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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.

For a uniform contraction from a comparison function on a bounded set, the closed set $M$ is complete and has diameter at most $2R$. Suppose first $R>0$. The limit condition supplies $r_0>0$ and $q_0<1$ such that $\psi (r)/r\le q_0$ for $0<r\le r_0$. On the remaining compact interval $[r_0,2R]$, if nonempty, the continuous ratio $\psi (r)/r$ attains a maximum $q_1<1$, since it is strictly below one at every point. Thus there is one $q<1$ with $\psi (r)\le qr$ for every distance $0\le r\le 2R$ arising in $M$.

Consequently $d(Tx,Ty)\le qd(x,y)$ throughout $M$. Banach's theorem applies and gives a unique fixed point in $M$. No linear structure or compactness of the ball is used; the compact object in the argument is only the scalar distance interval. If the closed ball is the singleton at radius zero, its self-map fixes that point. The estimate on the ratio near zero is what prevents the compact-interval supremum argument from approaching one at its missing endpoint.

The unit circle is the inverse image of the closed set $\{1\}$ under the continuous map $(x,y)\mapsto x^2+y^2$, so it is closed. It is also bounded, since every point has Euclidean norm one. The Heine-Borel theorem gives compactness.

The answers, in the order completeness, sequential compactness, connectedness, and path connectedness, areDefine $F(0)=0$ and $F(x)=\sqrt{x}\sin (1/x)$ for $x>0$. The estimate $|F(x)|\le \sqrt{x}$ proves continuity at zero by the squeeze theorem; elsewhere it follows from the usual continuous functions. Thus $X$ is the graph of a function $F$ on the closed set $[0,\infty )$.

To check closedness directly, if $(x_n,F(x_n))\to (x,y)$ in the Euclidean plane, then $x\ge 0$ and continuity gives $y=F(x)$. Therefore $X$ is a closed subset of the complete metric space ${\mathbb{R}}^2$, hence is a complete metric space. The sequence $(n,F(n))$ has no convergent subsequence, since its first Cartesian coordinates tends to infinity, so sequential compactness fails.

For any two points $(u,F(u)),(v,F(v))$, the mapis a continuous path in $X$ joining them. Thus $X$ is path-connected, and consequently connected. The factor $\sqrt{x}$ is crucial: the oscillations shrink to the single point $(0,0)$.

Assume condition (2). Fix $a\in X$ and $\epsilon >0$. The superlevel setis a closed set and does not contain $a$. Its complement is therefore an open neighbourhood of $a$, so some open ball $B(a,r)$ avoids $F$. Consequently $f(x)<f(a)+\epsilon$ throughout that ball, givingThe same bound holds for every smaller radius. Taking $r\downarrow 0$ and then $\epsilon \downarrow 0$ yieldsTogether with the previous implication this proves the equivalence. A supremum may equal $f(a)+\epsilon$ even when every individual value is strictly smaller; the non-strict estimate here handles that possibility.

First suppose $F$ is a closed set. If $x\notin F$, the definition in part (ii) gives an open ball $B(x,r)$ disjoint from $F$. Thus every $y\in F$ has $d(x,y)\ge r$, and the distance to a set satisfies $\mathrm{dist}(x,F)\ge r>0$. Its contrapositive is the required implication.

Conversely, suppose that zero distance to a set forces membership. For $x\notin F$ we have $\mathrm{dist}(x,F)>0$. Choose $0<r<\mathrm{dist}(x,F)$. Then $B(x,r)\cap F=\varnothing$, proving $X\setminus F$ is an open set. HenceFor $F=\varnothing$, use the extended-real convention $\inf \varnothing =+\infty$: both assertions hold, with the implication vacuous. Part (b) only uses nonempty sets, so all its distances to a set are ordinary finite real numbers.

A subset $F\subseteq X$ is a closed set in $X$ when its complement is an open set in the metric topology. Equivalently,The open ball is taken inside the ambient metric space $X$; closedness always depends on that ambient space.

Let $(F_n)$ be a Cauchy sequence for the Hausdorff distance, and putThe bounded set hypothesis makes ${\delta }_n$ finite, and the Cauchy sequence property gives ${\delta }_n\to 0$. Each $A_n$ is nonempty and a closed set, with $A_{n+1}\subseteq A_n$.

