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The closure of a subset $A$ of a topological space $X$ is the smallest closed set containing $A$. Equivalently, it is the intersection of all closed sets containing $A$, or the set of points whose every neighbourhood in $X$ meets $A$. The notation ${\mathrm{cl}}_{X}A$ makes the ambient topological space explicit. This definition works without a metric or any appeal to sequences. ^[ref]

For $A=[0,1)$ in the usual real line, $\overline{A}=[0,1]$: every neighbourhood of $1$ meets $A$, while a point outside $[0,1]$ has an open ball disjoint from $A$. Passing to the closure adds the missing endpoint and retains all original points, including isolated points when present. In general, the closure is $A$ together with its limit points; its individual points are the adherent points of $A$.

The closure turns an arbitrary subset into a closed set with no unnecessary added points. It satisfies $A\subseteq \overline{A}$ and $\overline{\overline{A}}=\overline{A}$; a subset is closed exactly when $A=\overline{A}$. In a metric space, $x\in \overline{A}$ exactly when some sequence of points of $A$ has limit of a sequence $x$: choose its $n$th point in $A\cap B(x,1/n)$. For a general topological space, keep the neighbourhood definition; sequences need not detect every point of the closure.

Enumerate all intervals with rational endpoints, using the countability of the rational numbers and listing pairs of enumeration indices by increasing sum. If every point $b$ of $B$ is an isolated point, some such interval contains $b$ and no other point of $B$. Assign to $b$ the least index of such an interval. Distinct points cannot have the same index, so this gives an injective function into $\mathbb{N}$ and proves that $B$ is a countable set. The argument uses isolation in the subspace topology, not closedness: the discrete subset $\{1/n:n\ge 1\}$ illustrates that an accumulation point outside $B$ is allowed.

Choose a strictly decreasing positive sequence $q_0=1,q_1,q_2,\dots$. Begin with $C_0=[0,1]$ and at step $n+1$ keep two closed end intervals of length $q_{n+1}/2^{n+1}$ from every component interval of $C_n$. This constructs a compact set $C={\bigcap }_n{C}_n$ withThe second equality is continuity from above of a measure. Every stage has positive gaps and maximum component length at most $2^{-n}$, so $C$ contains no nondegenerate interval and is a nowhere dense set. It has no isolated points: both endpoints of the stage-$n$ component containing a point remain in $C$, and at least one distinct endpoint is within $2^{-n}$. Thus its topological sparseness is compatible with positive Lebesgue measure.

Taking $q_n=(2/3{)}^n$ gives the usual Cantor set of measure zero. Taking $q_n=1/2+2^{-(n+1)}$ gives a generalized construction of measure $1/2$. This is one useful family of generalized constructions; the removed intervals need not have the same proportions at every stage.

Every nonempty countable complete metric space has an isolated point. Otherwise it is the countable union of its singleton subsets, each of which is nowhere dense, contradicting the Baire category theorem.

If such a space has infinitely many points, it has infinitely many isolated points. Indeed, after deleting any finite collection of isolated points, the remaining set is closed, complete, countable, and nonempty. It therefore has a point isolated in the remaining space; because the deleted set is finite, that point is also isolated in the original space.

A limit point of a subset $A$ of a topological space is a point $x$ whose every neighbourhood meets $A\setminus \{x\}$. Thus $1$ is a limit point of $[0,1)$ in the real line, even though $1\notin [0,1)$. In contrast, an isolated point of $A$ belongs to its closure but is not a limit point of $A$. This distinction separates limit points from the more inclusive notion of an adherent point.

The convention intended by this question is existence of a dense forward orbit, also called point transitivity. On a nonempty compact metric space, this means that some $x\in X$ satisfiesEquivalently, that forward orbit visits every nonempty open set.

The other common definition of topological transitivity uses intersections of iterated open sets, as in part (b). The distinction matters on spaces with isolated points: parts (b) and (c) ask precisely how the two conventions are related. We use point transitivity for the dense-orbit convention throughout this solution. As usual for this definition, the state space is assumed nonempty; on an empty space the open-set condition is vacuous and there is no point with a forward orbit.

Let $R$ be a nonconstant rational map of the Riemann sphere, and write $R^n$ for its $n$-fold iterate, with $R^0$ the identity. The appropriate notion of convergence is uniform convergence on compact subsets in the chordal metric, which treats poles as ordinary values at infinity. A family of meromorphic maps is a normal family if every sequence in it has a subsequence converging locally uniformly in that metric; the limit is meromorphic or constant, possibly the constant infinity.

