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Let $f$ be differentiable on an open interval containing $c$. If the finite limit of a function $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, then $f^{\prime }(c)=L$.

Choose $h_n>0$ tending to zero with $[c,c+h_n]$ in the interval. The mean value theorem gives $x_n\in (c,c+h_n)$ such thatThe right-hand side tends to $f^{\prime }(c)$ by differentiability. Since $x_n\to c$ with $x_n\ne c$, the left-hand side tends to $L$. Uniqueness of the limit of a sequence gives equality. The same reasoning excludes an infinite limit of a function for $f^{\prime }$ at $c$.

Thus derivatives may be discontinuous, but a removable discontinuity cannot be repaired by changing the value of a derivative at a single point of differentiability.

For $a<b$ and real-valued functions $f,g$ that are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, there exists $\xi \in (a,b)$ withApply Rolle's theorem to the linear combination $h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$, which has equal endpoint values. The displayed product identity permits zero derivatives. For the ratio form, additionally assume $g^{\prime }(x)\ne 0$ for every $x\in (a,b)$. Then Rolle's theorem ensures $g(b)\ne g(a)$, so both denominators below are nonzero:Neither derivative needs to be continuous, and $f^{\prime }$ may vanish. Choosing $g(x)=x$ recovers the mean value theorem; the ratio form also gives the zero-over-zero endpoint case of L'Hôpital's rule. ^[ref]

The circle $S^1$ and the closed interval $[0,1]$ are both compact and connected, so those two properties alone do not distinguish them. Instead examine connectedness after removing a point. Every punctured circle is connected: after rotating the missing point to $(1,0)$, the continuous map $t\mapsto (\cos t,\sin t)$ from the connected open interval $(0,2\pi )$ maps onto it. Continuous maps preserve connectedness. ^[ref]

If a homeomorphism $h:S^1\to [0,1]$ existed, choose $p=h^{-1}(1/2)$. Its restriction would give a homeomorphism between $S^1\setminus \{p\}$ and $[0,1]\setminus \{1/2\}$. But the latter splits into two nonempty disjoint open sets in its subspace topology, $[0,1/2)$ and $(1/2,1]$, contradicting connectedness of the punctured circle. Removing an endpoint of $[0,1]$ leaves a connected space; the proof deliberately chooses the preimage of an interior point. This illustrates how a simple connectedness test can rule out every possible homeomorphism.

A bounded closed interval includes both of its endpoints:If $a=b$, it consists of one point. Its length is $b-a$. In contrast, the open interval $(a,b)$ excludes both endpoints; half-open real intervals include just one.

Including the endpoints has a useful consequence for limits: a convergent sequence whose terms lie in $[a,b]$ has its limit of a sequence in $[a,b]$ too. Finite endpoints also make this a bounded set, so it is a compact set by the Heine-Borel theorem. These two properties support both the nested interval theorem and the extreme value theorem.

A closed subgroup is a subgroup which is a closed set in the ambient topology. It is a topological group in its subspace topology, but need not be an open subgroup.

The additive subgroup $\mathbb{Z}\subset \mathbb{R}$ of integers is a closed subgroup: every noninteger lies in an open interval $(n,n+1)$ avoiding $\mathbb{Z}$. It is also a discrete subgroup, since $(n-1/2,n+1/2)\cap \mathbb{Z}=\{n\}$. Nevertheless it is not an open subgroup of $(\mathbb{R},+)$: every open interval containing an integer contains nonintegers too. Thus having a discrete topology internally says nothing about being an open set in the ambient topology.

