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There is an essential qualification to the printed hypothesis. If compactly generated group means merely generated algebraically by a compact subset, the claim is false. In the additive group $\mathbb{Q}$ with its usual topology,is compact, being a convergent sequence together with its limit, and generates every rational number since any integer denominator divides some factorial. The group is metrizable. However, an invariant Radon measure has a common mass $c$ on each singleton. If $c>0$, the compact set $K$ has infinite measure, contradicting finiteness on compact sets. If $c=0$, countable additivity makes the entire countable group have measure zero. Thus it has no nonzero Haar measure. This proves that compact generation does not imply local compactness and disproves the literal unrestricted statement.

The intended existence statement is valid when local compactness is included, either explicitly or in the convention that the group has a compact generating neighbourhood. Here is a construction for a locally compact, compactly generated, metrizable Hausdorff group. No invariant measure is assumed during its construction.

Enlarge a compact generating set to a symmetric compact neighbourhood of the identity element $S$. Then $G={\bigcup }_n{S}^n$, and each $S^n$ is compact. Moreover $S^n$ lies in the interior of $S^{n+1}$, since multiplication by the open neighbourhood of the identity element contained in $S$ supplies an open neighbourhood of every point of $S^n$. This gives a compact exhaustion. Every compact metric space has a countable dense subset: take finite $1/k$-nets and unite them. Their union over this exhaustion is dense in $G$, so the metric space is separable and has a countable open basis.

We use nonnegative compactly supported continuous functions and cutoffs equal to one on specified compact sets. Their existence is covered by the source's permission concerning auxiliary functions. Fix a nonzero such function $h$. For nonzero $u\ge 0$ define the weighted covering quantityThe quantities are finite: an open set where $u$ is bounded below has translates covering the compact support of $f$, so finitely many sufficiently large weighted translates dominate it. The denominator is positive, since evaluation at a point where $h>0$ gives $(h:u)\ge h(y)/\Vert u{\Vert }_{\infty }>0$.

Each $I_u$ is nonnegative, homogeneous, monotone, left invariant and subadditive; these follow respectively by rescaling, comparing or combining covers, and translating their centers. Also $I_u(h)=1$. Composing a cover of $f$ by translates of $h$ with a cover of $h$ by translates of $u$ givesThis is a bound independent of $u$. On functions supported in a fixed compact set $C$, choose a cutoff $k\ge 1$ on $C$. Monotonicity and subadditivity implyThus the covering functionals are uniformly Lipschitz on each fixed support.

Choose a nonzero $u_n\ge 0$ with support in an neighbourhood of the identity element shrinking to $e$. The nonnegative functions supported in each compact member of the exhaustion form a separable metric cone in the supremum norm, as a subspace of continuous functions on that compact metric space. Select a countable dense family from each cone and include $h$. The uniform scalar bounds permit a diagonal subsequence along which $I_{u_n}$ converges on their countable union. The displayed Lipschitz estimate extends convergence to every nonnegative member of $C_c(G)$. Denote the limit by $I$. It inherits positivity, homogeneity, monotonicity, subadditivity, invariance and $I(h)=1$.

The crucial remaining step is additivity. Fix $f_1,f_2\ge 0$ with combined support $C$, and fix a compact neighbourhood of the identity element $V_0$. Choose a cutoff $k$ equal to one on $C{V}_0^{-1}{V}_0$. For $\delta >0$, putsetting the ratio to zero off the denominator's nonzero set. These ratios are continuous: $k=1$ on a neighbourhood of the supports, and off those supports the numerator vanishes. Their sum is at most one. Compactness and continuity give, for each $\eta >0$, an neighbourhood of the identity element $V$ small enough that each $r_i$ oscillates by at most $\eta$ on any translate of $V$ meeting $C$. To justify this uniform choice, such translates lie inside the fixed compact set $C{V}_0^{-1}{V}_0$ when $V\subset V_0$; continuity of $(y,v)\mapsto r_i(yv)$ and a finite cover of that compact set give a single small right-increment neighbourhood, then choose $V^{-1}V$ inside it.

