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For a finite bipartite graph with left vertex set $L$, Hall marriage theorem says that a matching in a graph covering $L$ exists exactly when $|N(S)|\ge |S|$ for every $S\subseteq L$. Necessity follows because distinct matched neighbors of $S$ lie in $N(S)$.

For sufficiency, induct on $|L|$, the empty case being immediate. If a nonempty proper subset $S$ has $|N(S)|=|S|$, apply the induction hypothesis to the graph on $S,N(S)$. For the remaining graph, every $T\subseteq L\setminus S$ satisfiesso induction matches the remaining vertices too. If there is no such tight subset and $|L|>1$, every nonempty proper subset has at least one extra neighbor. Select an edge and remove its endpoints. Each subset of the remaining left vertices loses at most one neighbor, so it still satisfies the required condition; induction supplies the rest of the matching in a graph. The case $|L|=1$ follows directly from the condition.

For a doubly stochastic matrix $A=(a_{ij})$, make a bipartite graph joining row $i$ to column $j$ when $a_{ij}>0$. For a row subset $S$,Hall marriage theorem therefore gives a perfect matching, or a permutation $\sigma$, with $\boxed{a_{i,\sigma (i)}>0\text{ for every }i}$.

Let $\mathbf{r}$ be position relative to the center of mass, and let the new origin have vector $\mathbf{c}$ relative to it. The inertia tensor about this origin isExpanding and using $\int \mathbf{r}dm=0$ proves the tensor form of the parallel axis theorem:For a uniform cube of side $L$ and mass $M$, centered coordinate integrals give $I_0=(M{L}^2/6)I$. The displacement to a vertex is $(L/2,L/2,L/2)$ up to signs. Using axes along its incident edges givesThe principal moments of inertia are $\boxed{M{L}^2/6,11M{L}^2/12,11M{L}^2/12}$. The first principal axis is the body diagonal through the vertex; every direction perpendicular to it has the repeated moment.

Choose a branch of $t=1-z$ with $|\arg t|<\pi$ and write $d=c-a-b\notin \mathbb{Z}$. The Gauss hypergeometric equation isChanging to $t$ turns this into the same equation with third parameter $a+b+1-c=1-d$. Its solution regular at $t=0$ is $y_1=F(a,b,1-d;t)$. Substitution of $y=t^{d}v$ gives the Gauss hypergeometric equation for $v$ with parameters $c-a,c-b,1+d$. Thus $y_2=t^{d}F(c-a,c-b,1+d;t)$ is also a solution. Their distinct nonintegrally separated characteristic exponents at a regular singular point $0,d$ imply linear independence near $z=1$, soThis also follows from the exponent transformation in the Papperitz symbol.

Initially assume $\Re c>\Re b>0$, so the Euler integral for the hypergeometric function applies. If $\Re d>0$, taking $z\to 1$ gives an integral evaluated by the beta function and determinesIndeed $y_1\to 1$ and $y_2\to 0$ there. To find $B$, instead work in the nonempty parameter region $\Re d<0$, put $z=1-ϵ$ and scale the integration variable by $1-s=ϵu$. The singular contribution isThe defining integral of the beta function evaluates this integral to $\Gamma (c-b)\Gamma (-d)/\Gamma (a)$. Since ${ϵ}^{-d}{y}_1\to 0$ and ${ϵ}^{-d}{y}_2\to 1$, it determines $B$. Analytic continuation in the parameters then yields the hypergeometric connection formulaThe usual power-series definition requires $c\notin \{0,-1,-2,\dots \}$; exceptional parameters need a separately specified analytic limit. Reciprocal gamma functions make vanishing coefficients at denominator poles meaningful. The branch must remain fixed when continuing in $z$.