We need estimates that survive taking the closure. For every nonempty $F$, the distance to a set is Lipschitz continuous with constant one: the triangle inequality gives $\mathrm{dist}(x,F)\le d(x,y)+\mathrm{dist}(y,F)$, and interchanging $x,y$ provesFor every $m\ge n$, points of $F_m$ have distance at most ${\delta }_n$ from $F_n$. The same holds throughout their union and its closure, by this continuity. Since $F_n\subseteq A_n$,More generally, if $m\ge n$, every point in ${\bigcup }_{k\ge n}{F}_k$ is within ${\delta }_n$ of $F_m\subseteq A_m$. Passing to its closure again, and using $A_m\subseteq A_n$, givesChoose strictly increasing indices $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. Then $(A_{n_j})$ satisfies part (ii), so it converges in Hausdorff distance to the nonempty closed set $G={\bigcap }_j{A}_{n_j}={\bigcap }_n{A}_n$, with $\rho (A_{n_j},G)\le 2^{1-j}$.

For $n\ge n_j$, the triangle inequality gives the explicit full-sequence boundIt tends to zero as $j\to \infty$. ThereforeThis proves Completeness of the Hausdorff hyperspace using the proposed tail closures, rather than assuming their diameters shrink.

The intersection $G$ is a closed set because arbitrary intersections of closed sets are closed. Its nonemptiness needs a construction: bounded closed sets need not be compact sets in this setting.

Fix $n$, $x_n\in F_n$, and $\eta >0$. Recursively choose $x_{k+1}\in F_{k+1}$, for $k\ge n$, using the approximation property of an infimum so thatThis is possible because $x_k\in F_k$ and $\mathrm{dist}(x_k,F_{k+1})\le \rho (F_k,F_{k+1})\le 2^{-k}$. The two bounds form convergent geometric series, so the triangle inequality shows $(x_k{)}_{k\ge n}$ is a Cauchy sequence. By completeness it converges to some $x\in X$.

For each fixed $j$, all sufficiently late $x_k$ lie in $F_j$, by nesting. Closedness of $F_j$ gives $x\in F_j$, and hence $x\in G$. In particular, starting with any $x_1\in F_1$ proves $G\ne \varnothing$. Moreover,For each starting point $x_n$, let $\eta \downarrow 0$ in this bound on its distance to a set $G$. This does not assert that a nearest point exists. It proves $\mathrm{dist}(x_n,G)\le 2^{1-n}$. Since $G\subseteq F_n$, the other directed Hausdorff distance is zero, andThis is the Summably nested Hausdorff intersection theorem; the summable movement replaces any need for compactness.

Write $h(A,B)={\sup }_{a\in A}\mathrm{dist}(a,B)$, so that $\rho (A,B)=\max \{h(A,B),h(B,A)\}$ is the Hausdorff distance. Since $X$ is a bounded set, there is a finite $M$ with $d(x,y)\le M$. Nonemptiness of $A,B$ therefore gives $0\le h(A,B)\le M$, so $\rho$ is finite. Its symmetry and nonnegativity are immediate, as is $\rho (A,A)=0$.

If $\rho (A,B)=0$, then every $a\in A$ has zero distance to a set $B$. Since $B$ is a closed set, part (a)(iii) gives $A\subseteq B$. Interchanging the sets gives $B\subseteq A$. Thus $\rho (A,B)=0$ implies $A=B$. Closedness is essential here: without it a set and its closure have Hausdorff distance zero.

For the triangle inequality, fix $a\in A$ and $ϵ>0$. The approximation property of an infimum supplies $b\in B$ with $d(a,b)<\mathrm{dist}(a,B)+ϵ$, and then $c\in C$ with $d(b,c)<\mathrm{dist}(b,C)+ϵ$. Consequently,Let $ϵ\downarrow 0$ and take the supremum over $a$ to obtain $h(A,C)\le h(A,B)+h(B,C)$. Applying this inequality in the reverse direction as well yieldsAll metric axioms follow. Boundedness was used to ensure finiteness, and closedness to distinguish different sets; neither was needed for the triangle inequality.