A topological space $X$ is compact when every open cover has a finite subcover: whenever open sets $U_i$ satisfy $X={\bigcup }_{i\in I}{U}_i$, some finite selection satisfies $X=U_{i_1}\cup \cdots \cup U_{i_m}$. A subset $K$ of an ambient topological space is compact when its subspace topology is compact; equivalently, every open cover of $K$ by ambient open sets has a finite subcover. The word every matters: finding one finite subcover of one open cover is not enough to establish compactness. ^[ref]

The closed interval $[0,1]$ is compact by the Heine-Borel theorem. For a concrete open cover, take the three open intervals $(-0.1,0.45)$, $(0.35,0.8)$, and $(0.7,1.1)$. Their union is $(-0.1,1.1)$, so they cover every point of $[0,1]$, including both endpoints. The overlaps have lengths $0.1$ and $0.1$, leaving no gaps. Each member is necessary for this particular open cover: removing them respectively leaves $0$, $0.5$, or $1$ uncovered.

Above, hollow markers denote the excluded endpoints of the covering open intervals, while the solid endpoints belong to the covered closed interval. Below, each displayed open interval is one member of an infinite open cover of $(0,1)$; finitely many rows always leave points near zero uncovered. The rows are vertically separated for readability and all refer to the same real line.

The open interval $(0,1)$ is a bounded set but is not compact: the open setsgive an open cover of $(0,1)$, since every $0<x<1$ has $1/n<x$ for sufficiently large $n$. These are also open sets for the subspace topology on $(0,1)$. Any nonempty finite selection has a largest index $N$, so its union is just $(1/N,1)$ and misses the point $1/(N+1)$, which still belongs to $(0,1)$. Thus no finite subcover exists, however many members we select. The same family is not an open cover of $[0,1]$, since it misses both endpoints.

In a metric space, compactness is equivalent to sequential compactness: every sequence in the subset has a convergent subsequence whose limit of a sequence belongs to that subset. This equivalence does not extend to all topological spaces. For example, $x_n=1/n$ for $n\ge 2$ has limit of a sequence zero in $[0,1]$. The same sequence lies in $(0,1)$, but every subsequence still has limit of a sequence zero on the real line, so no subsequence converges to a point of $(0,1)$. Merely approaching a boundary is harmless when the required point remains available. ^[ref]

Both panels show the sequence $x_n=1/n$ for $n\ge 2$ approaching zero. In the closed interval, zero is an allowed limit of a sequence; in the open interval, the hollow marker identifies the missing point. This single sequence disproves sequential compactness of $(0,1)$; compactness of $[0,1]$ still requires control of every open cover or every sequence, supplied by the Heine-Borel theorem.

An infinite set with the discrete topology similarly fails compactness, using the open cover by all its singletons. Every finite set is compact in any topology: select one covering open set for each point. The empty set is compact because the empty selection is a finite subcover. This open cover definition does not require the topological space to be Hausdorff; in algebraic geometry it is often called quasi-compactness.

The beauty of compactness is that information valid around each point can often be controlled using finitely many regions. A finite subcover lets us combine finitely many local bounds, or choose the smallest of finitely many positive tolerances, to obtain control across the whole compact space. This is why compactness repeatedly turns approximation into existence and why it is so useful in optimization through the extreme value theorem.

Every closed set in a compact space is compact, and a continuous image of a compact space is compact. The latter proof pulls an open cover back by inverse images, selects a finite subcover, and maps forward. This makes compactness useful for proving existence: the extreme value theorem guarantees that a continuous function from a nonempty compact space to the real line attains both a minimum and a maximum. For example, $q(x)=x(1-x)=1/4-(x-1/2{)}^2$ on $[0,1]$ has maximum $1/4$ at $1/2$ and minimum zero at its endpoints. On $(0,1)$ its infimum is still zero but is never attained. ^[ref]

A topological space $X$ is connected when it has no separation of a topological space: no partition of a set into two nonempty open subsets $U,V$ satisfyingHere the union covers all of $X$, and the intersection is the empty set, so no point belongs to both pieces. For a subset $X$ of a larger topological space, "open" means open in the subspace topology: each piece is the intersection of $X$ with an ambient open set. Both pieces must be open; merely dividing a drawing into two colours does not establish a separation of a topological space. ^[ref]