The Fatou set $F(R)$ consists of points with a neighborhood on which the iterate family is normal. The complementary Julia set is $J(R)=\hat{\mathbb{C}}\setminus F(R)$. Thus the sphere is the disjoint union of an open stability region $F(R)$ and a closed instability region $J(R)$. These definitions concern all subsequences of iterates, rather than assuming that the iterates themselves converge on every stable region.

Equicontinuity makes this distinction precise. On a compact subset $K$ of a neighborhood, the family is equicontinuous if, for every $ϵ>0$, some $\delta >0$ makes $\chi (R^n(z),R^n(w))<ϵ$ whenever $z,w\in K$ and $\chi (z,w)<\delta$, uniformly in $n$. The Arzelà-Ascoli theorem says that equicontinuity of continuous maps from a compact domain to a compact metric target gives relative compactness for uniform convergence. Apply it to a compact exhaustion and take a diagonal subsequence to obtain local normality. Locally uniform spherical limits of meromorphic maps are again meromorphic or constant, by using finite target charts and the locally uniform limit theorem for holomorphic functions. Conversely, if equicontinuity failed, there would be maps $f_j$ and points $z_j,w_j\to z$ with $\chi (f_j(z_j),f_j(w_j))$ bounded below by a positive number. A subsequence converging uniformly on a compact neighborhood contradicts this. Hence local normality and local equicontinuity are equivalent here.

Both sets are completely invariant:For the inverse direction, if the family is normal near $R(z)$, composition with $R$ makes $R^{n+1}$ normal near $z$, and adding the identity does not affect normality. For the forward direction, suppose the iterates are equicontinuous near $z$. A nonconstant holomorphic sphere map is open. Thus each sufficiently small neighborhood of $z$ maps onto a neighborhood of $R(z)$; every nearby target point $w$ has a preimage $x$ as close to $z$ as desired. Thenis uniformly small. Applying this at each point of the image neighborhood proves equicontinuity there. This argument works at rational critical points and requires no single-valued inverse branch. Surjectivity of $R$ supplies the asserted image equalities. Also $F(R^k)=F(R)$ and $J(R^k)=J(R)$ for $k\ge 1$: one direction follows by taking a subfamily, and the other by splitting $R^n$ into the finitely many residue classes $R^r\circ (R^k{)}^m$, $0\le r<k$.

For degree $d\ge 2$, the Julia set is nonempty. Otherwise the family would be equicontinuous on the whole compact sphere, giving a subsequence $R^{n_j}$ converging uniformly to a continuous sphere map $G$. Maps sufficiently close in chordal distance are homotopic, using the unique short spherical geodesic between their values. They therefore have equal topological degrees. This is impossible because $\mathrm{deg}{R}^{n_j}=d^{n_j}\to \infty$ whereas $G$ has one fixed degree.

It cannot be a nonempty finite set either. Complete invariance would make $R$ a permutation of that finite set, with just one sphere preimage for each of its points. That preimage would have local degree of a holomorphic map $d$. Every holomorphic cycle in this set would consequently be superattracting for its return map, hence lie in $F(R)$, a contradiction. In particular $J(R)$ contains at least three points.

The Montel theorem in the form used here says that a family of meromorphic maps on a domain, each omitting the same three distinct points of the sphere, is normal for spherical convergence. It also shows that $J(R)$ is perfect. If $a$ were isolated in $J(R)$, choose a disc $U$ with $U\cap J(R)=\{a\}$ and four distinct points of $J(R)$. For each $n$, complete invariance makes $R^n(U)\cap J(R)$ have at most the single point $R^n(a)$. Partition the iterates into finitely many families according to which of the four points, if any, that single value equals. Each family omits at least three fixed points, so each is normal by Montel; their finite union is normal. This would put $a$ in $F(R)$, a contradiction. Therefore the degree-at-least-two Julia set is compact, infinite and has no isolated points.

The same omitted-value argument explains its global reach. If an open $U$ meets $J(R)$, then ${\bigcup }_{n\ge 0}{R}^n(U)$ can omit at most two sphere points, since three omissions would imply normality on $U$. If $J(R)$ had an interior and were not the whole sphere, these image sets would remain in $J(R)$ and their complement would contain the nonempty open set $F(R)$, impossible. Thus a proper Julia set has empty interior and equals the boundary of $F(R)$. If the Julia set is the whole sphere, the Fatou set is empty instead.