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 a set $X$ with a function $d:X\times X\to [0,\infty )$ satisfying $d(x,y)=0⟺x=y$, symmetry and the triangle inequality. A subset $U$ is open if every $x\in U$ has some open open ball $B(x,r)$ contained in $U$.

For finitely many open sets $U_1,\dots ,U_k$ and $x$ in their intersection, choose radii $r_j>0$ with $B(x,r_j)\subseteq U_j$. The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is $X$, also open.

For distinct points put $\delta =d(x_1,x_2)>0$. The balls $B(x_1,\delta /3)$ and $B(x_2,\delta /3)$ are disjoint: a common point would make $d(x_1,x_2)<2\delta /3$ by the triangle inequality. This proves the Hausdorff condition.

A singleton $\{x\}$ is closed because each $y\ne x$ has a ball of radius $d(x,y)/2$ avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.

The map is continuous when the preimage $f^{-1}(U)$ is open in $X$ for every open set $U$ in $Y$. Equivalently the preimage of every closed set in $Y$ is closed in $X$, since preimages commute with complementation.

The family $\mathcal{F}$ of closed sets contains $\varnothing$ and $X$, is closed under arbitrary intersection, and is closed under finite union. Conversely these axioms define a topology by declaring the complements of the members of $\mathcal{F}$ to be the open sets.

Set $S=\{x_n:n\in \mathbb{N},x_n\ne \ell \}$. This is a countable set, hence a closed set in the stated topology, and $\ell \notin S$. Consequently $X\setminus S$ is an open neighbourhood of $\ell$. By convergence of a sequence, all sufficiently late terms lie outside $S$. A sequence term outside $S$ must equal $\ell$. Thus the sequence is eventually constant, and only finitely many indices have $x_n\ne \ell$.

Let $S=\{x:f(x)\ne f(a)\}$, which is countable. For any closed set $C\subseteq \mathbb{R}$, consider its preimage. If $f(a)\in C$, then $a\in f^{-1}(C)$ and that preimage is closed by definition. If $f(a)\notin C$, then $f^{-1}(C)\subseteq S$, so it is countable and again closed. Therefore every closed set has a closed preimage, proving continuity.

Both $\varnothing$ and $X$ satisfy the condition. For an arbitrary intersection of members, either every member contains $a$, in which case the intersection contains $a$, or one does not, in which case that member is countable and so is the intersection. The empty intersection is $X$.

For a finite union, either a member contains $a$, so the union does too, or all members omit $a$ and are countable, so their finite union is countable. Thus the closed set axioms hold and define a topology. This is a Fortissimo space: every point other than $a$ is isolated, while the open neighbourhoods of $a$ have countable complement.

A Hausdorff space separates any two distinct points by disjoint open neighborhoods. For a subspace $A\subseteq X$, intersect such neighborhoods in $X$ with $A$ to obtain disjoint neighborhoods in the subspace topology. Thus subspaces inherit the Hausdorff property.

If $(z_1,z_2)$ lies off the diagonal of a Hausdorff space's square, choose disjoint open neighborhoods $U$ of $z_1$ and $V$ of $z_2$. The open rectangle $U\times V$ contains the pair and avoids the diagonal. Therefore its complement is open and the diagonal is closed.

The map $F:X\times Y\to Z\times Z$ given by $F(x,y)=(f(x),g(y))$ is continuous by the coordinate criterion from E3. The equality locus is $F^{-1}({\Delta }_Z)$ and is closed as an inverse image of a closed set.

Let $q:X\to X/\sim$ be the quotient map. If the quotient is Hausdorff, apply the same argument to $q$ in both coordinates:is closed in $X\times X$.

Conversely suppose $R$ is closed and $q$ is open. Take two distinct quotient points, with representatives $x,y$. Since $(x,y)\notin R$, an open rectangle $U\times V$ around it avoids $R$. The images $q(U)$ and $q(V)$ are open quotient neighborhoods. They are disjoint, because a quotient point in their intersection would have equivalent representatives $u\in U$, $v\in V$, giving $(u,v)\in R$. Hence the quotient is Hausdorff. Thus closedness of the relation and openness of the quotient map suffice for a Hausdorff quotient.