For example, the closed interval $I=[0,1]$ with its usual subspace topology from the real line is connected. The proposed split $[0,1/2)\cup [1/2,1]$ fails: the second subset is not open in $I$, since every neighbourhood of $1/2$ in $I$ contains points less than $1/2$. More generally, a connected subset of the real line is precisely a real interval. In contrast, for the unionthe open intervals $(-3,0)$ and $(0,3)$ exhibit $U,V$ as nonempty open subsets in the subspace topology of $D$. Their intersection is empty and their union is $D$, so $D$ is not connected. Although $U,V$ are closed intervals in the real line, they are also open within $D$.

Why is connectedness useful? A continuous image of a connected space is connected: a proposed separation of a topological space in the image would pull back through the continuous map to a forbidden separation of a topological space in its source. Thus a continuous map from a real interval to the real line cannot skip values between two attained values. This is the intermediate value theorem. For a computed example, the continuous function $f(x)=x^2$ on $[0,1]$ has image $[0,1]$; the value $1/2$ occurs at $x=1/\sqrt{2}\approx 0.707107$. Connectedness guarantees an input attaining that value even when we have no formula for it.

Path connectedness is a stronger condition: it supplies a continuous path between every pair of points. The topologist's sine curve below shows that a connected space need not be a path-connected space.

A subset $A$ of a topological space $X$ is a connected subset when $A$, with the subspace topology inherited from $X$, is a connected space. Thus connectedness is tested inside $A$: a forbidden separation of a topological space consists of nonempty subsetswhere $O_1,O_2$ are open sets in $X$. The ambient open sets need not be disjoint outside $A$.

For example, in the real line with its usual topology, $[0,2]$ is a connected subset, whereas $\{0,2\}$ is not: its two singleton sets are open in its subspace topology. A subset of a connected space need not be connected. Conversely, $[-2,-1]$ is a connected subset of the disconnected topological space $[-2,-1]\cup [1,2]$. The ambient topological space need not itself be connected.

A finite set of distinct points in the real line inherits the discrete topology: a sufficiently small open interval around each point isolates that point. Its only nonempty connected subsets are singleton sets. This depends on the topology: a nonempty finite set with the indiscrete topology is a connected space, since its only open sets are itself and the empty set. Here, as in the definition of connected space, the empty set is also connected because it admits no separation of a topological space.^[ref]

The distinction matters when finding connected components: we search among connected subsets of the specified topological space, rather than judging whether the surrounding drawing looks like one piece.

In the usual subspace topology from the real line, a subset $A\subseteq \mathbb{R}$ is connected exactly when it has the defining property of a real interval:Every real number between two members must be present. Thus open intervals, closed intervals, intervals with only one endpoint included, and unbounded real intervals are connected. Singleton sets are connected too. With the no-separation of a topological space convention used here, the empty set satisfies the same property; some texts reserve the word connected for nonempty topological spaces.^[ref]

Removing an interior point changes the answer:These two real intervals are exactly its connected components and its path components. Removing an endpoint instead leaves a real interval, for example $[-2,2]\setminus \{2\}=[-2,2)$, which stays connected. This is why an interior point of a closed interval is a cut point, whereas an endpoint is not. A nonempty finite set of points in the real line is connected exactly when it is a singleton set.

Proof outline: if $a,b\in A$ but $c\in (a,b)$ is missing, the subsetsgive a separation of a topological space, since the two open intervals make both pieces open in $A$. Conversely, a nonempty real interval contains the continuous path $t\mapsto (1-t)a+tb$ between any two of its points, so it is path-connected and hence connected. Together with continuous image of a connected space, this characterization gives the intermediate value theorem: a continuous function on a real interval cannot jump over a value lying between two attained values.

