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Take the circular cylinderIts first fundamental form is $R^{2}d{\theta }^2+d{z}^2$. The universal covering parametrization ${\mathbb{R}}^2\to S$ is therefore a local isometry from the Euclidean plane, after scaling the first coordinate. By geodesic preservation by a local isometry, every geodesic on the cylinder lifts to a straight line defined for all $t\in \mathbb{R}$. Thus the cylinder is geodesically complete. It is noncompact because $z$ is unbounded, and it is not a plane because one of its principal curvatures is $1/R$.

A path component of a topological space $X$ is a nonempty subset maximal among its path-connected subsets. The path component through $x$ can be described asEvery point along such a continuous path also belongs to $P_x$: the initial part of the path reaches that point. Thus the paths witnessing membership stay inside $P_x$. This makes path components useful for deciding which allowed positions can be reached from one another by continuous motion.

Every path component lies inside one connected component, because every path-connected space is connected. Thus each connected component is a union of path components. In a locally path-connected space the two classifications agree, by connected locally path-connected spaces are path-connected. For a subset of the real line with its usual subspace topology, the classifications also agree: every connected component is a real interval, and interpolation supplies a continuous path within it.

The topologist's sine curve shows why the distinction matters in general. Use specifically the closed versionwith the subspace topology from the usual Euclidean plane. It has one connected component, $T$, but two path components, $G$ and $L$. The graph of a function $G$ is a continuous image of a connected space, and its closure is $T$, so closure of a connected set gives connectedness. Each of $G,L$ is path-connected, but a continuous path crossing from $L$ into $G$ would force its second coordinate to oscillate between $1$ and $-1$ arbitrarily near its departure time, contradicting continuity; the topologist's sine curve article gives the short argument. Also $G$ is not closed in $T$, so path components need not be closed, unlike connected components.^[ref]

Being joined by a continuous path is an equivalence relation. Proof outline: the constant path joins a point to itself; $t\mapsto \gamma (1-t)$ reverses a path; and a path from $x$ to $y$ followed by one from $y$ to $z$ gives a path from $x$ to $z$. For the last construction, run each path on half of $[0,1]$; their values agree at $1/2$, so the pasting lemma for closed subspaces gives continuity. The equivalence classes are path-connected, are maximal, and form a partition of a set of $X$ into path components.^[ref]

A connected open subset of the Euclidean plane. For a bounded polygonal example, the Dirichlet Laplacian is the self-adjoint operator associated with the Dirichlet energy on $H_0^1(\Omega )$; its eigenfunctions are smooth in the interior but need not be smooth at reentrant boundary corners.

A plane graph is a planar graph with a specified graph drawing whose edges meet only at common endpoints, also called an embedding in the Euclidean plane. Its faces of a plane graph and facial boundary walks refer to that embedding. This geometric use of embedding differs from a graph embedding, which maps one abstract graph into another.

Two triangles are similar if a Euclidean similarity maps one onto the other. Their corresponding angles agree, and corresponding side lengths have a common positive ratio.

Specify the correspondence first: $A$ matches $A^{\prime }$, $B$ matches $B^{\prime }$ and $C$ matches $C^{\prime }$. For similar triangles, the common scale factor $k>0$ satisfiesThe ratios compare matching sides in the same order. A Euclidean similarity can also move, turn or reflect the triangle; its orientation on the page is irrelevant. In the Euclidean plane, matching just two interior angles is enough: their sums force the third to match, and equal corresponding angles force the side ratios to agree. This is the angle-angle (AA) criterion for similar triangles. Matching just one angle is insufficient.

Scale the right triangle with vertices $(0,0)$, $(4,0)$ and $(0,3)$ by $k=2$. Its new vertices are $(0,0)$, $(8,0)$ and $(0,6)$, and its side lengths are $6,8,10$. The matching side ratios are $6/3=8/4=10/5=2$. Both triangles have a right angle and the same two other interior angles, approximately $36.87^{\circ }$ and $53.13^{\circ }$. However, their areas are $3\cdot 4/2=6$ and $6\cdot 8/2=24$: doubling lengths multiplies area by four.