For a discrete random variable with probabilities $p_i$, its Shannon entropy in bits is $H(X)=-{\sum }_i{p}_i{\log }_2{p}_i$, with $0{\log }_{2}0=0$. Gibbs inequality states thatwith equality exactly when $p=q$. If $q_i=0<p_i$, the left side is infinite. Otherwise apply $-\ln u\ge 1-u$ to $u=q_i/p_i$ for $p_i>0$:Equality requires $q_i=p_i$ on the support and no remaining mass outside it. In particular, comparison with the uniform probability distribution on two outcomes gives binary Shannon entropy at most $1$.

Write $q=1-p_1$. For $q>0$, grouping the last two outcomes givesEquality holds if $p_2=p_3$; if $q=0$ it also holds, and then both are zero. Thus equality holds exactly when $p_2=p_3$.

For the discrete memoryless channel, assume $\alpha ,\beta \ge 0$ and $\alpha +\beta \le 1$, and set $w=1-\alpha -\beta$. Both conditional output probability distributions have Shannon entropy $H(w,\alpha ,\beta )$. The third output has probability $\beta$ regardless of the input; the first two have total probability $1-\beta$. Their Shannon entropy is maximized by making them equal, which a uniform input achieves. Hence the channel capacity isand thereforeThe usual zero-times-log convention includes $\beta =1$, when $C=0$. If $w=\alpha$, the rows coincide and every input achieves zero mutual information; otherwise the uniform input is the unique maximizing input.

The Baire category theorem says that in a nonempty complete metric space, every countable intersection of open dense sets is dense. To prove it, start in an arbitrary nonempty open set $O$ and let $G_n$ be open dense. Inductively choose closed balls $\overline{B}(x_n,r_n)$ with $0<r_n\le 2^{-n}$, contained in $G_n$ and in the interior of the previous ball, with the first also contained in $O$. Density provides a point at each stage, and openness permits the smaller closed ball. The centers are a Cauchy sequence; completeness supplies a limit lying in every closed ball, hence in $O\cap {\bigcap }_n{G}_n$. The proof also works at isolated points, where sufficiently small balls are singletons.

The incomplete metric space $\mathbb{Q}$ is a counterexample: it is the countable union of its closed nowhere dense singletons, so the complements are open dense with empty intersection.

For the sequence in the problem, putEach $E_N$ is closed by continuity. Pointwise convergence makes each point's values a Cauchy sequence, so $[0,1]={\bigcup }_N{E}_N$. By the Baire category theorem, one $E_N$ has nonempty relative interior, which contains a nonempty open interval $J$ in $(0,1)$. This proves the required common tail bound.

If continuous $g_n$ converged pointwise to the rationality indicator, take $ϵ=1/4$ and let $m\to \infty$ in this bound. On $J$, $|g_n-g|\le 1/4$ for $n\ge N$. Hence a fixed continuous $g_N$ is at least $3/4$ at every rational point and at most $1/4$ at every irrational point of $J$. The density of both sets contradicts continuity. Thus the rationality indicator is not a pointwise limit of continuous functions.

Nevertheless $h_n(x)=x^n$ are continuous and converge pointwise to $h(x)=0$ for $x<1$, $h(1)=1$. This limit is discontinuous at $1$: a pointwise limit of continuous functions can be discontinuous.

For comoving galaxies their relative comoving vector $\mathbf{\chi }$ is constant, and the physical displacement is $\mathbf{r}(t)=a(t)\mathbf{\chi }$, where $a$ is the scale factor. Differentiating gives the Hubble lawPeculiar velocity relative to this comoving frame is excluded in this relation.

Transport a wave crest between neighbouring comoving observers separated by a small physical distance $d\ell$. Their recession speed is $Hd\ell$, while the light travel time is $dt=d\ell /c$ to first order. The local Doppler shift therefore changes its wavelength by $d\lambda /\lambda =Hdt$. Multiplying these infinitesimal shifts along the ray, or integrating their logarithms, yieldsThe use of the local first-order Doppler law is exact in the limit of infinitesimal steps: the accumulated second-order errors tend to zero. A single special-relativistic recession velocity assigned to widely separated galaxies would not justify this cosmological redshift relation.

Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.