Let $X,Y$ be topological spaces, let $X$ be compact, and let $f:X\to Y$ be a continuous map. Then its image $f(X)$ is compact with the subspace topology inherited from $Y$. More generally, a continuous map sends every compact subset $K\subseteq X$ to a compact subset $f(K)\subseteq Y$, by applying the same statement to its restriction to $K$. Neither $X$ nor $Y$ needs to be Hausdorff.^[ref]

For example, the polynomial function $f(x)=x^2$ sends the compact closed interval $[0,1]$ onto $[0,1]$. The open intervals $V_1=(-1/10,16/25)$ and $V_2=(9/25,11/10)$ give an open cover of the image. Their inverse images in the domain areBoth are open sets in the subspace topology of $[0,1]$, even though neither is an open set in the ambient real line. They cover the domain, so the corresponding two target open sets cover the image.

Matching colors identify the target open sets and their inverse images. Filled endpoints are included and hollow endpoints excluded. This particular open cover illustrates the mechanism; compactness requires a finite subcover for every open cover.

For the general proof, write an arbitrary open cover of $f(X)$ as $\{f(X)\cap V_i:i\in I\}$ with each $V_i$ an open set in $Y$. By continuity, the inverse images $f^{-1}(V_i)$ form an open cover of $X$. Its compactness provides indices $i_1,\dots ,i_m$ such thatThis gives the required finite subcover of $f(X)$. The argument uses inverse images because continuous maps need not send open sets to open sets. For the empty set, the empty selection suffices.

The image of a compact Hausdorff space under a continuous map need not itself be Hausdorff. Take $D=\{0,1\}$ with the discrete topology and $S=\{0,1\}$ with the Sierpiński space topology $\{\varnothing ,\{1\},S\}$. The identity map $D\to S$ is a continuous bijection, since the inverse image of every target open set is an open set in $D$. Its image is compact, but the two points of $S$ have no disjoint open neighborhoods. If the target $Y$ is Hausdorff, its subspace topology does make $f(X)$ Hausdorff, and the compact subset of a Hausdorff space theorem also makes $f(X)$ closed in $Y$.

The practical payoff is existence: the extreme value theorem uses the compact image of a nonempty compact space under a real-valued continuous function to obtain an attained minimum and maximum. When $f$ is also a bijection onto a Hausdorff target, the compact-to-Hausdorff continuous bijection theorem supplies continuity of its inverse without a separate formula.

A map $f:X\to Y$ between topological spaces is continuous when every open set $V\subseteq Y$ has an inverse image $f^{-1}(V)=\{x\in X:f(x)\in V\}$ that is an open set in $X$. The notation denotes a subset, so $f$ need not have an inverse function.

For the continuous map $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, with the usual metric topology,The two open intervals form an open set; an inverse image can split into separate pieces without violating continuity. The defining condition concerns inverse images, rather than images: here $f((-1,1))=[0,1)$ is not an open set in the real line. Checking this one inverse image illustrates the definition; it alone does not prove continuity for all target open sets.

If $X$ is an infinite set with its cofinite topology and $Y$ is a Hausdorff space, every continuous map $f:X\to Y$ is a constant function. Otherwise two distinct image points would have disjoint open neighbourhoods. Their inverse images would be nonempty disjoint open sets of $X$, contradicting its hyperconnected space property.

For example, $f(n)=n$ from ${\mathbb{N}}_{>0}$ with its cofinite topology into the real line with its usual Euclidean metric is not continuous: the inverse image of the open interval $(1/2,3/2)$ is $\{1\}$, which is not open in the infinite domain. A continuous real-valued readout on that domain can only report one value. Finite cofinite domains are different: their discrete topology allows arbitrary functions into any topological space.

An orthonormal basis of the polynomial space ${\Pi }_n$ need not be indexed by polynomial degree. Permuting the normalized Legendre polynomials, for example, preserves orthonormality but changes the degrees attached to the labels. Results about the number of polynomial roots of $p_k$ require the extra condition $\mathrm{deg}{p}_k=k$, or that the polynomials arise by applying the Gram-Schmidt process to $1,x,x^2,\dots$ in that order. Under this degree ordering, $p_k$ is orthogonal to every polynomial of lower polynomial degree because those earlier basis members span ${\Pi }_{k-1}$.

