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A real additive function $f:\mathbb{R}\to \mathbb{R}$ that is continuous at one point $x_0$ is continuous everywhere and has form $f(x)=xf(1)$. Indeed $f(0)=0$, and $f(h)=f(x_0+h)-f(x_0)\to 0$ as $h\to 0$. Then $f(x+h)-f(x)=f(h)\to 0$ for every $x$. Additivity gives $f(q)=qf(1)$ for every rational number; continuity and the density of the rational numbers extend this identity to every real argument.

The step from the rational numbers to the real numbers can be made explicit. Put $c=f(1)$. For an arbitrary real number $x$, the density of the rational numbers supplies a sequence of rational numbers $q_n\to x$. Because continuity at the original point already implied continuity everywhere,This is where continuity enters: knowing all the rational values alone does not justify the first equality.

For an original numerical example, use the floor function to define the rational lower approximants $q_n=\lfloor 10^n\sqrt{2}\rfloor /10^n$. They obey $0\le \sqrt{2}-q_n<10^{-n}$. When $f(1)=3$ and $f$ is continuous at one point, rational homogeneity of an additive function and the preceding argument giveThe error bound here follows from the resulting linear map, rather than from a quantitative bound on the initially assumed continuity.

A family of pairwise disjoint real intervals of positive length is countable. Each interval contains a rational number in its interior by the density of the rational numbers. Fix an enumeration of the rational numbers and select the first member lying inside each interval. Disjointness makes this selection an injection into a countable set. The argument also applies to disjoint nonempty open subsets of the real line.

Every family of pairwise disjoint nonempty open disks in the plane is countable. The density of the rational numbers gives each disk a point of the countable set ${\mathbb{Q}}^2$. Fixing an enumeration and taking the first point in each disk yields an injection from disks to labels. This argument requires no countable choice of representatives and is useful when geometric separation would otherwise produce an uncountable family.

Every nonempty open real interval contains an irrational number. Choose a rational number $q$ strictly inside the interval using density of the rational numbers; then $q+\sqrt{2}/N$ remains inside for sufficiently large positive integers $N$ and is irrational. Thus the irrational numbers form a dense subset of the real line. Together with rational density, this gives approaching sequences of both types at every real point, useful for testing continuity of functions defined differently on the two sets.

For a nonnegative measurable function $f:X\to [0,\infty ]$ on a measure space, defineThe supremum may be infinite. A simple function assigns finitely many heights to measurable sets, so its integral is height times measure, summed over those pieces. This definition groups points by their values rather than requiring a partition into short intervals. See MIT lecture 11, Definition 108.

A real-valued function is Lebesgue integrable when ${\int }_X|f|d\mu <\infty$. Its integral is ${\int }_X{f}_{+}d\mu -{\int }_X{f}_{-}d\mu$, where $f_{+}=\max (f,0)$ is its positive part and $f_{-}=\max (-f,0)$. This finiteness requirement avoids subtracting two infinities. For functions taking values in the complex numbers, integrate the real part and imaginary part separately.

Consider the Dirichlet function $f=1_{\mathbb{Q}\cap [0,1]}$ on $[0,1]$. It is a simple function, andIts Riemann integral does not exist. For every partition of an interval $0=x_0<\cdots <x_n=1$, each nondegenerate $[x_{i-1},x_i]$ contains both rational numbers and irrational numbers, by their respective density of the rational numbers and density of the irrational numbers. Thus its infimum there is $0$ and its supremum is $1$. The lower Darboux sum is ${\sum }_{i}0(x_i-x_{i-1})=0$, whereas the upper Darboux sum is ${\sum }_{i}1(x_i-x_{i-1})=1$. Refinement never closes this gap. A finite scatterplot cannot represent this behavior: the assertion concerns every subinterval and every partition of an interval.

Another view builds the Lebesgue integral from horizontal layers. For any nonnegative measurable function, the layer cake representation statesBoth sides may be infinite. The right side integrates the measure of a superlevel set against Lebesgue measure in the height variable $t$. See Gautam Iyer's measure theory notes, Theorem 9.5.

For $f(x)=x^2$ on $[0,1]$ with Lebesgue measure, the superlevel set at $0\le t<1$ is $(\sqrt{t},1]$, of Lebesgue measure $1-\sqrt{t}$; for $t\ge 1$ it is empty. ConsequentlyAt height $t=1/4$, the layer extends from $x=1/2$ to $x=1$, a width of $1/2$. The picture shows why the same area can be accumulated vertically or horizontally. In a probability space, the identical formula expresses the expected value of a nonnegative random variable through its exceedance probabilities.

