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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.

Ixodes is a genus of hard ticks. Members take blood meals from vertebrate hosts across their life cycle. Identifying a tick as Ixodes does not establish which microorganisms it carries or whether transmission occurred.

Argas is a genus of soft ticks. Its members lack the conspicuous dorsal shield of hard ticks such as Ixodes. The genus identifies a taxonomic group, while host choice and disease-transmission competence require species-specific evidence.

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.

Four positive-radius circles with pairwise disjoint interiors, successive external tangencies, and four distinct contact points have those contact points on a circle. Apply the reciprocal Möbius transformation $w=1/(z-T)$ at one contact $T$. The two circles through $T$ become disjoint parallel lines; the other two disks lie between them and avoid zero. Collinearity for tangent circles between parallel lines places the three finite transformed contacts on a straight line. That straight line misses zero: otherwise its segment across the strip would pass through zero, while that segment is contained in the two transformed closed disks. Its inverse is therefore a genuine circle through $T$, containing all four original contacts. Allowing internal tangency without the pictured disjoint-interior hypotheses admits collinear degeneracies instead.

The pork tapeworm. Human intestinal adults and tissue cysticerci follow different routes of infection.

The beef tapeworm.

Taenia multiceps is a tapeworm whose coenurus larvae commonly develop in the central nervous system of an intermediate host. Sheep are important hosts, while adults develop in canids. The historical name Coenurus cerebralis refers to the larval stage.^[ref]

Taenia serialis is a tapeworm forming a coenurus in the tissues of an intermediate host. Canids serve as definitive hosts, while rabbits are frequent intermediate hosts. The adult and tissue larva are stages of the same species.^[ref]

An explicit contradiction in a node label, such as both $A$ and $\neg A$, or $\perp$. Opposite universal restrictions alone cause no clash when there are no successors.

A test that stops further witness generation at a node when an appropriate earlier ancestor already represents its required pattern of labels. The exact test depends on the logic and tableau calculus. An isolated root has no ancestor and is not directly blocked.

For a symplectic matrix $M$ of size $2m$ with determinant one, the defining identity $M^{T}JM=J$ gives $M^{-1}=J^{-1}{M}^{T}J$. Since similarity and transposition preserve the characteristic polynomial, ${\chi }_{M^{-1}}={\chi }_M$. Factoring $xI-M=M(x{M}^{-1}-I)$ then proves the displayed reciprocal identity for $x\ne 0$. If ${\chi }_M(z)=(z-\lambda {)}^{d}R(z)$ with $R(\lambda )\ne 0$, this identity gives ${\chi }_M(x)=x^{2m-d}(1-\lambda x{)}^{d}R(1/x)$ near $x={\lambda }^{-1}$. Thus reciprocal eigenvalues have the same algebraic multiplicity. The fixed points $\pm 1$ require the stronger parity argument in even multiplicities at fixed points of reciprocal root pairing.

Suppose distinct Sylow subgroups for $p$ in a finite group have trivial intersections. Conjugation by one of them, $P$, fixes $P$. The stabilizer of any other $Q$ is $H=P\cap N_G(Q)$. Since $H$ normalizes $Q$, the product $HQ$ is a subgroup of $p$-power order; maximality of $Q$ gives $H\le Q$. Thus $H\le P\cap Q=1$. All other orbits have size $|P|$, proving the congruence.