_1,y_2,y_3,y_4\}$. Join $x_1$ only to $y_1,y_2$, and join each of $x_2,x_3,x_4$ to all of $y_2,y_3,y_4$. There is a perfect matching using $x_1{y}_1$ and a matching of the remaining complete bipartite graph. The unique minimum-degree vertex is $x_1$, of degree two, but $x_1{y}_2$ cannot occur in a perfect matching because $y_1$ has no other neighbour. Thus no minimum-degree vertex is flexible, while the other three vertices are flexible.

The recursive ordinal addition definitions are $a+0=a$, $a+(b+1)=(a+b)+1$, and $a+\lambda ={\sup }_{b<\lambda }(a+b)$ for a nonzero limit ordinal. Ordinal multiplication is $a\cdot 0=0$, $a\cdot (b+1)=a\cdot b+a$, and $a\cdot \lambda ={\sup }_{b<\lambda }a\cdot b$. For $a>0$, ordinal exponentiation is $a^0=1$, $a^{b+1}=a^b\cdot a$, and $a^{\lambda }={\sup }_{b<\lambda }{a}^b$; the exceptional zero base has $0^0=1$ and $0^b=0$ for $b>0$.

Transfinite induction gives ${\omega }^a\ge a$. At zero this is immediate; at a successor, ${\omega }^{b+1}={\omega }^b\omega \ge {\omega }^b+1\ge b+1$. At a limit, continuity and the induction hypothesis give ${\sup }_{b<a}{\omega }^b\ge {\sup }_{b<a}b=a$. Also $a\mapsto {\omega }^a$ is strictly increasing: its successor values strictly increase, and continuity preserves the strict inequality between any two distinct indices.

For $\alpha >0$, ${\omega }^{\alpha +1}>{\omega }^{\alpha }\ge \alpha$, so there is a least $\delta$ with ${\omega }^{\delta }>\alpha$. It cannot be zero and cannot be a limit: at a limit, a supremum greater than $\alpha$ has a term already greater than $\alpha$, contradicting minimality. Thus $\delta ={\alpha }_0+1$, andStrict increase proves uniqueness, and ${\alpha }_0\le {\omega }^{{\alpha }_0}\le \alpha$ proves the requested index bound.

Choose the largest positive integer $a_0$ with ${\omega }^{{\alpha }_0}{a}_0\le \alpha$. It exists because $\alpha <{\omega }^{{\alpha }_0}\omega ={\sup }_{m<\omega }{\omega }^{{\alpha }_0}m$. The remaining tail of the ordinal interval has an order type $\rho$ with $\alpha ={\omega }^{{\alpha }_0}{a}_0+\rho$; maximality implies $\rho <{\omega }^{{\alpha }_0}$. If $\rho >0$, repeat the construction on it. The leading exponents strictly decrease, and there is no infinite strictly decreasing sequence of ordinals: its set of values would have a least member followed by a smaller one. Hence the process terminates, yielding the Cantor normal formThe first exponent is determined by the preceding band, the first coefficient by maximality, and the tail by the ordinal interval left after that initial segment. Recursion therefore proves uniqueness of every exponent and coefficient. The PDF's second coefficient is printed ${\alpha }_1$ rather than $a_1$; the stated natural-number coefficients and the construction identify the intended expression.

For the final assertion, powers of $\omega$ are additively indecomposable: if $\tau <{\omega }^{\eta }$, then $\tau +{\omega }^{\eta }={\omega }^{\eta }$ for $\eta >0$. One proves this by induction on $\eta$. At a successor, bound $\tau$ by a finite multiple of ${\omega }^{\eta -1}$ and use the supremum over all such multiples; at a limit, bound $\tau$ by a smaller power and use continuity. For $\eta =0$ the only possible $\tau$ is zero. Thus a smaller leading exponent is absorbed when a larger-leading ordinal is added on its right. If $\beta ,\gamma$ have unequal leading exponents, either $\beta +\gamma =\gamma$ or $\gamma +\beta =\beta$. If their leading exponents coincide, the leading coefficient of either sum is the sum of their two positive leading coefficients; it exceeds each coefficient, so neither absorption equality holds. Zero is never commensurable with an ordinal. Consequently commensurability is exactly equality of the leading exponent, equivalently membership of both positive ordinals in one of the stated bands.

By the gravitational shell theorem, the inward acceleration at radius $r$ is $Gm(r)/r^2$, where $m(r)=4\pi {\int }_0^r\rho (s)s^{2}ds$. Force balance for a thin fluid shell gives $-\nabla P=\rho \nabla \Phi$, and henceAssembling shells from infinity gives $E_{\mathrm{grav}}=-{\int }_0^{R}Gm(r)dm(r)/r=-4\pi G{\int }_0^{R}m\rho rdr$. Using the pressure equation and integration by parts yieldsFor a star with negligible external pressure, $P(R)=0$, this is $E_{\mathrm{grav}}=-3⟨P⟩V$. Nonzero external pressure would add the displayed surface term.

For a nonrelativistic ideal gas, the virial relation is $E_{\mathrm{grav}}=-2E_{\mathrm{kin}}$ and total energy is negative. Under adiabatic compression its kinetic energy scales as $R^{-2}$, faster than the magnitude $R^{-1}$ of gravity, providing restoring support; its adiabatic index $5/3$ exceeds $4/3$. For an ultrarelativistic gas, $E_{\mathrm{grav}}=-E_{\mathrm{kin}}$ and both terms scale as $R^{-1}$, giving marginal stability at adiabatic index $4/3$ in this idealized Newtonian model. Corrections such as general relativity can then destabilize it.

Integrating the kinetic energy $p^2/(2m_p)$ over filled momentum states givesSince $p_F=h[3n/(4\pi g_s){]}^{1/3}$, the pressure isFinally $|E_{\mathrm{grav}}|\sim G{M}^2/R$, whereas $PV\sim (h^2/m_p)[M/(m_p{R}^3){]}^{5/3}{R}^3$. Equating these by the virial theorem givesThis is an order-of-magnitude estimate for nonrelativistic degeneracy support, not a uniform-density exact stellar solution.