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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.

Continuous functions $X\to [0,1]$ on a normal Hausdorff space separate points and closed sets by Urysohn's lemma. The evaluation map into the product of their codomains is injective and continuous. For each point in an open set, a coordinate equal to one there and zero outside the open set detects a smaller coordinate neighborhood, proving the inverse topology.

Let $C$ be closed in a normal topological space and $g:C\to [-M,M]$ continuous. Separate its closed level sets $g\le -M/3$ and $g\ge M/3$ by Urysohn's lemma to obtain $h:X\to [-M/3,M/3]$ taking the corresponding endpoint values. Checking the outer and middle ranges gives the displayed bounds. Iteration on residuals gives a uniformly convergent series with total supremum norm at most $M$. The uniform limit theorem gives the bounded Tietze extension theorem.

A continuous function on a compact subset of a Tychonoff space extends to the whole space. Embed the space into a cube of closed unit intervals. The Tychonoff theorem makes the cube compact Hausdorff and thus normal. The compact subset is closed in that cube, so the Tietze extension theorem extends its bounded function to the cube. Restrict this extension to the embedded original space. Normality of the original space is unnecessary.

Suppose $Y$ is closed in a normal space $X$ and $|f|<1$ on $Y$ for a continuous real-valued $f$ on $X$. The closed set $Z=\{|f|\ge 1\}$ misses $Y$. Urysohn's lemma supplies $u:X\to [0,1]$ equal to zero on $Y$ and one on $Z$. Then $(1-u)f$ agrees with $f$ on $Y$ and has range in $(-1,1)$. This permits the inverse transform $s\mapsto s/(1-|s|)$ in the proof of unbounded Tietze extension.

Two distinct positive-radius circles have external tangency when their closed disks meet at exactly one point. The Euclidean distance between their centres equals the sum of their radii. The contact point divides the segment of centres internally in the ratio $r_1:r_2$.

Two distinct positive-radius circles have internal tangency when their boundaries meet at exactly one point and one disk contains the other. Their centre separation equals the absolute difference of their radii.

Let two positive-radius circles have external tangency at $Q$ and lie in the closed strip between two parallel lines, touching opposite boundaries at $P,R$. Then $P,Q,R$ are collinear. To prove this, choose coordinates with the boundaries $x=0,d$, put $P=(0,0)$ and $R=(d,h)$, and write the centres as $O_1=(r_1,0)$, $O_2=(d-r_2,h)$. The contact point isThe closed line segment $PR$ is contained in the union of the two closed disks: its portions from $P$ to $Q$ and from $Q$ to $R$ are chords. The strip and external tangency hypotheses are essential to this particular formulation.