For a positive integration weight on a nondegenerate interval, this implies that $p_k$ has $k$ distinct polynomial roots in the open interval. If fewer than $k$ roots change sign there, multiply $p_k$ by the product of their linear factors. The product has constant nonzero sign except at its finitely many zeros, so its weighted integral cannot vanish, contradicting orthogonality to this lower-degree polynomial. The usual Gaussian quadrature construction therefore uses the degree-ordered family; arbitrary relabelling is not enough.

A real-valued function that is continuous on a nonempty compact space attains both a global minimum and a global maximum. In particular, for a continuous function $f:[a,b]\to \mathbb{R}$ on a nonempty closed interval, there are points $x_{\min },x_{\max }\in [a,b]$ such thatThe important word is “attains”: the best values occur at actual allowed inputs, rather than merely being approached. The interval version is ^[ref].

For a worked example, take $f(x)=(x-1{)}^2$ on $[-1,2]$. Its global minimum is $f(1)=0$, and its global maximum is $f(-1)=4$. Nonnegativity gives the first answer; $|x-1|\le 2$ on the specified closed interval gives the second. The endpoint $x=-1$ is essential: looking only for a zero derivative would find the minimum and miss the maximum.

For comparison, the continuous function $g(x)=x$ on the open interval $(0,1)$ has infimum $0$ and supremum $1$, but neither is a value of $g$. Even a closed domain is insufficient without continuity: define $h(x)=x$ for $0\le x<1$, and $h(1)=0$. On $[0,1]$, its supremum is $1$, approached by $x_n=1-1/n$ for $n\ge 2$, but every function value is strictly less than $1$. This endpoint change makes $h$ discontinuous, preventing attainment. On the unbounded real line, the continuous function $x\mapsto x$ has neither a global minimum nor a global maximum. These examples show what can fail when a hypothesis is dropped; functions outside the hypotheses may still have extrema.

The extreme value theorem is an existence tool for optimization: first establish that a continuous objective function on a nonempty compact feasible set has a best value, then investigate where it occurs. It neither locates the optimizing point nor promises uniqueness or differentiability. For example, the absolute value function $|x|$ on $[-1,2]$ has its global minimum at zero although no derivative exists there; a constant function has every input as a minimizer. This role in optimization is explained in University of Toronto MAT237 notes, section 1.4: compactness and the extreme value theorem.

A short proof outline identifies the mechanism. The continuous image of a compact space is compact; a nonempty compact subset of the real line contains its infimum and supremum. For a closed interval, one can instead choose inputs whose outputs approach the supremum, extract a convergent subsequence by the Bolzano-Weierstrass theorem, and use continuity to make its limiting output equal that supremum. The closed interval retains the limiting input. Repeating the argument for the infimum supplies the minimum.

For every finite $d\ge 1$, a subset $K\subseteq {\mathbb{R}}^d$ of Euclidean space is compact if and only if it is a closed set and a bounded set. Here closed means closed in the ambient Euclidean space, not merely in $K$ itself, and the bounded set condition uses the Euclidean norm. The empty set is included: it is a closed set, a bounded set, and compact. ^[ref]

This makes an apparently infinite test practical. To verify compactness of a closed interval $[a,b]$ with finite endpoints, we need not examine every open cover: it is closed in the real line and a bounded set. By contrast, the open interval $(0,1)$ is a bounded set but is not closed in the real line, as the missing-endpoint figure in compact space illustrates. The whole real line is closed but is not a bounded set, so it too fails compactness. These examples show why both hypotheses are needed.

The same criterion holds in any real or complex finite-dimensional vector space equipped with a norm: all norms there are equivalent norms, so they give the same open sets, closed sets, and bounded sets. Since compactness depends only on the topology, choosing a different norm does not change the conclusion. Without an appropriate topology, a bare vector space alone has no notion of compactness.