An attracting periodic point and its holomorphic cycle has a contracting return map in a small disc, so its basin is in the Fatou set. Repelling periodic points belong to the Julia set: return iterates fix the point but have derivatives equal to unbounded powers of their periodic-orbit multiplier, contradicting derivative bounds for a normal subsequence near that fixed value. A standard further global property is that repelling periodic points are dense in the Julia set. Fatou components can also support convergence towards a parabolic cycle on the boundary, or rotation behavior in a Siegel disc or a Herman ring; normality need not mean attraction. For example, $R(z)=z^d$ has $J(R)=\{|z|=1\}$, with the interior converging to zero and the exterior to infinity. On the unit circle, arbitrarily close radii have incompatible limiting behavior, preventing normality there.

Degree one has a different picture. A rational map of degree one is a Möbius transformation. Up to Möbius conjugacy, an elliptic map is $z\mapsto az$ with $|a|=1$; its iterates are rotations and are equicontinuous on the sphere, so $F$ is the whole sphere and $J$ is empty (including the identity). For $z\mapsto az$ with $|a|>1$, all compact subsets away from zero converge to infinity, while $z_n=a^{-n}\to 0$ and $R^n(z_n)=1$ prevent equicontinuity at zero. Thus $J=\{0\}$; for $|a|<1$ the reciprocal coordinate gives $J=\{\infty \}$. A nonidentity parabolic map is conjugate to $z\mapsto z+1$. Its iterates converge locally uniformly to infinity on the finite plane, but $z_n=-n\to \infty$ has $R^n(z_n)=0$, so $J=\{\infty \}$. Holomorphic conjugacy carries these sets to the corresponding fixed points of the original map. These finite Julia sets explain why the degree restriction in many global Julia-set properties is essential.

A set is countable if it admits an injective function into $\mathbb{N}$, equivalently if it is finite or can be listed in a sequence. The empty set is included; an infinite countable set has a bijection with $\mathbb{N}$.

Every rational number has a representation $p/q$ with $p\in \mathbb{Z}$ and $q\ge 1$. For each $k\ge 1$, there are only finitely many such pairs with $|p|+q=k$. List these finite levels successively, ordering each level by $p$, and discard repeated fractions when they first recur. Every rational number appears in some finite level, proving $\boxed{\mathbb{Q}\text{ is countable}}$.

For the uncountability proof, suppose all numbers in $(0,1)$ could be listed as $r_1,r_2,\dots$. Choose each decimal expansion not to end in an infinite string of nines, and denote its $j$th digit by $d_{ij}$. Set $b_i=1$ when $d_{ii}\ne 1$, and $b_i=2$ otherwise. Then the decimaldefines a number in $(0,1)$ with a unique such expansion, since its digits are only $1$ and $2$. It differs from $r_i$ at digit $i$ for every $i$, contradicting the enumeration. This Cantor diagonal argument proves $\boxed{\mathbb{R}\text{ is uncountable}}$ because its subset $(0,1)$ already is.

For the union of two countable sets $C,D$, take injective functions ${\iota }_C:C\to \mathbb{N}$ and ${\iota }_D:D\to \mathbb{N}$. The mapis injective: it is injective within each branch, and their images have different parity. Thus $\boxed{C\cup D\text{ is countable}}$, regardless of overlap.

The property of $A$ says that both $A$ and its complement are dense subsets of the real line. Both possible cardinalities occur. To establish the needed density directly, given $x\in \mathbb{R}$ and $\epsilon >0$, choose an integer $q>1/\epsilon$. Then $\lfloor qx\rfloor /q$ is rational and differs from $x$ by less than $1/q<\epsilon$. For an irrational approximation, choose a rational $r$ within $\epsilon /2$ of $x$ and a positive integer $q$ large enough that $\sqrt{2}/q<\epsilon /2$. The number $r+\sqrt{2}/q$ is irrational and lies within $\epsilon$ of $x$; irrationality follows from that of $\sqrt{2}$, since otherwise subtracting $r$ and multiplying by $q$ would make $\sqrt{2}$ rational. To recall the latter fact, a reduced fraction $p/q=\sqrt{2}$ would satisfy $p^2=2q^2$, forcing $p$ even and then $q$ even, contrary to coprimality. ThereforeThe second set is uncountable because, if it were countable, its union with $\mathbb{Q}$ would make $\mathbb{R}$ countable. The same two density arguments give the required approximations from $A$ and from its complement in either example.