A proof outline avoids extra assumptions on the measure space. For a nonnegative simple function $s={\sum }_j{a}_j{1}_{A_j}$ with disjoint measurable sets $A_j$, the superlevel set has measure ${\sum }_{a_j>t}\mu (A_j)$. Integrating this finite sum gives ${\sum }_j{a}_j\mu (A_j)=\int sd\mu$. Next use approximation by nonnegative simple functions $s_n\uparrow f$. For every $t$, the superlevel sets $\{s_n>t\}$ increase to $\{f>t\}$. Continuity from below of a measure and the monotone convergence theorem on the height variable give the formula for $f$.

The density of the rational numbers gives a countable way to separate two unequal values:Indeed, $f(x)<g(x)$ holds exactly when some rational number lies strictly between them. Every set inside the union is a Lebesgue measurable set, by the definition of a measurable function and the closure properties of a sigma-algebra. Since $\mathbb{Q}$ is countable, the union is measurable.

A set is countable if it is finite or admits an injection into the natural numbers; an infinite countable set can equivalently be listed as a sequence. Let $A={\bigcup }_{i\ge 1}{A}_i$ with each $A_i$ countable. Choose an enumeration of each nonempty $A_i$, allowing repetition if it is finite. List the pairs $(i,j)\in {\mathbb{N}}^2$ in successive finite diagonals of constant $i+j$ and output the $j$th member of $A_i$. This lists every member of $A$, so $A$ is countable; empty sets contribute nothing, and repetitions can be removed by retaining first occurrences. This is the countable union of countable sets theorem, in the usual setting with axiom of countable choice for selecting the enumerations.

Each positive-length interval $J_i$ has a nonempty interior containing a rational number by the density of the rational numbers. Fix one enumeration of the rational numbers and let $q_i$ be its first member in the interior of $J_i$. Disjointness gives $i\ne j\Rightarrow q_i\ne q_j$, so $i\mapsto q_i$ is an injection of $I$ into a countable set. Hence $I$ is countable. This proof of the countability of disjoint positive-length intervals uses a fixed enumeration and needs no arbitrary simultaneous choice of rationals.

For the additive function, setting both arguments to zero gives $f(0)=0$, and then $f(-x)=-f(x)$. Continuity at the given point $x_0$ impliesFor any real $x$, $f(x+h)-f(x)=f(h)$, so $f$ is continuous at every point. This proves the continuity part of additive function continuous at one point is linear.

Let $c=f(1)$. Repeated addition and negation give $f(m)=mc$ for every integer $m$. For integers $p$ and positive $q$,so $f(r)=cr$ for every rational number $r$. Given any real $x$, choose rational $r_n\to x$. Continuity and density of the rational numbers giveConversely each real multiple of $x$ satisfies the Cauchy functional equation.

For the multiplicative equation, assume first that $g$ is not identically zero and choose $x_1$ with $g(x_1)\ne 0$. Then $g(x_1)=g(x_1)g(0)$ forces $g(0)=1$. Nextmakes every value nonzero, andmakes every value positive. Thus $h(x)=\log g(x)$ is a continuous real additive function, since $h(x+y)=h(x)+h(y)$. The proved classification gives $h(x)=kx$ for a real constant $k$. This solves the continuous exponential functional equation:Both types satisfy the equation and continuity, and every nonzero solution is everywhere strictly positive.

At zero, $|f(x)-f(0)|=|x|\to 0$, so the function is continuous. At a nonzero $x_0$, approach along rational and irrational sequences; density of the rational numbers and density of the irrational numbers give limiting values $x_0$ and $-x_0$, which are different. Thus the continuity set is $\boxed{\{0\}}$.

At zero, the difference quotient is one for rational nonzero increments and minus one for irrational increments, so the derivative does not exist. Elsewhere differentiability implies continuity, and continuity already fails. Thus the differentiability set is $\boxed{\varnothing }$.

For the first additional construction, modify the rational/irrational example toAt zero, $|g(h)/h|=|h|\to 0$, so $g^{\prime }(0)=0$. At any nonzero point, density of the rational numbers and density of the irrational numbers give incompatible limiting values $x^2$ and $-x^2$. Thus $g$ is discontinuous, and hence not differentiable, there. Its differentiability set is exactly $\{0\}$.

For the second construction, use a summable absolute-value cusp series:On $[-R,R]$, each summand is bounded by $2^{-n}(R+1/2)$. The Weierstrass M-test therefore gives uniform convergence on every bounded interval, and the uniform limit theorem makes $G$ continuous everywhere.

