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Suppose continuous functions $f_n:\mathbb{R}\to \mathbb{R}$ converge pointwise to $f$. For each $ϵ>0$, some nonempty open interval $I$ and integer $N$ satisfy $|f_n(x)-f(x)|\le ϵ$ for every $x\in I$ and $n\ge N$. Indeed the closed sets $Q_N={\bigcap }_{n,m\ge N}\{|f_n-f_m|\le ϵ\}$ cover $\mathbb{R}$. The Baire category theorem gives an interval inside one $Q_N$, and passing $m\to \infty$ proves the assertion. The interval may depend on $ϵ$; this does not assert uniform convergence on one fixed interval.

The mean value theorem says that an average rate of change is attained as an instantaneous rate somewhere between the endpoints. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Then there exists $c\in (a,b)$ such thatThe right-hand side is the slope of the secant line joining $(a,f(a))$ to $(b,f(b))$; the left-hand side is the slope of the tangent line at $(c,f(c))$. The derivative need not be continuous, and the function need not be differentiable at the endpoints. ^[ref]

For the polynomial $f(x)=x^2$ on $[1,3]$, the secant line has slope $(9-1)/(3-1)=4$. Since the derivative is $f^{\prime }(x)=2x$, the required interior point is $c=2$. The tangent line there is $y=4+4(x-2)=4x-4$, whereas the secant line is $y=1+4(x-1)=4x-3$. Their equal slopes make them parallel, but they are distinct straight lines: the theorem equates rates, not heights.

The practical strength is that the unknown point $c$ can disappear from an estimate. Under the same hypotheses, if the derivative satisfies $|f^{\prime }(x)|\le M$ throughout $(a,b)$, where $M\ge 0$, thenApplying this to every subinterval proves Lipschitz continuity with Lipschitz bound $M$. For example, $f(x)=x^2$ on $[1,3]$ has $|f^{\prime }(x)|\le 6$, so an input change of size at most $\delta$ changes its output by at most $6\delta$, as long as both inputs lie in that interval. A local derivative bound therefore controls finite changes without computing an integral. The sign of the derivative similarly gives the derivative criterion for monotonicity.

Taking $M=0$ shows that a function with zero derivative throughout a real interval is a constant function, provided it is continuous at any included endpoints. The interval condition is essential: on the domain of a function $(-2,-1)\cup (1,2)$, the function equal to zero on the left interval and one on the right interval has zero derivative everywhere. It is a locally constant function, yet not a constant function on the whole domain of a function. The mean value theorem cannot cross the gap.

A concise proof subtracts the secant line: set $m=(f(b)-f(a))/(b-a)$ and $h(x)=f(x)-f(a)-m(x-a)$. Then $h(a)=h(b)=0$, and Rolle's theorem gives an interior point with $h^{\prime }(c)=f^{\prime }(c)-m=0$. The geometric idea is to remove the average trend and find a horizontal tangent line in what remains.

A sequence of nested intervals keeps every new enclosure inside the previous one:Each step adds a restriction without discarding an earlier restriction. This is useful when an exact real number is described by increasingly precise bounds rather than by a finite decimal expansion. The bisection method, for instance, produces nested intervals that keep enclosing a zero of a function.

For a simple example, the closed intervals $I_n=[1-1/n,1+1/n]$ are nested and all contain $1$. Their lengths $2/n$ tend to zero, so the nested interval theorem leaves exactly that point. The family $J_n=[1-1/n,2+1/n]$ is also nested, but its intersection is the whole closed interval $[1,2]$. Nesting alone does not force uniqueness.

The open intervals $K_n=(0,1/n)$ illustrate a different possibility. Every finite collection has a nonempty intersection, but the intersection of all of them is empty: any positive candidate eventually exceeds $1/n$, and zero is excluded from every member. Each finite row in the diagram contains points; the full infinite family need not.

If $I_1\supseteq I_2\supseteq \cdots$ are nonempty closed intervals with finite endpoints, then at least one real number belongs to every member. If their lengths tend to zero, exactly one real number belongs to every member. The result turns an unlimited sequence of consistent enclosures into an existence guarantee.

