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An adherent point of $A$ is a point whose every neighbourhood meets $A$. The set of adherent points is the closure of $A$. This includes all points of $A$, even isolated ones; a limit point in the accumulation-point convention instead requires each neighbourhood to meet $A\setminus \{x\}$.

A subset $F$ of a topological space $(X,\tau )$ is closed when its complement in that same ambient space is open: $X\setminus F\in \tau$. For example, the closed interval $[0,1]$ is closed in the usual real line because its complement is the open set $(-\infty ,0)\cup (1,\infty )$. Always specify the ambient topological space: a subset can be closed in a subspace topology without being closed in the surrounding space. ^[ref]

Being closed is not the logical negation of being open. In the usual real line, all four combinations occur: $(0,1)$ is open but not closed; $[0,1]$ is closed but not open; the half-open interval $[0,1)$ is neither open nor closed; and both the empty set and the whole real line are open and closed, hence clopen. Nor does being closed imply being a bounded set: $[0,\infty )$ is closed because its complement $(-\infty ,0)$ is open, but it is not a bounded set. The symbol $\infty$ is not an extra point that this interval must contain.

The coloured strokes mark membership; filled endpoint circles mean included points, and hollow endpoint circles mean excluded points. The empty set has no coloured points, while arrows at both ends of the coloured real line indicate that membership continues indefinitely. These examples use the usual topology of the real line.

In a metric space $X$, a subset $F$ is closed exactly when every convergent sequence $(x_n)$ in $F$ whose limit of a sequence $x$ lies in $X$ has $x\in F$. For example, $x_n=1-1/n$ lies in $[0,1)$ for $n\ge 1$ and has limit of a sequence $1$, which is missing; thus this half-open interval is not closed in the real line. This criterion is useful for keeping limiting constructions inside a prescribed set. ^[ref]

In every topological space, the general criterion is $F=\overline{F}$, using closure; equivalently, $F$ contains all its limit points. A closed set is always a sequentially closed set, but the converse can fail outside metric spaces. Arbitrary intersections and finite unions of closed sets are closed, by taking complements in the topology axioms.

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.

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.

At a zero $a$, Taylor's theorem gives $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$ near $a$. If some coefficient is nonzero, let $m$ be the least such index. Then $f(z)=(z-a{)}^{m}q(z)$ with $q(a)\ne 0$, and continuity makes $q$ nonzero on a smaller disc. Thus $a$ is an isolated zero. A limit point of zeros is also a zero by continuity and cannot be isolated, so all Taylor coefficients there vanish and $f$ vanishes on a nonempty open disc.

Let $A$ be the set of points having a neighborhood on which $f$ vanishes. It is nonempty and open. A point in its relative closure either already belongs to $A$ or is a limit point of distinct zeros; the preceding argument puts it in $A$ in either case. Thus $A$ is also relatively closed. Connectedness of $D$ gives $A=D$, proving $f$ is identically zero.

For conjugate reflection, the disc is invariant under conjugation, andHence $h$ is holomorphic, with derivative $\overline{g^{\prime }(\bar{z})}$. If $g$ is real on the real diameter, then for real $x\in (-1,1)$, $h(x)=\overline{g(x)}=g(x)$. The holomorphic difference $g-h$ therefore has zeros accumulating at zero, and the identity theorem just proved gives $g=h$ on the entire disc.

Interpret boundedness above uniformly: there is an $M<\infty$ such that $B^k⪯MI$ for every $k$. Since $B^k$ is a positive-definite symmetric matrix, the spectral theorem implies $(B^k{)}^{-1}⪰M^{-1}I$. ThereforeCombining the Armijo condition with ${\alpha }^k\ge C>0$ yieldsIf $f\ge L$, a telescoping sum givesThe nonnegative series converges, so its terms tend to zero:This proves first-order global convergence in the sense of stationarity from uniformly bounded quasi-Newton matrices. It does not imply that every nonconvex objective function reaches a global minimum, or that the iterates themselves converge. Every finite limit point is a stationary point, by continuity of the gradient.

The convention in this question counts a point whenever every open neighbourhood meets $A$, without excluding the point itself. These are adherent points, whose set is the closure $\overline{A}$; they differ from limit points under the convention that requires a different point of $A$ in each neighbourhood.

If $A$ is a closed set and $x\notin A$, the open set $X\setminus A$ is a neighbourhood of $x$ disjoint from $A$. Thus every adherent point lies in $A$. Conversely, if every adherent point lies in $A$, each $x\notin A$ has an open neighbourhood disjoint from $A$. Their union is $X\setminus A$, which is open. Hence $A$ is closed exactly when it contains all the points specified in the question.

The interior of a set and closure areThe neighbourhood description of the closure follows because failure to be adherent is equivalent to lying in an open set disjoint from $A$.

For closure of a connected set, suppose $\overline{A}=U\bigsqcup V$ were a separation into two nonempty relatively open sets. Their intersections with the connected subset $A$ would separate $A$ unless one intersection were empty. Say $A\subseteq U$. Choose $v\in V$ and an open set $O\subseteq X$ with $v\in O$ and $O\cap \overline{A}\subseteq V$. Since $v\in \overline{A}$, the set $O$ meets $A$, contradicting $A\subseteq U$. Thus the closure of a connected set is connected. The empty-set case holds directly.

A point is in the closure $\overline{E}$ exactly when every neighbourhood meets $E$. This is equivalent to belonging to $E$ or being a limit point of $E$.

If $p\notin \overline{E}$, choose $\epsilon >0$ with $(p-\epsilon ,p+\epsilon )\cap E=\varnothing$. Every $q$ with $|q-p|<\epsilon /2$ has a smaller neighbourhood contained in this interval and disjoint from $E$. Thus $(p-\epsilon /2,p+\epsilon /2)$ misses $\overline{E}$, and the complement of $\overline{E}$ is open. Hence $\overline{E}$ is closed.