To justify differentiability, each summand's difference quotient has absolute value at most $2^{-n}$, by the reverse triangle inequality. The sum of these bounds is finite, uniformly in the increment. Thus the tail of the difference quotient is uniformly small, and limits may be passed through the sum by first retaining finitely many terms and then making the tail small. If $x\ne 1/n$ for every $n\ge 2$, each individual summand is differentiable, givingIn particular $G^{\prime }(0)=-{\sum }_{n=2}^{\infty }{2}^{-n}=-1/2$. The accumulation of cusp locations at zero does not destroy the derivative, because their weights are summable.

At $x=1/m$, the same tail argument applies to both one-sided derivatives. Every term except the $m$th has the same derivative from both sides; the $m$th has derivatives $-2^{-m}$ and $2^{-m}$. ConsequentlyTherefore $G$ fails to be differentiable exactly at $1/2,1/3,1/4,\dots$, and is differentiable everywhere else.

The integers are countable: $j(z)=2z+1$ for $z\ge 0$ and $j(z)=-2z$ for $z<0$ is an injection $\mathbb{Z}\to \mathbb{N}$. Hence $\mathbb{Z}\times \mathbb{N}$ is a countable set. The map $(a,b)\mapsto a/b$ is a surjection onto the rational numbers, so the image result already proved gives $\mathbb{Q}$ is countable. Unique reduced fractions are not required for this surjection argument.

For the final unheaded geometry request, suppose for contradiction that no two distinct circles in the collection intersect. Restrict to the subcollection ${\mathcal{C}}_0$ of circles tangent to the $x$-axis. Each such nondegenerate circle has a unique tangency point, and the tangency-point map ${\mathcal{C}}_0\to \mathbb{R}$ is a surjection by the hypothesis.

The closed disk of a circle tangent at $(a,0)$ meets the $x$-axis only at $(a,0)$. Two distinct members of ${\mathcal{C}}_0$ cannot have nested disks: if one disk were inside the other, its tangency point would belong to the larger disk, so both tangency points would be equal and both circle boundaries would pass through it. That already contradicts nonintersection. Circles with disjoint boundaries and no nesting have disjoint disks. Thus their open disks are pairwise disjoint.

By the density of the rational numbers, every nonempty open disk contains a point of ${\mathbb{Q}}^2$, which is a countable set. Fix an enumeration of ${\mathbb{Q}}^2$ and assign to each disk the first enumerated point inside it. Disjointness makes this assignment an injection. This is the countability of pairwise disjoint open disks, and proves that ${\mathcal{C}}_0$ is a countable set. Its surjection onto $\mathbb{R}$ would make the real numbers countable, contrary to the Cantor diagonal argument. ThereforeThe proof allows circles on either side of the axis and counts tangential contact as intersection; it does not assume that all circles in the original collection are tangent to the axis.

The martingale convergence theorem says that if a discrete-time martingale $(M_n)$ satisfies $C={\sup }_n\mathbb{E}|M_n|<\infty$, then there is an integrable random variable $M_{\infty }$ such thatThe hypothesis does not by itself ensure convergence in L1.

For each pair of rational numbers $a<b$, the Doob upcrossing inequality gives $\mathbb{E}{U}_N[a,b]\le (C+|a|)/(b-a)$. The monotone convergence theorem therefore gives a finite expected value for $U_{\infty }[a,b]={\lim }_N{U}_N[a,b]$, so this upcrossing count is finite almost surely. There are only countably many such pairs, hence all their upcrossing counts are simultaneously finite outside one event of probability zero.

If ${\mathrm{liminf}}_n{M}_n<{\mathrm{limsup}}_n{M}_n$, the density of the rational numbers supplies $a<b$ strictly between them, forcing infinitely many upcrossings. Consequently $M_n$ has a limit in the extended real numbers almost surely. By the Fatou lemma,A limit of either $+\infty$ or $-\infty$ would make this lower limit infinite, so the limit is finite almost surely, and the same inequality proves its integrability.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

For a increasing function $F$, define its one-sided limits byThey exist because of monotonicity. The function is discontinuous at $x$ only if the jump intervalis nonempty. By the density of the rational numbers, choose $q_x\in J_x\cap \mathbb{Q}$.

If $x<y$, monotonicity gives $F(x+)\le F(y-)$, so $J_x$ and $J_y$ are disjoint. Consequently $q_x\ne q_y$, and $x\mapsto q_x$ injects the set of discontinuities into $\mathbb{Q}$. Since the rationals are countable by part (ii), the set of discontinuities is countable.

The printed assertion needs an endpoint qualification: the enumeration includes zero, but $1_{[0,0)}$ is identically zero. In fact $f$ is continuous at zero; the claimed discontinuities occur at every $r_j>0$.