Write $I_n=[a_n,b_n]$. Nesting makes the left endpoints nondecreasing and the right endpoints nonincreasing. Their bounds give finite supremum and infimum, withIndeed, every $b_n$ is an upper bound for all the $a_j$, so the least-upper-bound property gives ${\sup }_j{a}_j\le b_n$ for every $n$. This proves the displayed interval is nonempty. If $b_n-a_n\to 0$, its two endpoints must coincide. This is the connection to the completeness of the real numbers discussed in Günther Hörmann and David Langer, Einführung in die Analysis, §§3.15–3.16: nested intervals and order completeness.

The distinction between existence and uniqueness is useful in computation. If an unknown point $x_{\ast }$ lies in every $[a_n,b_n]$, the midpoint $m_n=(a_n+b_n)/2$ satisfiesThus the enclosure itself certifies an absolute error without knowing the answer. The bisection error bound is the special case where each update halves the width.

Closedness and a finite initial enclosure matter. The open intervals $(0,1/n)$ have no common point, while the closed but unbounded intervals $[n,\infty )$ also have empty intersection. A picture of a few intervals cannot certify the intersection of an infinite family; the hypotheses of the nested interval theorem do that work.

An open arc is a topological space homeomorphic to an open interval. Unlike an arc, which is homeomorphic to a closed interval, an open arc does not include its two endpoints.

If a nonzero continuous real function on an interval is orthogonal to every polynomial of degree below $n$, it has at least $n$ distinct interior zeros. If it had only finitely many zeros and fewer than $n$ sign changes, multiply it by the product of linear factors at its sign-changing zeros. The product has constant sign and is nonzero on an open interval, so its integral cannot vanish. Orthogonality to this degree-below-$n$ polynomial contradicts that conclusion. Infinitely many interior zeros already satisfy the assertion. Orthogonal polynomials of respective degrees $0,\dots ,n-1$ form a basis of the needed polynomial space.

Let $U_{\alpha }=\{x:m_u(f)(x)>\alpha \}$. It is open: it is the union of all open intervals $I$ for which ${\int }_I|f|>\alpha |I|$. This also proves measurability of the uncentered maximal function. Triangle inequality inside each average proves sublinearity and absolute homogeneity.

Take a compact subset $K\subset U_{\alpha }$, and choose finitely many such intervals covering it. The one-dimensional Wiener covering lemma selects pairwise disjoint intervals $I_1,\dots ,I_N$ whose concentric triples cover $K$. For a finite family this follows by choosing a longest remaining interval and discarding all intervals intersecting it: each discarded interval has no greater length and lies in the triple of the chosen one. ThusTaking the supremum over compact subsets, by inner regularity of Lebesgue measure, givesThis establishes weak type (1,1). It is the one-dimensional uncentered maximal weak-type inequality for the finite measure $|f(t)|dt$; the constant need not be optimal for this question.

A horseshoe for an interval map consists of an open interval $J$ containing two disjoint open subintervals each mapped onto $J$. Glendinning chaos means that some positive iterate has a horseshoe.

Solving on the appropriate branches of the tent map givesHere $F(x_{-2})=x_{-1}$ and $F(x_{-1})=x_0$. On $J=[x_{-1},x_0]$ the second iterate isBoth endpoints map to $x_0$, and the minimum is $v=\mu -{\mu }^2/2$. Both monotone branches cover $J$ when $v\le x_{-1}$, equivalently $(\mu -1)({\mu }^2-2)\ge 0$. Restricting them to preimages of the interior proves a horseshoe for $F^2$ when $\mu \ge \sqrt{2}$.

For $1<\mu \le \sqrt{2}$, $G(J)\subseteq J$. Set $d=(\mu -1)/(\mu +1)$ and $h(x)=(x_0-x)/d$. This maps $J$ onto $[0,1]$; substitution in each linear piece gives $hG{h}^{-1}=F_{{\mu }^2}$. The renormalization of the tent map near its fixed point therefore makes $F^4$ conjugate on $J$ to $F_{{\mu }^2}^2$. Applying the preceding threshold proves its horseshoe when $2^{1/4}\le \mu <\sqrt{2}$.

For $\mu \ge \sqrt{2}$, the inverse branch of $G$ fixing $x_0$ contracts distance to $x_0$ by ${\mu }^{-2}$. Compose $m$ copies of it with the other inverse branch. This contraction has a fixed point with a periodic itinerary, distinct from $x_0$ and at distance $O({\mu }^{-2m})$. Hence nontrivial periodic points accumulate at $x_0$.