Generally, a Euclidean similarity multiplies both a base and its perpendicular height by $k$, so the area becomes $\frac{1}{2}(kb)(kh)=k^2\frac{1}{2}bh$. This explains why an enlarged drawing needs four times as much surface when each length doubles. It also enables indirect measurement: under assumptions of level ground, vertical objects and parallel light rays, an object of height $3$ casting a shadow of length $4$ provides a right triangle. Another object's shadow of length $8$ gives a similar triangle and height $3(8/4)=6$. The shared angles justify the inference; visual resemblance alone does not.

Euclid's Elements VI, Definition 1 describes similarity using corresponding angles and proportional sides. VI.4 establishes the implication from matching angles to matching side ratios, while VI.19 states the squared relation between side ratios and areas. These are documented ancient formulations of the principles used above, without establishing a first discoverer.^{[ref][ref][ref]}

The solution set of a mathematical condition $P$ in a specified domain $D$ collects every allowed value that satisfies it:It describes the answer without committing to the steps used to find it. For the equation $x^2=1$ over the real numbers, the solution set is $\{-1,1\}$; restricting to positive real numbers leaves $\{1\}$. For $2x=1$, the solution set is $\{1/2\}$ over the rational numbers and the empty set over the integers. The domain is part of the question.

An equation can have one solution, several solutions, no solutions, or an entire domain of solutions. Over the real numbers, $2x+3=9$ gives the singleton set $\{3\}$, $x^2+1=0$ gives the empty set, and the mathematical identity $(x+1{)}^2=x^2+2x+1$ gives the whole real line. Changing that second equation's domain to the complex numbers gives $\{-i,i\}$. The elementary definition by values that satisfy an equation and the classification by solution sets appear in College Algebra 2e, §2.2: Solving Linear Equations in One Variable.

Requiring several conditions simultaneously takes the intersection of their solution sets. The squared equation $x=(x-2{)}^2$ has solution set $\{1,4\}$. Adding the restriction $x\ge 2$ giveswhich is also the solution set of $\sqrt{x}=x-2$. The extraneous solution example explains why the sign restriction is needed. Requiring at least one of several conditions instead takes their union. This turns logical combinations of constraints into reusable operations on sets.

For several variables, a solution is an ordered assignment, usually represented by a tuple. The real equation $x+y=1$ has solution set $\{(t,1-t):t\in \mathbb{R}\}$, a straight line in the Euclidean plane with coordinates $(x,y)$. Adding $x-y=0$ intersects that straight line with another, leaving the single tuple $(1/2,1/2)$. A system of linear inequalities has no solution exactly when the intersection of its individual solution sets is the empty set. For optimization constraints, this common solution set is the feasible region.

Because a solution set is a set, the order in which its members are listed is irrelevant. The multiple root of $(x-1{)}^2=0$ is still the one solution $1$; its multiplicity of a root is additional polynomial information, rather than a second occurrence inside the solution set. Equivalent equations preserve the solution set on the stated domain, even if their algebraic forms differ.

Two equally spaced families of perpendicular lines partitioning the Euclidean plane into congruent squares. A general affine map can turn this grid into parallelograms, or collapse directions if its linear part of an affine map is not injective.

For $n\ge 1$, a topological $n$-manifold with boundary is a Hausdorff second-countable space locally homeomorphic to relatively open sets in the half-space ${\mathbb{H}}^n$. The openness here is in the subspace topology of ${\mathbb{H}}^n$, not necessarily in ${\mathbb{R}}^n$. Its intrinsic boundary $\partial M$ consists of points sent to $x_n=0$; the remaining points have ordinary coordinate charts with images open in Euclidean space. The intrinsic boundary is independent of the chosen chart.^[ref]

The closed interval $[0,1]$ illustrates the distinction: its interior points have open interval charts, while $0$ and $1$ have charts modeled on $[0,\epsilon )\subseteq {\mathbb{H}}^1$, using $x$ and $1-x$. Its intrinsic boundary is $\{0,1\}$. The circle has empty intrinsic boundary even though its boundary of a set as a subset of the Euclidean plane is the whole circle. A topological manifold without boundary is therefore a special case with $\partial M=\varnothing$, and a smooth atlas on the half-space charts gives a manifold with boundary in the smooth sense.