If $u$ has support in $V$ and $F\le {\sum }_j{c}_j{T}_{x_j}u$, discard translates not meeting $C$ when covering $f_i$. Let $a_{ij}$ be the supremum of $r_i$ on the remaining translate's support. Then $f_i\le {\sum }_j{a}_{ij}{c}_j{T}_{x_j}u$, and comparing both suprema with the ratios at one common point gives $a_{1j}+a_{2j}\le 1+2\eta$. Taking covering infima yieldsAlong the subsequence this holds eventually, since its supports shrink. Pass to $I$, then let $\eta \downarrow 0$ and $\delta \downarrow 0$; the uniform bound on $I_u(k)$ gives $I(f_1)+I(f_2)\le I(f_1+f_2)$. Subadditivity gives the reverse inequality. Thus $I$ is additive.

Extend $I$ linearly to real functions using positive and negative parts, and then complex-linearly to $C_c(G)$. The compact-support bounds above make it a positive, locally bounded functional. The Riesz representation on compactly supported continuous functions gives a Radon measure $\mu$ with $I(f)=\int fd\mu$. Translation invariance of $I$ and uniqueness in that representation give $\mu (xE)=\mu (E)$ for Borel $E$. It is nonzero because $I(h)=1$, and finite on compact sets because a cutoff dominating their indicators has finite integral. Finally every nonempty open set has positive measure: otherwise its translates cover the compact support of $h$, a finite subcover would give that support measure zero, contradicting $I(h)=1$. Hence $\mu$ is a Haar measure under the stated local-compactness qualification. This is Haar measure from normalized covering functionals, with the limiting additivity proved explicitly.

There is an indexing issue in the printed topology. Its cumulative sums omit $a_0$, whereas the post-slot workload of a queue in part (a) includes $a_0$. To see the problem, take all $a_t=\lambda$ for $t\le -1$, with $\lambda <c$, and compare two inputs with $a_0=\lambda$ and $a_0=c+1$. Their distance according to the printed cumulative-sum expression is zero, yetThus the post-slot workload is not continuous in the literal printed topology. On two-sided sequences that expression is a seminorm, since it also ignores future coordinates; the fixed-mean input class is not a vector space (if signed inputs are allowed, it is an affine set).

The appropriate repair for the post-slot convention is to use one-sided past inputs including time zero and the weighted cumulative-input topology for a slotted queue:This is a finite metric on the fixed-mean class, since cumulative differences are $o(n)$; equality of all cumulative sums implies equality of all past coordinates. It is equivalent to augmenting the printed cumulative-difference distance by $|a_0-b_0|$: their cumulative sums differ only by this extra coordinate and a one-step shift.

Here is the full corrected continuity proof. Fix $a$ with $A_n(a)/n\to \lambda <c$, and let $\epsilon =(c-\lambda )/4$. Choose $N\ge 1$ such that $A_n(a)/n\le \lambda +\epsilon$ whenever $n\ge N$. If $d_{\#}(a,b)<\epsilon /2$, then, for all such $n$,Both queue workloads therefore maximize over the same finite collection $0\le n<N$, containing the value zero. For any two finite lists, the difference of their maxima is at most the maximum absolute difference of corresponding entries. Thusin this neighborhood. In particular the queue workload is locally Lipschitz continuous and hence continuous.

An alternative indexing repair retains the printed topology but interprets the time-zero function as the workload before slot $(-1,0)$, namely ${\sup }_{n\ge 0}\{{\sum }_{j=1}^n{a}_{-j}-cn\}$. The same finite-window argument proves its continuity. One must then shift the time labels consistently in the later recursions. The remaining answers use the post-slot convention and $d_{\#}$.