For $\mu <\sqrt{2}$, the conjugate tent parameter $\nu ={\mu }^2<2$ has every nonzero periodic point in $[\nu -{\nu }^2/2,\nu /2]$. This interval is forward invariant; below it, a nonzero point increases until entering it. The positive lower bound separates those periodic points from zero, which corresponds to $x_0$. Thus no nontrivial periodic points are sufficiently close to $x_0$ from the left. A point sufficiently close on the right maps to such a left-hand point; periodicity would pass to its image and is impossible. Thus $x_0$ is isolated among periodic points, even though a fourth-iterate horseshoe exists elsewhere.

The nonzero leading coefficients and distinct degrees make $P_0,\dots ,P_{n-1}$ a basis of the polynomials of degree at most $n-1$. Consequently the given orthogonality implies ${\int }_a^{b}f(x)Q(x)dx=0$ for every such polynomial $Q$.

If $f$ has infinitely many distinct zeros in $(a,b)$, there is nothing to prove. Otherwise suppose it has fewer than $n$. Among those zeros, list the sign-changing ones as $t_1<\cdots <t_m$, with $m<n$, and put $Q(x)={\prod }_{j=1}^m(x-t_j)$, using $Q=1$ when $m=0$. On each interval between zeros, $f$ has a fixed sign by continuity. Crossing a sign-changing zero flips both $f$ and $Q$; crossing any other zero flips neither sign. Thus $fQ$ has one constant sign wherever nonzero. Since $f$ is not identically zero, continuity gives an open interval on which $fQ$ is strictly of that sign. Hence its integral is nonzero, contrary to orthogonality to $Q$. Therefore $f$ has at least $n$ distinct zeros in $(a,b)$. This is the reusable orthogonality forces many interior zeros argument.

Fix $ϵ>0$. For each $N$, the set $Q_N={\bigcap }_{n,m\ge N}\{x:|f_n(x)-f_m(x)|\le ϵ\}$ is closed, because each $f_n-f_m$ is continuous. At every $x$, pointwise convergence makes $(f_n(x))$ a Cauchy sequence, so $\mathbb{R}={\bigcup }_{N\ge 1}{Q}_N$. The Baire category theorem supplies an $N_0$ for which $Q_{N_0}$ contains a nonempty open interval $I$. For $x\in I$ and $n\ge N_0$, pass $m\to \infty$ in $|f_n(x)-f_m(x)|\le ϵ$ to obtainThis local uniform Cauchy control for pointwise convergent continuous functions allows both $I$ and $N_0$ to depend on $ϵ$.

The direct-sum assertion fails in both degrees when the pieces overlap. Take genuine open covers, so the failure does not depend on any subtle choice of smooth structures on subsets.

For degree zero, let $M=\mathbb{R}$, $U=(-\infty ,1)$ and $V=(-1,\infty )$. By degree-zero de Rham cohomology, $H_{\mathrm{dR}}^0$ consists of locally constant functions. Each of $M,U,V$ is connected and nonempty, soThey are not isomorphic as real vector spaces. The restriction map is $c\mapsto (c,c)$; the overlap forces equality instead of allowing two independent constants.

For degree one, let $M=S^1$, and remove distinct points $a,b$ to form $U=S^1\setminus \{a\}$ and $V=S^1\setminus \{b\}$. Each is diffeomorphic to an open interval. Every one-form $f(t)dt$ on an interval has the smooth primitive ${\int }_{t_0}^{t}f(s)ds$, so both pieces have zero first de Rham cohomology. On the circle, the form $\eta =xdy-ydx$ is closed because all two-forms on a one-dimensional manifold vanish. The parameterization $(x,y)=(\cos t,\sin t)$ givesAn exact form has zero integral around this loop: the integral of $dF$ is the difference of the endpoint values of the periodic function $F$. Thus $[\eta ]\ne 0$, while the proposed right side is zero, disproving the isomorphism.

In fact $H_{\mathrm{dR}}^1(S^1)\cong \mathbb{R}$. For a periodic one-form $f(t)dt$, subtract $cdt$ where $c=(2\pi {)}^{-1}{\int }_0^{2\pi }f(t)dt$. The remainder has the periodic primitive ${\int }_0^t(f(s)-c)ds$, and hence is exact. This also explicitly identifies the nonzero class used above.