A topological tripod is three copies of the closed interval $[0,1]$ with their zero endpoints identified to a single junction $o$. Equivalently, embed three line segments with distinct directions in the Euclidean plane and use the subspace topology. It is a compact, connected metric space, and is Hausdorff, second countable, and paracompact. Every interior arm point has a coordinate chart to an open interval, and each outer endpoint has a half-interval chart as in a topological manifold with boundary. The junction has neither kind of chart.

Here is a precise local obstruction. Let $S$ be a small open neighbourhood of $o$ consisting of initial segments of the three arms, excluding their outer endpoints. Every connected open neighbourhood $N$ of $o$ contained in $S$ contains an initial segment on every arm, because $N$ is open at $o$. Its intersection with each arm has no gaps: if a point between $o$ and a point of $N$ were missing, cutting there would separate the part beyond it from the rest of $N$, contradicting connectedness. Thus $N\setminus \{o\}$ has exactly three connected components, one interval per arm.

Suppose a coordinate chart $h$ to the real line existed at $o$. Choose a small open interval $I$ about $h(o)$ lying inside $h(S\cap \mathrm{domain}(h))$, and restrict the chart to $N=h^{-1}(I)$. This is a connected open neighbourhood of $o$ contained in $S$. Removing $o$ leaves three connected components in $N$, while removing $h(o)$ leaves two in $I$. A homeomorphism preserves connected components, so this is impossible. This restriction step is essential: the count is proved for any sufficiently small connected chart neighborhood, not merely one chosen drawing.

A higher dimension of a manifold cannot repair the junction. For $n\ge 2$, repeat the argument with an open ball about $h(o)$: a punctured open ball in ${\mathbb{R}}^n$ is path-connected, whereas $N\setminus \{o\}$ still has three connected components. A zero-dimensional chart would make $o$ an isolated point, which it is not. A chart at a boundary point of a topological manifold with boundary similarly gives a punctured relative half-ball: one connected component when $n=1$, and a path-connected space when $n\ge 2$. Consequently the topological tripod fails the local definition even when intrinsic boundary is allowed.

The circle provides a useful warning about scale: deleting a point from the whole circle leaves a connected space, but deleting it from a sufficiently small connected arc leaves two connected components. Local chart tests use the latter. The circle and interval are not homeomorphic article uses deletion from the whole spaces for a different, global obstruction.

Consider the graph of a function $x\mapsto \sin (1/x)$, using the sine of the multiplicative inverse of $x$, together with a vertical closed interval:Give $T$ the subspace topology from the Euclidean plane. The symbol $\times$ denotes a Cartesian product. The right endpoint $(1,\sin 1)\approx (1,0.841471)$ belongs to $G$. The continuous map $x\mapsto (x,\sin (1/x))$ sends the connected real interval $(0,1]$ onto $G$. Hence $G$ is connected, and its closure $\overline{G}=T$ is connected by closure of a connected set. Every height in $[-1,1]$ is approached as $x$ tends to zero; for example,For $n=1$, these inputs are $p_1=2/(5\pi )\approx 0.127324$ and $q_1=2/(7\pi )\approx 0.090946$. For any other height $y\in [-1,1]$, choose $\theta \ge 1$ with $\sin \theta =y$; then the sequence $(1/(\theta +2\pi n),y)$ lies in $G$ and has limit of a sequence $(0,y)$.

Nevertheless, $T$ has exactly two path components, $G$ and $L$. Within $G$, join inputs $a,b\in (0,1]$ by the continuous path $t\mapsto ((1-t)a+tb,\sin (1/((1-t)a+tb)))$. Within $L$, use $t\mapsto (0,(1-t)c+td)$ for heights $c,d\in [-1,1]$.

