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Evaluation at zero is a bounded homomorphism on $C[0,1]$. Its kernel $I$ is a closed self-adjoint ideal and hence a C-star subalgebra. It has no identity: applying a proposed identity to $f(t)=t$ would force that identity to equal one at every positive point, contradicting continuity and its value zero at zero. The map $(f,\lambda )\mapsto f+\lambda 1$ from the unitization of a C-star algebra is a bijective star homomorphism, with inverse $h\mapsto (h-h(0),h(0))$. Pulling back the supremum norm verifies the C-star identity, so this is the prescribed unitization norm. Thus adjoining the missing constant direction recovers all continuous functions on the interval.

For a Unitary element of a C-star algebra $u$, the displayed element is also unitary. Its denominator has a multiplicative inverse by the Neumann series, since $\Vert \overline{\lambda }u\Vert <1$. The spectrum of $u$ lies on the unit circle, where $|t+\lambda |=|1+\overline{\lambda }t|$. The resulting continuous function $f(t)=(t+\lambda )/(1+\overline{\lambda }t)$ therefore has $\overline{f}f=f\overline{f}=1$ on the spectrum. The continuous functional calculus preserves products and the involution, giving $f(u{)}^{\ast }f(u)=f(u)f(u{)}^{\ast }=1$. The complex conjugate in the denominator is necessary for the unit-modulus calculation.

A strictly convex norm permits at most one best approximation from a linear subspace, but does not force the distance infimum to be attained. In the Hilbert space ${\ell }^2$, the finite-support vector subspace $c_{00}$ is dense and omits $u=(2^{-j}{)}_{j\ge 1}$. Its distance from $u$ is zero and no member attains that distance. Finite-dimensionality restores existence by compactness of bounded sets in the vector subspace. If $Y$ is a bounded projection onto a vector subspace, the estimate $\Vert u-Yu\Vert \le (1+\Vert Y\Vert ){\inf }_q\Vert u-q\Vert$ follows directly by writing $u-Yu=(I-Y)(u-q)$ for each candidate $q$, even when the infimum is not attained.

On ${\ell }_1$, the unilateral shift operator $R$ and left shift operator $L$ satisfy $LR=I$ and $\Vert L\Vert =1$, while $L$ has nontrivial kernel. If $\Vert A\Vert <1$, the Neumann series makes $L(R+A)=I+LA$ invertible. Were $R+A$ invertible, this identity would express $L$ as a product of invertible operators, a contradiction. Thus the open unit ball about $R$ contains no invertible operator.

On the absolutely summable sequence space, this weighted shift has $T^m{e}_j=e_{j+m}/[j(j+1)\cdots (j+m-1)]$, so $\Vert T^m\Vert =1/m!$. The spectral radius formula gives radius zero. Its spectrum is exactly $\{0\}$ because it is not onto, while it has no eigenvalues: zero is excluded by injectivity, and any nonzero eigenvalue equation forces all coordinates to vanish starting at the first. Thus it is a nonzero quasinilpotent element of the operator algebra.

For a Poisson sample tested against mean ${\mu }_0=1$, the exact uniformly most powerful one-sided test thresholds the sample sum and randomizes on its boundary if necessary. A normal approximation gives critical sum $n+z_{1-\alpha }\sqrt{n}$. Equating its standardized alternative threshold to $-z_{1-\beta }$ gives the displayed design for approximate power $1-\beta$ at mean ${\mu }_1>1$. At size $0.05$ and target power $0.9$, the two quantiles are $1.645$ and $1.282$. Rounding up supplies an integer design; exact Poisson power still requires discrete evaluation.

In a uniform random graph process, realize edge arrivals by independent continuous uniform times. For a fixed set of $r$ vertices to form an isolated clique at some time, all $e=\left(\genfrac{}{}{0ex}{}{r}{2}\right)$ internal edges must arrive before the first of the $r(n-r)$ outgoing edges. Their relative arrival ordering is uniform, so the exact probability isA union bound over the vertex sets therefore bounds the probability that such an isolated clique ever appears by $\left(\genfrac{}{}{0ex}{}{n}{r}\right)/\left(\genfrac{}{}{0ex}{}{r(n-r)+e}{e}\right)$. For fixed $r\ge 4$ this is $O(n^{r-e})\to 0$, since $e>r$. This argument controls all process times at once, including the short interval during which an isolated component could exist, without discretizing time.

Refill the stock to a fixed positive size $N$ immediately whenever a step empties it. The resulting Markov chain has state space $\{1,\dots ,N\}$ and transitions $P(i,j)=1/i$ for $j<i$ and $P(i,N)=1/i$. It is irreducible and aperiodic, since every state reaches $N$, $N$ reaches every state, and $P(N,N)>0$. Its unique stationary distribution is ${\pi }_N=1/H_N$ and ${\pi }_k=1/((k+1)H_N)$ for $k<N$. Normalization is the harmonic number identity. For $k<N$, stationarity follows from ${\sum }_{j=k+1}^{N-1}1/(j(j+1))+1/N=1/(k+1)$; the same telescoping sum starting at $1$ gives stationarity at $N$. The finite-state convergence theorem therefore makes this stationary distribution the limiting distribution from every initial state.

