If $x\ne b$, then $d_a(b,x)=d(b,a)+d(a,x)\ge d(b,a)$. The strict open-ball inequality therefore excludes every such $x$, while including $b$. Hence $\boxed{B_{d_a}(b,d(b,a))=\{b\}}$. Every point other than $a$ is isolated in this metric.

Write $r_x=d(x,a)$. The prescribed function is symmetric and nonnegative. If $x\ne y$, both $r_x,r_y$ could be zero only if $x=y=a$, so $d_a(x,y)>0$. For the triangle inequality, if any two of $x,y,z$ coincide it is immediate. If all three are distinct,Thus $\boxed{d_a\text{ is a metric}}$. Also $d_a(a,x)=d(a,x)$, a fact useful in the following parts.

The map is continuous if it is continuous at every point of its domain. Equivalently the inverse image of every open subset of $Y$ is open in $X$. These equivalent formulations use the given metrics' induced topologies.

Continuity at $x$ means that for every $\epsilon >0$ there is $\delta >0$ such that $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\epsilon$ for every $y\in X$. Equivalently, the inverse image of each neighborhood of $f(x)$ is a neighborhood of $x$.

A subset $U$ of a metric space is open if for every $x\in U$ there exists $r>0$ with $B_d(x,r)\subseteq U$, where $B_d(x,r)=\{y:d(x,y)<r\}$. The radius may depend on the point. The empty set is open vacuously.

For $x\ge 1$, $(1+x{)}^2\le 4x^2$, so $x\log x/(1+x{)}^2\ge (\log x)/(4x)\ge 0$. The latter integral diverges because its primitive is $(\log x{)}^2/8$. Thus the absolute integral diverges and the answer is $\boxed{\text{not integrable on }(1,\infty )}$.

On $(0,1)$ its absolute value is at most $x|\log x|$, which has integral $1/4$, by integration by parts or the substitution $x=e^{-t}$. The function is measurable and this majorant is integrable, so $\boxed{x\log x/(1+x{)}^2\text{ is integrable on }(0,1)}$.

Integrating from one to $R$ gives $(R^{\alpha +1}-1)/(\alpha +1)$ for $\alpha \ne -1$. It has a finite limit as $R\to \infty$ exactly when $\alpha +1<0$, while at $\alpha =-1$ the logarithm diverges. Nonnegativity equates finiteness of this integral with Lebesgue integrability. Hence $\boxed{x^{\alpha }\in L^1(1,\infty )⟺\alpha <-1}$.

The function is nonnegative and measurable. For $\alpha \ne -1$, integrate from $\epsilon$ to one to get $(1-{\epsilon }^{\alpha +1})/(\alpha +1)$. The limit is finite exactly when $\alpha +1>0$. At $\alpha =-1$ the integral is $-\log \epsilon \to \infty$. Thus $\boxed{x^{\alpha }\in L^1(0,1)⟺\alpha >-1}$.

Normalize each optimal vector by its sum $1/4$. Player I's row strategy and Player II's column strategy are respectivelyAs a direct check $Aq=(4,4{)}^{\mathsf{T}}$ and $p^{\mathsf{T}}A=(4,5,4)$, so Player II guarantees a loss of at most four and Player I guarantees a gain of at least four.

Both objectives equal $1/4$: $1^{\mathsf{T}}x=1/12+1/6=1/4$ and $1^{\mathsf{T}}y=1/6+1/12=1/4$. Weak duality bounds the value of every primal feasible point above by every dual feasible objective. Equality for this feasible pair therefore proves both are optimal, without needing to solve either problem further.

The proposed vectors are nonnegative, and direct multiplication givesThus $Ax\le 1$ and $A^{\mathsf{T}}y\ge 1$, proving primal and dual feasibility.

The dual isFor positive payoffs and an optimal solution $y_{\ast }$, strong duality gives $1^{\mathsf{T}}{y}_{\ast }=1/v$. Normalizing gives Player I's optimal row strategy $p_{\ast }=y_{\ast }/(1^{\mathsf{T}}{y}_{\ast })$. Its expected payoff against every column is at least $v$. Thus the primal represents Player II's minimization of worst loss and the dual represents Player I's maximization of guaranteed gain.

Strictly positive entries imply $v>0$. If $q$ is an optimal strategy, set $x=q/v$. Then $Ax\le 1$ and $1^{\mathsf{T}}x=1/v$. Conversely, a nonzero feasible $x$ with $s=1^{\mathsf{T}}x$ gives a strategy $q=x/s$ satisfying ${\max }_i(Aq{)}_i\le 1/s$. Thus $v\le 1/s$, or $s\le 1/v$. The optimal objective is therefore $s_{\ast }=1/v$, andA nonzero feasible point exists by taking a sufficiently small positive multiple of a probability vector, and positivity of $A$ bounds the feasible objective, so the optimum is attained.

For general entries choose a constant $c$ so that $A^{\prime }=A+c11^{\mathsf{T}}$ is strictly positive. Every pair of strategies has its payoff increased by exactly $c$, so optimal strategies are unchanged. Solve the positive problem for $A^{\prime }$, recover $v^{\prime }=1/s_{\ast }^{\prime }$, and then $\boxed{v=v^{\prime }-c}$ for the original game.

A mixed strategy for Player II is a probability vector $q\ge 0$ with ${\sum }_j{q}_j=1$, specifying the random choice of a column. If Player I chooses row $i$, the expected payment is $(Aq{)}_i$. A mixture of rows cannot exceed the maximum of these row expectations, so the smallest payment Player II can guarantee isAn optimal strategy is a vector attaining that minimum, which exists by compactness of the probability simplex and continuity. The minimax theorem identifies this guaranteed loss with Player I's maximal guaranteed gain, the value of the finite zero-sum game.

No. At any cube vertex its three incident faces are pairwise adjacent along the three incident edges. They would therefore require three different colours. Equivalently the dual of the planar cube embedding contains a triangle, which obstructs a face two-colouring.

Yes. Colour an even-parity vertex one colour and an odd-parity vertex the other. Every cube edge changes parity, so this is a proper two-colouring. At least two colours are needed because edges exist.

Both are bipartite. Partition the vertices by parity of the number of one bits. A cube edge changes one bit and an interior diagonal changes three bits, so both types join opposite parity classes. In fact $Q_3^{\prime }$ is $K_{4,4}$: from any vertex its three cube neighbors and its opposite vertex exhaust the four vertices of opposite parity.

Both are Hamiltonian. A Hamiltonian cycle already in the cube isSuccessive triples differ in exactly one bit, and each of the eight vertices occurs exactly once before returning. Adding edges preserves this cycle in a graph.