The obstruction to a continuous path between $L$ and $G$ is persistent oscillation, not a visible gap. Proof outline: suppose such a path exists, directed from $L$ to $G$, and take its last time in $L$. Immediately afterwards its first coordinate $x(t)>0$ approaches zero and its second coordinate is $\sin (1/x(t))$. The intermediate value theorem forces the first coordinate to pass through inputs of both displayed types arbitrarily close to that departure time. The second coordinate therefore takes both values $1$ and $-1$ arbitrarily close to that time, contradicting continuity. ^[ref]

The sampled graph of a function illustrates the formulas; it does not prove connectedness or the failure of path connectedness. Taking a closure preserves connectedness, but this example shows that it can destroy path connectedness.

A triangle is the convex hull of three noncollinear points: the three vertices must not lie on one line. Its boundary consists of the three joining line segments; depending on context, "triangle" can mean this boundary or the filled region it encloses. Here a triangle's area refers to the filled region. The noncollinearity assumption excludes a collapsed shape with zero area, for which ordinary interior angles are not defined.

For example, the positions $(0,0)$, $(4,0)$ and $(0,3)$ form a right triangle with a right angle at the origin. Taking the side of length $4$ as its base gives perpendicular height $3$, so its area is $4\cdot 3/2=6$. The height is measured at a right angle to the chosen base; it is not generally the length of another side. In this example it happens to be a side because the triangle is a right triangle.

By contrast, side lengths $3,4,6$ define a valid triangle: each side is shorter than the sum of the other two, as required by the strict triangle inequality. But $3^2+4^2=25\ne 36=6^2$, so this is not a right triangle and the Pythagorean theorem cannot calculate its longest side from the other two. The law of cosines instead gives the opposite interior angle $C$ throughThis example separates the definition of a triangle from the additional assumption of a right angle.

Every triangle in the Euclidean plane has interior angles summing to $\pi$ radians, a result proved in Euclid's Elements I.32.^[ref] It lets two measured angles determine the third, and underlies the angle-angle criterion for similar triangles. The qualification "Euclidean plane" matters: the earlier spherical triangle demonstrates a different sum.

Take the circle $S^1=\{(x,y)\in {\mathbb{R}}^2:x^2+y^2=1\}$ with its subspace topology. Write $W=(-1,0)$ for its west point and $E=(1,0)$ for its east point. Two coordinate charts obtained by stereographic projection areBoth take values in the entire real line. Geometrically, join $W$ to $(x,y)$ and intersect the resulting line with $x=0$: the height is $t$. Projecting from $E$ instead gives height $u$. Each projection omits its own starting point, but the other coordinate chart includes that point. Hence $U\cup V=S^1$.^[ref]

To verify the homeomorphisms, solve the displayed equations for $(x,y)$. The inverse functions areEach formula lies on the circle, avoids its excluded point, and is continuous for every real argument. Substitution into the projection formulas returns $t$ or $u$; substitution in the other direction returns $(x,y)$. Thus these are continuous bijections with continuous inverse functions, as required for coordinate charts. Since the Euclidean plane is Hausdorff and second countable, the subset $S^1$ with its subspace topology is too. This checks all three conditions for a topological manifold, rather than just drawing a smooth-looking curve.

On the overlap $U\cap V$, both $x=\pm 1$ are excluded, so $y\ne 0$. Using $y^2=1-x^2$, the topological transition map isFor example, $(x,y)=(3/5,4/5)$ has $t=1/2$ and $u=2$. The two connected components of the overlap, the upper and lower arcs, correspond to the positive and negative parts of the punctured real line. The values $t=0$ and $u=0$ represent $E$ and $W$, respectively, and belong to only one chart each. The reciprocal coordinate change and its inverse function are smooth functions away from zero, so this topological atlas also supplies a smooth atlas.

Why can neither chart be enlarged to a single coordinate chart on the whole circle? The circle is a closed set and a bounded set in the Euclidean plane, hence compact by the Heine-Borel theorem. A global coordinate chart would give a homeomorphism to a nonempty open set $O\subseteq \mathbb{R}$. Its image would be compact and therefore closed in the real line. By connectedness of the real line, this forces $O=\mathbb{R}$, which is not compact. This is the one-dimensional instance of no compact manifold has a single Euclidean chart. Locally the circle is indistinguishable from the real line; globally compactness obstructs a single chart.

The unit square here is the Cartesian product $[0,1]\times [0,1]$ in the Euclidean plane, a square of side length one.