For operators on a complex Hilbert space, pointwise boundedness of $⟨Tx,x⟩$ over a family implies uniform boundedness of their operator norms. The complex polarization identity recovers each $⟨Tx,y⟩$ from four diagonal values. Apply the Uniform boundedness principle first to the functionals in $y$ for fixed $x$, then to the operators in $x$. On a real Hilbert space the claim fails: multiples of a nonzero skew-symmetric operator have zero diagonal forms but unbounded norms. For example, rotations through a quarter turn on ${\mathbb{R}}^2$ multiplied by integers suffice.

For each $m\ge 1$, enumerate the $m^2$ half-open intervals $[j/m^2,(j+1)/m^2)$ partitioning $[0,1)$ and use the functions $m$ times their indicator functions. Concatenate these finite blocks in increasing $m$. Every function in block $m$ has L1 norm $1/m$, which tends to zero. Every $x\in (0,1)$ is in one interval of every block, so the values at that point have a subsequence increasing to infinity. Thus small integral norm does not bound the pointwise upper limit, even when every function is nonnegative. In contrast, Fatou's lemma prevents a strictly positive lower limit everywhere on an interval when the norms tend to zero.

For each irrational real cut $r$, the displayed comparisons with named rationals are finitely satisfiable. Quantifier elimination for dense linear orders makes them determine a complete type, omitted by the named rational structure itself. Different cuts give different types, so there are continuum many. An omega-saturated model must realize all of them and cannot be countable, although the named rational structure is countable, atomic and prime.

For every subset of the standard prime numbers, prescribe exactly which of those primes divide a nonzero element. Every finite prescription is realized by the product of its finitely many positive divisibility requirements. Compactness theorem extends each prescription to a complete one-type of the complete theory of the integer ring. Different subsets give incompatible partial types. The countable first-order language supplies the matching upper bound.

In the consecutive-clearance inspection model, a run of $K$ consecutive good items is required to leave full inspection. With $q=1-p$, $0<p<1$, its Markov chain has transitions $P_{i0}=p$, $P_{i,i+1}=q$ for $i<K$, and $P_{K0}=p/r$, $P_{KK}=1-p/r$. Thus every defect in the clearance phase resets the count, rather than preserving the cumulative total.

Writing $D=1+(r-1)q^K$, the stationary distribution isTo prove this, balance at $i=1,\dots ,K-1$ gives ${\pi }_i=q^i{\pi }_0$, while balance at $K$ gives $(p/r){\pi }_K=q^K{\pi }_0$. Summing the geometric series and normalizing gives ${\pi }_0=p/D$; balance at zero follows as well. The inspected fraction is $1-(1-1/r){\pi }_K=1/D$. The fraction of all production that is both defective and uninspected is $p(r-1)q^K/D$; the fraction among defective items alone is $(r-1)q^K/D$. All these long-run limits follow from the Markov chain ergodic theorem and the independent inspection coins. The distinction from cumulative clearance matters already at $K=2$.

The class of torsion-free abelian groups has an infinite first-order axiomatization but no finite axiomatization. Add $\forall x(nx=0\to x=0)$ for every positive integer $n$ to the abelian group axioms. If an equivalent finite theory existed, compactness would make it follow from finitely many of these conditions. A cyclic group of prime order larger than the finitely many integers mentioned satisfies those conditions but has torsion, a contradiction.

For an abelian group, the scheme $\forall x(nx=0\to x=0)$ for every positive integer $n$ expresses absence of nonzero elements of finite order. Each fixed $n$ gives one sentence, while the entire scheme expresses the torsion-free abelian group property.

A closed subspace of a topological space $X$ is a closed subset of $X$ equipped with its subspace topology.

Every topological space, including one without the T0 space property, embeds into a power of this three-point space. For each open set $U$, use a coordinate equal to $2$ on $U$ and $0$ elsewhere. Its inverse image of the only proper nonempty open set is $U$. For each point $a$, add a coordinate equal to $1$ at $a$ and $0$ elsewhere. These coordinates are continuous because the subspace $\{0,1\}$ has the indiscrete topology, and they separate all points. The combined diagonal map is injective and induces precisely the original topology, hence is a topological embedding.

A family of maps with a common domain determines the displayed map into the Cartesian product of their codomains. It is injective exactly when the family separates points. For continuous coordinate maps it is continuous in the product topology. If inverse images of open sets under the coordinates generate the original topology, the inverse from the image is also continuous; thus the map is a topological embedding. A basis of such inverse images, as opposed to a subbasis, is a sufficient stronger hypothesis.

A subset $Y\subseteq X$ is C-embedded if every continuous real-valued function on its subspace topology extends continuously to $X$. Closed subsets of normal spaces have this property by the Tietze extension theorem. The condition also makes sense for nonclosed subsets.