Yes. There is a computable isomorphism, obtained by an effective back-and-forth construction. Decidability of the two orders makes the usual existence argument into an algorithm.

Maintain a finite order-preserving partial isomorphism of structures $f_s$ from $(\mathbb{N},{<}_A)$ to $(\mathbb{N},{<}_B)$, starting with the empty map. At a forth step, take the least ordinary natural number $a$ outside its domain. Look at the finitely many mapped elements below and above $a$ in ${<}_A$. Their images impose an open interval in ${<}_B$: it lies above all the lower images and below all the upper images, with a missing bound interpreted as an unbounded side.

This interval contains a fresh element. If both bounds exist they are ordered correctly by the induction hypothesis; density supplies an element strictly between them. If there is just one bound, the absence of endpoints supplies an element beyond it; if there are no bounds, choose any element. Density and the absence of endpoints moreover give infinitely many points in every such open interval, so finitely many already used images can always be avoided.

Enumerate $b=0,1,2,\dots$ in the ordinary presentation order, test whether $b$ is unused and satisfies every required ${<}_B$ inequality, and choose the first successful candidate. All tests are decidable, and the existence argument proves termination. Extend $f_s$ by $a\mapsto b$.

At a back step, take the least natural number outside the range, interchange the roles of $A$ and $B$, and carry out exactly the same search for a fresh preimage. Alternate forth and back steps. Every stage is an effective terminating finite computation, and every stage remains an order-preserving partial isomorphism of structures.

Let $f={\bigcup }_s{f}_s$. The least-unused scheduling puts every element of $A$ into the domain and every element of $B$ into the range. For example, after $n+1$ forth steps, all ordinary presentation numbers at most $n$ have been included, since each step removes the least missing one. Thus $f$ is a bijection preserving and reflecting the orders. To compute $f(n)$, simulate the construction until $n$ appears in its domain; this terminates. The corresponding range search computes its inverse. HenceThe searches use the decidable presentations, rather than a possibly ineffective choice of points in an abstract dense linear order without endpoints.

No: the Lebesgue measure can decrease when separate components of the Minkowski sum merge. A one-dimensional example is enough to disprove the assertion in general. TakeThe Minkowski sum $A+tB$ consists of four open intervals, with left endpoints $0,10,2t,10+2t$ and common length $(1+t)/10$. At $t=4$ the four open intervals are disjoint, soAt $t=5$ the middle two open intervals coincide and the remaining three are disjoint, soFor a counterexample in any prescribed dimension $n>1$, take Cartesian products of these open sets with $(0,1)$ in the remaining coordinates. Compare $t_2=5$ with $t_1=5-\delta$, where $\delta ={\delta }_n>0$ is sufficiently small, and choose the interval width $\epsilon >0$ so small that $6\epsilon <2\delta$. The four intervals remain disjoint at $t_1$, givingChoosing $(1-\delta /6{)}^n>3/4$ makes the latter smaller. Here $A_n=((0,\epsilon )\cup (10,10+\epsilon ))\times (0,1{)}^{n-1}$ and $B_n=((0,\epsilon )\cup (2,2+\epsilon ))\times (0,1{)}^{n-1}$. The Brunn–Minkowski inequality gives a lower bound at each value of $t$; it does not give the claimed monotonicity for arbitrary open sets.

An interval map $F:I\to I$ has a horseshoe when there are an open interval $J\subseteq I$ and two disjoint open intervals $K_0,K_1\subseteq J$ such that

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Fréchet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

For $n\ge 0$, putEach $S_n$ is closed, the hypothesis gives $\mathbb{R}={\bigcup }_n{S}_n$, and $E_n\subset E_{n+1}$ because a function that vanishes on an open set has derivative zero there.

The open setis dense. Indeed, inside any nonempty open interval choose a nondegenerate closed subinterval $J$. The complete metric space $J$ is covered by the closed sets $J\cap S_n$, so the Baire's theorem makes one of them contain a relative open interval, which lies in some $E_n$.

On each connected component $I$ of $\Omega$, the function $F$ is one polynomial. To see this, any compact subinterval $K\subset I$ is covered by the increasing family $(E_n)$. A finite subcover therefore gives $K\subset E_N$ for some $N$, and $F^{(N)}=0$ on $K$. Hence $F$ is polynomial on $K$; overlapping compact intervals force these polynomials to agree throughout $I$.