Finally, the condition on $B$ makes every one of its points an isolated point, so $B$ is a discrete subset of the real line. Enumerate all rational intervals $(p_j,q_j)$ with $p_j<q_j$; enumerate pairs of indices of the rational list by increasing sum and keep those whose endpoints have the required order. For each $b\in B$, its isolating neighbourhood contains a rational interval $(p_j,q_j)$ containing $b$ and no other point of $B$. Assign to $b$ the least such index $j$. Two distinct points cannot receive the same index, because that interval would then contain two members of $B$. This gives an injective function $B\to \mathbb{N}$ and proves that discrete subsets of the real line are countable:Using the least index makes the assignment explicit and requires no arbitrary selection of an uncountable family of intervals.

First construct the spectral measure of a normal operator and prove the open-set assertion needed for all three numbered conclusions. The algebra $A=C^{\ast }(I,T)$ generated by $T$ and $T^{\ast }$ is commutative because $T$ is a normal operator. The Gelfand transform identifies $A$ with $C({\Phi }_A)$. By spectral permanence for C-star algebras,is onto: its image is the spectrum of $T$ in $A$, which is also its spectrum in $\mathcal{B}(H)$. It is injective because a character is determined by its values on $T,T^{\ast }$ and $I$, and $\chi (T^{\ast })=\overline{\chi (T)}$. Thus $h$ is a homeomorphism. Apply part (a) and setThis is a projection-valued measure with $P(K)=I$ andThe associated continuous functional calculus is isometric: $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }$ for $u\in C(K)$.

If $U\subseteq K$ is nonempty and relatively open, choose $z_0\in U$ and a nonzero continuous $u$ supported inside $U$, for example a small continuous bump about $z_0$ in the metric of $K\subseteq \mathbb{C}$. If $P(U)=0$, then $u(T)={\int }_{K}udP=0$, contradicting $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }>0$. ThereforeThis proves the claim from faithfulness of the representation rather than assuming it as part of the spectral theorem.

Now take an isolated point $\lambda \in K$. Its singleton is relatively open, so $P(\{\lambda \})\ne 0$. Multiplication of spectral integrals givesThus its range is contained in the eigenspace $\ker (T-\lambda I)$. Conversely, for $v$ in that kernel, the positive scalar spectral measure ${\mu }_v(B)=⟨P(B)v,v⟩$ satisfiesIt follows that ${\mu }_v(K\setminus \{\lambda \})=0$, and hence $\Vert (I-P(\{\lambda \}))v{\Vert }^2=0$. ConsequentlyThe identification of this projection with the eigenspace holds even for a non-isolated $\lambda$; isolation guarantees that it is nonzero.

False. The rational numbers with the Euclidean metric are countable and have no isolated point.

The Baire category theorem says that a complete metric space cannot be a countable union of nowhere-dense closed sets. Equivalently, every countable intersection of open dense sets is dense.

A point $x$ of a metric space is an isolated point when some $\epsilon >0$ satisfies $B(x,\epsilon )=\{x\}$.

The limit of a function $f:E\to \mathbb{R}$ at a limit point $p$ of $E$ is $\ell$ whenThe function converges as $x\to p$ if such a finite $\ell$ exists. The puncture excludes any value assigned at $p$.

For $p\in E$, continuity at $p$ meansWhen $p$ is also a limit point, this says that the limit of a function is $f(p)$. At an isolated point of $E$, every function is continuous relative to $E$, even though the punctured limit does not have a unique forced value.

For two isolated point masses in a Newtonian bound orbit, $\boxed{P^2=4{\pi }^2{a}^3/[G(M_s+M_p)]}$, where $a$ is the relative orbit's semimajor axis. For a negligible planet mass, using years, astronomical units and solar masses gives $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/(M_s/M_{\odot })$. The mass in the denominator is the total gravitating mass, not the planet's mass alone.