The restriction to a finite-dimensional vector space is essential. In the l2 sequence space, an infinite-dimensional vector space that is also a Hilbert space, let $e_n$ be the vectors of its standard orthonormal basis. All lie in the closed unit ball $B=\{x:\Vert x\Vert \le 1\}$, which is a closed set and a bounded set in the norm topology. For distinct indices, their inner product is zero, givingThus distinct terms remain distance $\sqrt{2}$ apart. No subsequence is a Cauchy sequence, hence none is a convergent sequence. The equivalence of compactness and sequential compactness in a metric space proves that this closed unit ball is not compact. The conclusion concerns the norm topology.

A short proof outline in Euclidean space explains the role of a finite-dimensional vector space. For a sequence in a bounded set, apply the Bolzano-Weierstrass theorem to each of the Cartesian coordinates in turn, retaining a subsequence at each of the finitely many steps. The final subsequence is a convergent sequence in all Cartesian coordinates, and a closed set contains its limit of a sequence. This yields sequential compactness, hence compactness. Conversely, a compact subset of a metric space is a closed set and a bounded set. ^[ref]

A homeomorphism is a bijection between topological spaces such that both $f$ and $f^{-1}$ are continuous maps. Equivalently, for every subset $U\subseteq X$, $U$ is an open set if and only if $f(U)$ is an open set in $Y$. A homeomorphism therefore transports the entire topology, including connectedness and compactness. It need not preserve numerical metric values, and a bounded set need not map to a bounded set.

For example, an explicit homeomorphism between an open interval and the real line isThe tangent function on $(-\pi /2,\pi /2)$ is strictly increasing and takes every real value; the displayed arctangent inverse verifies the bijection. Both formulas define continuous maps on their stated domains. Computed values are $f(1/4)=-1$, $f(1/2)=0$, and $f(3/4)=1$. Points approach infinity as $x$ approaches an excluded endpoint. Thus a bounded set can have the same topology as an unbounded one; neither of these topological spaces is compact.

A useful shortcut is the compact-to-Hausdorff continuous bijection theorem: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Its images of closed sets are compact sets, hence closed sets in the target, which proves continuity of the inverse. The indiscrete topology example shows why the Hausdorff assumption matters. ^[ref]

An interval map $F:I\to I$ has a horseshoe when there are an open interval $J\subseteq I$ and disjoint open intervals $K_0,K_1\subseteq J$ such thatIterating inverse branches produces full two-symbol itinerary dynamics.

An implicit solution describes a solution of a differential equation through a relation $F(x,y)=C$, instead of isolating $y$ in an explicit formula. The relation describes a level set; an actual scalar solution of a differential equation must be a differentiable function $y(x)$ on an interval whose function graph lies in the equation's domain. A whole level set need not be one such function graph.

For example, considerImplicit differentiation of $x^2+y^2=C$ gives $2x+2y{y}^{\prime }=0$. For $C>0$, the circle contains two solutions of a differential equation,Both satisfy the equation on the stated open interval, and their function graphs stay in $D$. An initial condition $(x_0,y_0)\in D$ fixes $C=x_0^2+y_0^2$ and selects the sign of $y_0$. In particular, $y(0)=2$ gives $C=4$ and the upper branch on $(-2,2)$. The constant alone does not select the branch.^[ref]

The precise local criterion uses the implicit function theorem. Let $F$ be continuously differentiable on an open subset of the plane, with $F(x_0,y_0)=C$. If its partial derivative $F_y(x_0,y_0)\ne 0$, the nearby level set is a unique continuously differentiable function graph $y=g(x)$ through that point. The chain rule givesThis proves that the local function graph solves $y^{\prime }=-F_x/F_y$ there. For a different prescribed equation $y^{\prime }=f(x,y)$, one must also check $F_x+F_{y}f=0$ along that branch and that it stays in the domain of $f$. A level set by itself does not verify an arbitrary ordinary differential equation.^[ref]