Let $D=\mathbb{R}\setminus \Omega$. This closed set has no isolated points. Otherwise, on the two sides of an isolated point $x$, $F$ would equal polynomials. Smoothness makes all their one-sided derivatives agree at $x$, so the two polynomials are identical and extend across $x$. Some derivative would then vanish on a neighborhood of $x$, contrary to $x\in D$.

Suppose that $D$ is nonempty. It is complete andis a countable closed cover. A second application of Baire's theorem gives an index $k$ and an open interval $U_0$ such thatChoose $x_0\in D\cap U_0$ and a bounded open interval $U$ centred at $x_0$ whose closure lies in $U_0$.

Because $D$ has no isolated points, every $x\in D\cap U$ is approached by distinct points of $D\cap U$. Difference quotients first give $F^{(k+1)}(x)=0$, and induction gives

No component of $\Omega$ meeting $U$ can cross $x_0$, so it has an endpoint in $D\cap U$. If its polynomial had degree $d\ge k$, then its $d$th derivative would approach a nonzero constant at that endpoint, contradicting the preceding display. Its degree is therefore less than $k$, so $F^{(k)}$ also vanishes on the part of that component in $U$. We conclude that $F^{(k)}=0$ throughout $U$, which says $U\subset E_k\subset \Omega$ and contradicts $D\cap U\ne \varnothing$.

A limit point $p$ of $E\subseteq \mathbb{R}$ is a point such that, for every $\epsilon >0$, the punctured neighbourhood $(p-\epsilon ,p+\epsilon )\setminus \{p\}$ meets $E$. A closed set contains all of its limit points.

Every nonempty open interval contains a rational number; applying this to $(p,p+\epsilon )$ shows that every $p\in \mathbb{R}$ is a limit point of $\mathbb{Q}$. Thus the limit points of $\mathbb{Q}$ form $\mathbb{R}$, and $\mathbb{Q}$ is not closed.

If $p\in \mathbb{Z}$, radius $1/2$ excludes all other integers. If $p\notin \mathbb{Z}$, its distances to the two neighbouring integers have a positive minimum, and a smaller radius excludes every integer. Thus $\mathbb{Z}$ has no limit points in $\mathbb{R}$ and is closed.

The moment generating function is $M_X(t)=\mathbb{E}(e^{tX})$ wherever that expectation is finite. A useful continuity theorem says that if the moment generating functions $M_n$ are finite on a common open interval containing zero and converge pointwise there to the moment generating function of a probability distribution, then the corresponding variables converge in distribution to that law. Equivalently one may assume a finite limiting function on such a neighborhood, with the usual continuity at zero, and obtain the uniquely determined limiting law.

For the gamma density, integrating the exponential against the density and changing variable to $(\lambda -t)x$ givesFor the waiting time to one head, independence of successive tosses gives $\mathrm{Pr}(Y=j)=p(1-p{)}^{j-1}$ for $j\ge 1$. Summing the geometric series givesThe waiting times between consecutive heads are independent copies of $Y$, so their sum $N$ has moment generating function $M_Y(t{)}^k$. ConsequentlyFor fixed $t<1/2$, numerator and denominator have leading terms $p$ and $p(1-2t)$ as $p\downarrow 0$. A fixed neighborhood of zero is inside the convergence domain for all sufficiently small $p$, so the continuity theorem applies. The limiting distribution isequivalently a chi-squared distribution with $2k$ degrees of freedom.

The moment generating function is $\boxed{M_X(\theta )=\mathbb{E}[e^{\theta X}]}$ for real $\theta$ where this expectation is finite. Existence on an open interval about zero determines the distribution and permits differentiation there to obtain moments, when the usual local integrability conditions hold.

The inclusion of $X$ into $\mathbb{R}$ is continuous, so a connected component $K$ of $X$ is an interval in $\mathbb{R}$. For $x\in K$, openness of $X$ gives $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq X$. This interval is connected and meets $K$, so their union is connected and maximality forces the interval into $K$. Thus $K$ is open in $\mathbb{R}$. A nonempty open connected subset of the line is an open interval, allowing an infinite endpoint or the whole line. Hence every component is an open interval.