The weighted binary-sequence metric is finite because $0\le d(ϵ,\eta )\le {\sum }_{n\ge 1}{2}^{-n}=1$. It is symmetric and nonnegative. If $d(ϵ,\eta )=0$, every nonnegative summand is zero, so every coordinate agrees; conversely agreement makes the sum zero. For three sequences, apply $|{ϵ}_n-{\zeta }_n|\le |{ϵ}_n-{\eta }_n|+|{\eta }_n-{\zeta }_n|$ in each coordinate and sum. This proves the triangle inequality, so $d$ is a metric.

If two sequences differ in one of their first $n$ coordinates, their distance is at least $2^{-n}$. Thus the open ball $B_d(\eta ,2^{-n})$ is contained in $C_n(\eta )$. For every $\eta \in C_n(ϵ)$, the latter cylinder equals $C_n(\eta )$, so each cylinder set is open in the metric topology. Every set in the specified topology, being a union of cylinder sets, is therefore metric-open.

Conversely, agreement in the first $n$ coordinates givesGiven a metric-open set containing $ϵ$, choose a ball $B_d(ϵ,r)$ inside it, then choose $n$ with $2^{-n}<r$. The bound gives $C_n(ϵ)\subseteq B_d(ϵ,r)$, establishing the required prefix-neighbourhood condition. The two topologies coincide. The strict inequality $2^{-n}<r$ avoids incorrectly treating the closed tail bound as a strict ball bound.

The sum of the two metrics is nonnegative and symmetric. It vanishes exactly when both coordinate distances vanish, hence exactly when the two points agree. Adding the two component triangle inequalities gives the triangle inequality on the Cartesian product, so the displayed function is a metric. Moreover,so the projection map is $1$-Lipschitz and therefore continuous.

To prove sequential compactness directly from compactness, take a sequence $(x_n)$ in $X$. If every point $x$ had a neighborhood containing sequence terms at only finitely many indices, these neighborhoods would form an open cover with a finite subcover, implying that only finitely many sequence terms exist, a contradiction. Thus some $x\in X$ has infinitely many indices in every neighborhood. Choose increasing indices $n_k$ with $d_X(x_{n_k},x)<1/k$; then $x_{n_k}\to x\in X$.

Now let $F\subset X\times Y$ be closed and suppose $y_n\in \pi (F)$ converges to $y$. Choose $x_n$ such that $(x_n,y_n)\in F$. A convergent subsequence $x_{n_k}\to x$ exists by the preceding argument. In the product metric, $(x_{n_k},y_{n_k})\to (x,y)$, and the fact that $F$ is closed gives $(x,y)\in F$. Thus $y\in \pi (F)$, proving that the projection with a compact factor is closed. The sequential criterion for a closed set in a metric space follows by choosing points at distance less than $1/k$ from any point of its closure.

For a counterexample without a compact factor, take $X=Y=\mathbb{R}$ with their usual metrics andThe continuous product function makes $F$ closed, but $\pi (F)=\mathbb{R}\setminus \{0\}$ is not closed. Thus compactness of the discarded factor is essential to the general conclusion.

Let $x$ have a dense forward orbit $O=\{f^j(x):j\ge 0\}$, and suppose that $f$ is surjective. Every tail $O_k=\{f^j(x):j\ge k\}$ is then dense. Indeed, continuity givesand the left side equals $f^k(X)=X$ by surjectivity.

Given nonempty open sets $U,V$, choose $k\ge 0$ with $f^k(x)\in U$. The dense tail $O_{k+1}$ supplies $j\ge k+1$ with $f^j(x)\in V$. Taking $n=j-k\ge 1$ gives $f^j(x)\in f^n(U)\cap V$. This proves the required positive-time intersection property.

The result can fail without surjectivity. Take the compact metric spacewith the metric induced from the real line. All nonzero points are isolated, so continuity there is automatic; at zero, $f(1/j)\to 0=f(0)$ proves continuity. The forward orbit of $1$ is $\{1,1/2,1/3,\dots \}$, whose closure is $X$, so this is point transitivity. But $f(X)=X\setminus \{1\}$. The singletons $U=\{1/2\}$ and $V=\{1\}$ are nonempty open sets, and for every $n\ge 1$,The isolated initial point can be visited only at the start of the dense forward orbit; deleting that initial segment destroys density.