Define the finite step function sumsThe assumed convergence of ${\sum }_{k}1/k^2$ and the bound on each indicator function giveThus the defining series converges everywhere and has uniform convergence. Part (b) proves that $f$ is Riemann integrable; it also gives, if desired,For $r_j>0$ and $0\le t<r_j$, every indicator function is nonincreasing, and the $j$th drops by one between $t$ and $r_j$. HenceTaking inputs from the left tending to $r_j$ proves a discontinuity there.

For completeness, continuity at zero follows directly from a finite-head and tail argument. Given $\epsilon >0$, choose $N$ with ${\sum }_{k>N}1/k^2<\epsilon$. Choose $\delta >0$ below every positive $r_k$ with $k\le N$. For $0\le t<\delta$, the first $N$ indicator functions have exactly their values at zero; the one with $r_k=0$, if present, is always zero. Thus $|f(t)-f(0)|\le {\sum }_{k>N}1/k^2<\epsilon$. This confirms that the endpoint exception is real. Replacing the printed enumeration of $[0,1)$ by one of $(0,1)$ would remove it.

Finally, the density of the rational numbers supplies a positive rational $r_j$ in every nonempty open subinterval of $[0,1]$. Thus a Riemann integrable function can have a discontinuity in every open subinterval: dense jumps do not prevent Riemann integrability.

Take an irrational number $a\in (0,1)$. The density of the rational numbers provides a sequence in $\mathbb{Q}\cap [0,1]$ converging to $a$, which is outside that subset. Thus the subset is not closed in $\mathbb{R}$. By the Heine-Borel theorem, it is not compact, despite being bounded.

Write $C^{\prime }=C\setminus \{(1,1)\}$ and $V=\{1\}\times [0,1)$. Let $L$ and $R$ be the portions of $C^{\prime }$ with first coordinate respectively less than and greater than one. Each is path-connected: move along a rational tooth to its corresponding open half of the top line, then horizontally there. By the density of the rational numbers, every point of $V$ is a limit point of both $L$ and $R$.

Use the elementary fact that adjoining any collection of limit points to a connected set preserves connectedness. Indeed, in a proposed separation, the connected subset lies on one side; relative openness would prevent any point on the other side from being a limit point of it. Therefore $L\cup V$ and $R\cup V$ are connected. They intersect in $V$, so their union $C^{\prime }$ is connected.

There is, however, no path from $(1,0)$ to $(0,1)$. If $\gamma (t)=(x(t),y(t))$ were one, let $t_0$ be the first time at which $y(t_0)=1$. Such a first time exists by continuity and compactness of $[0,1]$, and $t_0>0$. For $0\le t<t_0$, membership in the comb forces $x(t)\in \mathbb{Q}$. The continuous image of a connected space is connected, and a connected subset of $\mathbb{Q}$ is a singleton: an irrational number between any two distinct rational values would separate it. Hence $x(t)=1$ for $t<t_0$. By continuity, $\gamma (t_0)=(1,1)$, which was deleted.

True. Choose the irrational number $\alpha =\sqrt{2}$, and partition the rational numbers intoNo endpoint is rational, so these disjoint sets cover $\mathbb{Q}$. Each is nonempty by the density of the rational numbers, and each is clopen in the subspace topology on $\mathbb{Q}$.

Let two rational numbers be equivalent exactly when they belong to the same $C_n$. The bijectionis well-defined. For every subset $B\subseteq \mathbb{Z}$, the preimage of $h^{-1}(B)$ under the canonical projection of a quotient set is ${\bigcup }_{n\in B}{C}_n$, which is open in $\mathbb{Q}$. Thus every subset of $\mathbb{Q}/\sim$ is open: its quotient topology is the discrete topology. The usual subspace topology on the integers is also the discrete topology, because $\{n\}=\mathbb{Z}\cap (n-1/2,n+1/2)$. Consequently $h$ is a homeomorphism.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.

The displayed subspace of the Euclidean plane is path-connected: each vertical rational tooth joins the horizontal line at height one. Delete $(r,1)$ for rational $r$. The residual tooth $V=\{r\}\times [0,1)$ cannot be joined by a path to the top line. Before a path first reaches height one, its first coordinate is a continuous rational-valued function on an interval and must therefore be constant; it would reach the deleted point.

The punctured space remains connected. Its portions $L,R$ on either side of $x=r$ are path-connected, and $V\subseteq \overline{L}\cap \overline{R}$ by the density of the rational numbers. Adjoining limit points to a connected set preserves connectedness: a separation would place the connected subset on one side and prevent the other side from containing its limit points. Hence $L\cup V$ and $R\cup V$ are connected and intersect in $V$. Their union is connected. This example distinguishes connectedness from path-connected using the loss of a single attachment point.