Every closed set $F$ containing $E$ contains all limit points of $E$: a limit point outside $F$ would have a neighbourhood missing $F$, and hence missing $E$. Therefore $\overline{E}\subseteq F$. Since $\overline{E}$ is itself one of these closed sets,

Suppose $E$ is a closed set and $p\notin E$. Then $p$ is not a limit point of $E$, so some punctured neighbourhood of $p$ misses $E$. As $p\notin E$, the entire neighbourhood misses $E$. Thus $\mathbb{R}\setminus E$ is an open set.

Conversely, suppose every point $p\notin E$ has a neighbourhood disjoint from $E$. No such point can be a limit point of $E$, so every limit point lies in $E$. This proves the equivalence between containing all limit points and having an complement that is an open set.

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 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.

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.

The identity theorem states: if $V\subseteq \mathbb{C}$ is a nonempty connected open set and $p,q$ are holomorphic functions on $V$, and their equality set has a limit point in $V$, thenThe interior location of the limit point and connectedness are essential hypotheses.

For the proof put $h=p-q$. At any $z_0\in V$, its convergent Taylor series either has every coefficient zero, making $h$ zero on a neighbourhood, or has a first nonzero coefficient, of degree $m\ge 0$. In the latter case,where $k$ is holomorphic near $z_0$. By continuity, $k$ is nonzero on a smaller disc. Thus any zero at $z_0$ is isolated unless $h$ vanishes on a neighbourhood. An accumulation of distinct zeros at the assumed interior limit point therefore forces the neighbourhood alternative.

Let $E$ be the set of points of $V$ with a neighbourhood on which $h=0$. We have just proved $E\ne \varnothing$, and $E$ is an open set. It is also a closed set relative to $V$: a point of its closure outside $E$ would be approached by distinct points of $E$, hence by zeros, and the same Taylor series argument would place it in $E$. Since $V$ is connected, $E=V$. Thus $h=0$ on $V$, proving the theorem.

A metric space is sequentially compact if every sequence has a convergent subsequence whose limit belongs to the space. A limit point $\ell$ of $S$ is a point for which every open ball about $\ell$ contains a point of $S$ other than $\ell$.

Let $(x_n)$ lie in a compact metric space. If one value occurs infinitely often, take a constant subsequence. Otherwise its set of values $S$ is infinite. If $S$ had no limit point, every point of the space would have a neighborhood containing at most one member of $S$. A finite subcover of these neighborhoods would force $S$ finite, a contradiction. Thus $S$ has a limit point; the stated subsequence result supplies a convergent subsequence. This proves sequential compactness.

Assume now that $X$ is nonempty, as is necessary for the requested minimum and fixed point. The function $r(x)=d(f(x),x)$ is continuous, as a composition of $x\mapsto (f(x),x)$ with the metric. The extreme value theorem on a compact space implies that $r$ attains its infimum, say at $p$. If $f(p)\ne p$, strict decrease of distances givescontradicting minimality. Thus $f(p)=p$. If $p,q$ were distinct fixed points, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Hence there is exactly one fixed point. This is the fixed point theorem for a strictly distance-decreasing map on a compact metric space.

Compactness is essential. On $\mathbb{R}$, takeHere $f(x)-x=\log (1+e^{-x})>0$, so there is no fixed point. Nevertheless $0<f^{\prime }(x)=e^x/(1+e^x)<1$ everywhere, and the mean value theorem gives $|f(x)-f(y)|<|x-y|$ for distinct $x,y$. There is no uniform Lipschitz constant less than $1$, since $f^{\prime }(x)\to 1$ as $x\to \infty$.

A sequentially compact space is one in which every sequence has a subsequence converging to a point of the space. A limit point $\ell$ of $S$ means every open ball around $\ell$ meets $S\setminus \{\ell \}$.

In a compact metric space, a sequence with a finite set of values has a constant subsequence. If its value set is infinite and had no limit point, each point of the space would have a neighbourhood containing at most one point of that value set. A finite subcover would then make the value set finite, a contradiction. It therefore has a limit point in the space, and the permitted subsequence result gives a convergent subsequence. This proves sequential compactness.

For the remaining assertions assume the compact space is nonempty, as required for a fixed-point claim. The displacement function $q(x)=d(f(x),x)$ is continuous by the stated continuity of the metric and of $f$. The extreme value theorem gives a minimum $\delta =q(p)$. If $\delta >0$, then $p\ne f(p)$, and strict decrease of distances impliescontradicting minimality. Thus $\delta =0$ and $p$ is a fixed point. Two distinct fixed points would satisfy $d(p,q)=d(f(p),f(q))<d(p,q)$, also impossible. Hence the fixed point is unique.

On the noncompact real line, takeIts derivative $e^x/(1+e^x)$ is strictly between zero and one. The mean value theorem therefore gives $|f(x)-f(y)|<|x-y|$ for each distinct pair. Yet $f(x)-x=\log (1+e^{-x})>0$, so it has no fixed point. Compactness, rather than merely pointwise strict shrinkage, supplies attainment of a zero displacement in the preceding proof.

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.

If $X\subseteq \mathbb{R}$ is not compact, there is an unbounded continuous function $X\to \mathbb{R}$. By the Heine-Borel theorem, either $X$ is unbounded, when $f(x)=x$ works, or it is not closed. In the latter case choose $a\in \overline{X}\setminus X$; the continuous function $f(x)=|x-a{|}^{-1}$ is unbounded because points of $X$ approach the missing limit point $a$. This proof uses the specific compactness characterization of real subspaces; boundedness of a general metric space does not imply compactness.