Use the separated-set definition of topological entropy. For a compatible metric $d$ and $n\ge 1$, define the Bowen metric and separated-set count byFor fixed $n$, the metric $d_n$ generates the original topology: it includes the $k=0$ term and all finitely many iterations of a map are continuous. Thus its space is compact and $s_n(d,ϵ)$ is finite. With natural logarithms, setThe last limit exists, possibly with value infinity, because decreasing $ϵ$ can only increase separated-set counts.

Let $d^{\prime }$ be another compatible metric. The identity $(X,d)\to (X,d^{\prime })$ is a continuous function on a compact domain, hence uniformly continuous. For every $ϵ>0$, choose $\delta >0$ such that $d(x,y)<\delta$ implies $d^{\prime }(x,y)<ϵ$. The same implication at every iterate givesBy its contrapositive, every $ϵ$-separated set for $d_n^{\prime }$ is $\delta$-separated for $d_n$. Consequently $s_n(d^{\prime },ϵ)\le s_n(d,\delta )$ for every $n$, andTaking $ϵ\downarrow 0$ gives $h_{d^{\prime }}(f)\le h_d(f)$. Interchanging the two compatible metrics gives the reverse inequality. ThereforeCompactness is used for uniform continuity, not merely for pointwise equivalence of the two metrics.

Write the hyperbolic toral automorphism as $T_A([v])=[Av]$ on ${\mathbb{T}}^2={\mathbb{R}}^2/{\mathbb{Z}}^2$, where $A\in \mathrm{GL}(2,\mathbb{Z})$. Its determinant is $\pm 1$. The eigenvalues cannot be a nonreal conjugate pair: their product would have absolute value one, forcing each to have absolute value one, contrary to hyperbolicity. The same product condition and hyperbolicity imply two distinct real eigenvalues, labelled so thatWe will proveThis is the entropy of a hyperbolic toral automorphism; the absolute value is important when the expanding eigenvalue is negative.

Choose real eigenvectors $v_s,v_u$ and the adapted norm $\Vert a{v}_s+b{v}_u{\Vert }_{\ast }=\max (|a|,|b|)$. Its quotient metric on the torus isIt is a compatible metric, so part (a) allows its use. Let $\rho ={\min }_{0\ne \ell \in {\mathbb{Z}}^2}\Vert \ell {\Vert }_{\ast }>0$ and $D=\Vert A{\Vert }_{\ast }=|{\lambda }_u|$. Fix $ϵ>0$ small enough that $2ϵ<\rho$ and $(D+1)ϵ<\rho$.

If $[v]$ lies in the Bowen ball $B_n(0,ϵ)$, each $T_A^k[v]$ has a unique lift $z_k$ with $\Vert z_k{\Vert }_{\ast }<ϵ$. Uniqueness follows because two such lifts differ by a lattice vector of norm less than $2ϵ<\rho$. Moreover, $z_{k+1}-A{z}_k$ is a lattice vector, andIt must be zero. Thus $z_k=A^k{z}_0$: there is no hidden lattice jump between successive lifted iterates.

Write $z_0=a{v}_s+b{v}_u$. The simultaneous bounds $\Vert A^k{z}_0{\Vert }_{\ast }<ϵ$ for $0\le k<n$ are exactlyThe stable coordinate is largest at time zero and the unstable coordinate at time $n-1$. Conversely, every vector in this rectangle satisfies all the Bowen ball bounds. The quotient is injective on this rectangle by $2ϵ<\rho$. If $J=|\det (v_s,v_u)|$, its normalized Haar measure is thereforeThe quotient metric is translation invariant and $T_A$ is linear, so $d_n(x,y)=d_n(0,y-x)$; every Bowen ball has this same measure, independently of its centre.