The only isolated point of the subset $[0,1]$ is $1$. Indeed,so $\{1\}$ is open in the subspace topology. If $0\le x<1$, every basic neighbourhood $[a,b)$ containing $x$ also contains a point strictly between $x$ and $\min (b,1)$, hence another point of $[0,1]$. Such an $x$ is not an isolated point of the subset. Here isolation refers to $[0,1]$ with its subspace topology; $1$ is not an isolated point of the ambient Sorgenfrey line.

False, with “disjoint union” understood as a topological disjoint union. Take the rational numbers $\mathbb{Q}$ with the usual subspace topology from the real line.

Every connected subset of $\mathbb{Q}$ has at most one point. Indeed, if it contained $r<s$, choose an irrational number $t\in (r,s)$. The two relatively open setspartition $\mathbb{Q}$ and separate any subset containing both $r$ and $s$. Hence the connected components of $\mathbb{Q}$ are singletons.

If a topological disjoint union of nonempty connected spaces were homeomorphic to $\mathbb{Q}$, the image of each summand would be connected, hence a singleton. Every summand would therefore be a one-point topological space. Their topological disjoint union is a discrete space, since each summand is open. But $\mathbb{Q}$ is not a discrete space: for each $r\in \mathbb{Q}$, the distinct rational numbers $r+1/n$ approach $r$, so $r$ is not an isolated point. This is a contradiction. The set decomposition into connected components always exists; giving those pieces the topological disjoint union topology need not recover the original topology.

Let $o=(0,0)$ be the common point of the two circles. For sufficiently small $\epsilon >0$, the open neighbourhoodhas the following local structure: $W\setminus \{o\}$ consists of four disjoint open arcs, an upper and a lower branch from each circle. These are its four connected components, each open and closed relative to $W\setminus \{o\}$.

Suppose a coordinate chart of some dimension of a manifold $n$ existed at $o$. The case $n=0$ would make $o$ an isolated point, which it is not. For $n\ge 1$, shrink the chart inside $W$ so that its image is an open ball centred at the image of $o$. Its domain of a function $V$ is an open neighbourhood of $o$, so it meets all four arms of $W$. Intersecting those arms with $V\setminus \{o\}$ gives four disjoint nonempty clopen sets partitioning $V\setminus \{o\}$; hence there are at least four connected components.

But a punctured ball in $\mathbb{R}$ has exactly two connected components, and a punctured ball in ${\mathbb{R}}^n$ for $n\ge 2$ is path-connected. A homeomorphism preserves connected components, giving a contradiction in every dimension of a manifold. Thus $X$ is not a topological manifold. The argument uses arbitrarily small coordinate charts; the fact that the whole $X\setminus \{o\}$ has two pieces alone would not be a local obstruction.

The answer is no, because the number of non-isolated points differs. In $X$, every $1/n$ is an isolated point: a sufficiently small open interval around it contains no other member of $X$. The only non-isolated point is $0$, since every neighbourhood of $0$ contains some $1/n$.

In the product topology, $(x,y)\in X\times X$ is an isolated point exactly when both coordinates are isolated points of $X$. One direction follows because the product of two open singletons is open. For the other, any basic open rectangle containing $(x,y)$ contains another point if either coordinate has no open singleton. Hence the set of non-isolated points of $X\times X$ iswhich is infinite. A homeomorphism takes open singletons to open singletons, so it induces a bijection between the sets of non-isolated points. A space with just one such point cannot be homeomorphic to a space with infinitely many.

A subset of the plane can have three successive strict changes under taking its set of accumulation points. For example letPut $B=\{(1/n+1/(n^{2}m),0):n,m\ge 1\}$ and $C=\{(1/n,0):n\ge 1\}$. Then $A^{\prime }=B\cup C\cup \{0\}$, $A^{\prime \prime }=C\cup \{0\}$ and $A^{\prime \prime \prime }=\{0\}$, where $0$ denotes the origin. For fixed $n,m$, increasing $k$ gives a point of $B$; increasing $m$ gives a point of $C$; increasing $n$ gives the origin. These are all possible limits: in any convergent sequence of distinct points, either $n$ is unbounded, or a subsequence fixes $n$ and has unbounded $m$, or a subsequence fixes both and has unbounded $k$. The first-coordinate cluster intervals $[1/n,1/n+1/n^2]$ are disjoint and separated away from the origin. Repeating this classification after each differentiation proves the displayed derived sets. Each derivative removes one layer of isolated points, and all three displayed changes are strict.

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.