For $F=x^2+y^2$, the upper branch through $(1,\sqrt{3})$ has $F_y=2\sqrt{3}$ and tangent line slope $-1/\sqrt{3}$. At $(2,0)$, however, $F_y=0$ and $F_x=4$. The circle remains a smooth regular level set with a vertical tangent line: it can be written locally as $x=\sqrt{4-y^2}$, but not as a single function graph $y=g(x)$ over an open interval containing $2$. Moreover, $-x/y$ is undefined there, so the scalar solution of a differential equation cannot include this point in its domain. Changing the parameter draws the curve through the point but does not extend that scalar initial value problem.

When both partial derivatives vanish, even the geometry needs separate inspection. The level $x^2+y^2=0$ is an isolated point, not a function graph on any nonempty open interval. The zero level set of $G(x,y)=y^2-x^2$ instead consists of the two lines $y=x$ and $y=-x$ crossing at the origin, where $\nabla G=0$. Each line satisfies the implicit ordinary differential equation $y{y}^{\prime }=x$, including at the origin; the explicit form $y^{\prime }=x/y$ excludes the origin. Thus algebraically dividing by a vanishing coefficient changes the domain and can change which solutions of a differential equation are admissible. Vanishing $F_y$ alone is only failure of a sufficient test: $F(x,y)=y^3$ has the smooth zero function graph $y=0$ despite $F_y=0$ everywhere on it.

Implicit solutions are especially useful for exact first-order ordinary differential equations and for finding a first integral of an ordinary differential equation: a conserved relation can reveal allowed states without an explicit formula. For the normalized harmonic oscillator written as $\dot{x}=y$, $\dot{y}=-x$, the energy $H=(x^2+y^2)/2$ has $\dot{H}=x\dot{x}+y\dot{y}=0$. Its positive level sets are the same circles. Where $y\ne 0$, eliminating time gives $dy/dx=-x/y$ and the local implicit solution above. The harmonic oscillator's parametric curve continues through $y=0$ because its vector field governing evolution in time is defined there; treating $x$ as the independent variable fails at that turn. The origin is an equilibrium point of the harmonic oscillator, although it is excluded from the scalar equation $y^{\prime }=-x/y$.

The intermediate value theorem says that a real-valued continuous function cannot skip a value between two of its outputs. Let $a<b$ and let $f:[a,b]\to \mathbb{R}$ be a continuous function on the entire closed interval $[a,b]$. ThenThe word "between" includes both endpoint values: choose $c=a$ when $y=f(a)$, or $c=b$ when $y=f(b)$. If $y$ is strictly between them, every such $c$ lies in the open interval $(a,b)$. Continuity on $[a,b]$ means continuity at every interior point, together with the endpoint one-sided limits ${\lim }_{x\to a^{+}}f(x)=f(a)$ and ${\lim }_{x\to b^{-}}f(x)=f(b)$. No derivative is required.^[ref]

For finding a zero of a function, the useful special case isStrictly opposite endpoint signs certify a root bracket, because zero is strictly between $f(a)$ and $f(b)$. If an endpoint value is already zero, that endpoint supplies a zero of a function directly; an interior zero of a function need not exist. For example, $f(x)=x$ on $[0,1]$ has its only zero of a function at the left endpoint. The intermediate value theorem guarantees existence, but neither uniqueness nor a necessary sign-change test. Its numerical usefulness is that the bisection method can turn existence into successively tighter root brackets.

Suppose a interval map $F:A\to A$ maps every nondegenerate subinterval onto $A$ after finitely many iterates. Choose two separated closed intervals inside the interior of $A$ and a common iterate mapping each onto $A$. Between first-passage endpoint levels in each interval, take an open interval that maps exactly onto the interior of $A$. These two disjoint intervals are the branches of a horseshoe for an interval map for that iterate.

// Target: dynamical-systems.bigb