Take an $ϵ$-separated set of largest cardinality $s_n(d,ϵ)$. It is maximal under inclusion, so its radius-$ϵ$ Bowen balls cover the torus: a point outside all these balls could be added. The covering bound givesIts radius-$ϵ/2$ Bowen balls are disjoint, by the triangle inequality for $d_n$. Their measures sum to at most one, givingTaking logarithms, dividing by $n$, and letting $n\to \infty$ yields $h_d(T_A,ϵ)=\log |{\lambda }_u|$ for every sufficiently small $ϵ$. The limit as $ϵ\downarrow 0$ proves the answer. Contraction in the stable direction contributes no exponential growth; only expansion in the unstable direction does.

Write $v_2(m)$ for the 2-adic valuation of a nonzero integer $m$. Then $d(a,b)=2^{-v_2(a-b)}$ for unequal arguments. It is positive for $a\ne b$, vanishes exactly when $a=b$, and is symmetric because $v_2(-m)=v_2(m)$.

For the triangle inequality, if $2^r$ divides $a-b$ and $2^s$ divides $b-c$, then $2^{\min (r,s)}$ divides their sum $a-c$. ThusCoincident arguments also satisfy these inequalities immediately. Consequently $d$ is a metric, and indeed an ultrametric induced by the 2-adic absolute value.

For the sequence, $x_n-(-1)=2^n$, soThe same sequence diverges in the usual Euclidean metric; the convergence here depends on the 2-adic absolute value.

False. The real line with its usual metric is complete and is homeomorphic to $(0,1)$ with its usual metric, throughIts inverse is $h^{-1}(u)=\tan (\pi (u-1/2))$, so both maps are continuous. However, $(1/n{)}_{n\ge 2}$ is a Cauchy sequence in $(0,1)$ and has no limit there: its only possible real limit is zero. Hence $(0,1)$ is not a complete metric space. Completeness is a property of the metric, not just of its topology.

For rows $x_i,x_j$, three useful choices are Euclidean distance, standardized Euclidean distance, and Mahalanobis distance:The Euclidean choice is simple, rotation invariant and appropriate when the original coordinate scales carry comparable meaning. It is sensitive to scale changes, may be dominated by a high-variance variable and gives correlated or duplicate variables repeated influence. Squared Euclidean dissimilarity is sometimes used algorithmically, but removing the square root generally loses the triangle inequality.

The standardized Euclidean choice removes marginal units using fixed positive sample scales $s_r$. It gives equal marginal weight to variables, but ignores correlations and can magnify noise when an estimated variance is small. Constant variables require omission or an explicitly chosen positive external scale.

The Mahalanobis choice incorporates the full positive-definite sample covariance $S$, so it is Euclidean distance after whitening. It accounts for both scale and redundancy and is invariant under an invertible change of coordinates when $S$ is transformed consistently. However estimating $S$ requires enough observations and can be sensitive to extreme values; inversion is unstable near singularity and unavailable when $S$ is singular. With fixed positive scales or positive-definite $S$, all three square-root expressions are genuine metrics on the measurement space. Their different geometries encode different notions of which individuals should be considered similar.

For every binary vector $x$, $S_1(x,x)=1$, and henceThus $d_2$ is not a metric, because it fails identity of indiscernibles. It is nonnegative and symmetric. In fact it also satisfies the triangle inequality trivially: every value lies in $[1/2,1]$, so $d_2(x,z)\le 1\le d_2(x,y)+d_2(y,z)$. That does not repair its nonzero self-distance.

For fixed $p\ge 1$, the complement of the simple matching coefficient is the fraction of coordinates that disagree:This is the normalized Hamming distance. It is nonnegative, symmetric, and is zero exactly when every coordinate agrees. At each coordinate,Summing and dividing by $p$ proves the triangle inequality. Therefore $d_1$ is a metric on the binary vectors.