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A function $f:X\to Y$ between topological spaces is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$ for every subset $A\subseteq X$, where the bars denote closure. For a continuous $f$, pull back an open neighbourhood of $f(x)$ for $x\in \overline{A}$; its preimage meets $A$, so the neighbourhood meets $f(A)$. Conversely, apply the inclusion to $A=f^{-1}(C)$ with $C$ closed. Then $f(\overline{A})\subseteq C$ implies $\overline{A}\subseteq A$, so the preimage of every closed subset is closed and $f$ is continuous.

Choose an uncountable set $X$ and a point $a\in X$. Every point other than $a$ is isolated, and a set containing $a$ is open precisely when its complement is countable. Equivalently, the closed sets are all subsets containing $a$ together with all countable subsets. Arbitrary intersections and finite unions preserve this latter condition, so it defines a topology.

Every convergent sequence in this space is eventually equal to its limit. Indeed the set of its values different from its limit is countable and closed; its complement is an open neighbourhood of the limit. Eventual membership in this complement forces eventual equality. Nevertheless $a$ is not isolated, because its singleton has uncountable complement. In particular sequence convergence alone does not determine this topology.

A real-valued function on this space is continuous exactly when $\{x:f(x)\ne f(a)\}$ is countable. Sufficiency follows by considering the preimages of closed subsets of the real line: each either contains $a$ or is a subset of that countable exceptional set. For necessity, each set $\{x:|f(x)-f(a)|\ge 1/n\}$ is closed and excludes $a$, hence is countable; their countable union is the exceptional set.

An open neighbourhood of a point $x$ in a topological space $(X,\mathcal{T})$ is an open set $U$ containing $x$. A general neighbourhood need only contain an open neighbourhood; it need not itself be open. A finite intersection of open neighbourhoods of the same point is again an open neighbourhood, by the topology axioms.

The Cauchy integral formula says that if $f$ is holomorphic on the simply connected open set $U$, and $\gamma$ is a positively oriented simple closed curve in $U$, then, for every $z$ in its interior,Here the contour and its enclosed region lie in $U$; the contour can be taken piecewise continuously differentiable. More generally, the same formula needs only holomorphic function on an open neighbourhood of that closed region. The orientation is counterclockwise, so the winding number about an interior point is one.

The closed unit disk is a compact set, and $|f|$ is continuous on it because $f$ is holomorphic on an open neighbourhood. The extreme value theorem therefore supplies a point where its maximum is attained. If such a point is on the boundary, the conclusion follows. If it is in the interior, part (i) makes $f$ constant on the open disk; continuity then gives the same value on the boundary. ConsequentlyThe assumption that the closed disk lies in $U$ is what makes the boundary values and the compactness argument available.

The inverse function theorem states: if $f$ is continuously differentiable on an open subset of ${\mathbb{R}}^n$ and $Df(a)$ is invertible, then some open neighbourhood of $a$ is mapped bijectively onto an open neighbourhood of $f(a)$, with continuously differentiable inverse. Its derivative at $f(a)$ is $(Df(a){)}^{-1}$.

The present map is smooth, and at $a=(\pi /2,\pi )$,Also $f(a)=(0,0)$, so the theorem applies locally and givesThe inverse is defined near the image point $(0,0)$ and takes values near $a$; no assertion of a global inverse is required.

Convergence of a sequence to $\ell$ means that every open neighbourhood $U$ of $\ell$ contains all sufficiently late terms: there is $N$ such that $x_n\in U$ for all $n\ge N$. The threshold may depend on $U$.

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 $x_n\ne \ell$.

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 two-dimensional embedded submanifold is locally flattened by a continuously differentiable coordinate diffeomorphism of an ambient open neighbourhood onto a coordinate plane, or equivalently locally admits a regular two-parameter chart. A tangent vector at $x$ is the velocity ${\gamma }^{\prime }(0)$ of a differentiable curve in $M$ through $x$. All such vectors form the two-dimensional tangent space $T_{x}M$, the image of the differential of a regular parametrization. Normal vectors form its orthogonal complement $(T_{x}M{)}^{\perp }$ in ${\mathbb{R}}^3$, which has dimension one.

A two-dimensional embedded submanifold is locally flattened by a continuously differentiable coordinate diffeomorphism of an ambient open neighbourhood onto a coordinate plane, or equivalently locally admits a regular two-parameter chart. A tangent vector at $x$ is the velocity ${\gamma }^{\prime }(0)$ of a differentiable curve in $M$ through $x$. All such vectors form the two-dimensional tangent space $T_{x}M$, the image of the differential of a regular parametrization. Normal vectors form its orthogonal complement $(T_{x}M{)}^{\perp }$ in ${\mathbb{R}}^3$, which has dimension one.

An accumulation point of $A$ is a point $x\in X$ such that every open neighbourhood of $x$ contains a point of $A$ different from $x$:The point itself need not belong to $A$. Requiring a different point distinguishes accumulation from mere membership in the closure.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^n$ is $C^1$ on an open set and $Df(a)$ is invertible, then there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $f(a)$ such that $f:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse satisfiesIf $f$ is $C^k$ or smooth, the inverse has the same regularity.

The chain rule givesIts determinant is $-1$, so it is invertible. The map $G$ is smooth and $G(0,0)=(0,0)$. The inverse function theorem therefore yields an open neighbourhood $U$ of the origin for which $G(U)$ is open, contains the origin, and

Suppose $f$ is continuous and $x\in \overline{A}$. Every open neighbourhood $V$ of $f(x)$ has an open preimage containing $x$, which meets $A$. Hence $V$ meets $f(A)$, proving $f(x)\in \overline{f(A)}$.

Conversely, suppose the indicated inclusion holds for every subset. To prove continuity, take any closed set $C\subseteq Y$ and put $A=f^{-1}(C)$. Since $f(A)\subseteq C$ and $C$ is closed,Thus $\overline{A}\subseteq f^{-1}(C)=A$, so $A$ is closed. Every closed set has closed preimage, which is the closed-set characterization of continuity.This is the closure characterization of continuity; it requires no Hausdorff or metric hypothesis.

The closure $\overline{A}$ is the intersection of every closed subset of $X$ containing $A$. Thus it is the smallest such closed subset. Equivalently,For equivalence, if an open neighbourhood misses $A$, its complement is a closed superset of $A$ missing $x$. Conversely, a closed superset missing $x$ has an open complement that is a neighbourhood of $x$ disjoint from $A$.

First assume both factors are nonempty. If the product is compact, its continuous coordinate projections are onto, so each factor is compact by the continuous image of a compact space theorem.

For the converse, let $X,Y$ be compact and let $(W_{\lambda })$ be an open cover of their product. Fix $x\in X$. For every $y\in Y$, choose a basis rectangle $U_{x,y}\times V_{x,y}$ containing $(x,y)$ and lying in a cover member $W_{\lambda (x,y)}$. Compactness of $Y$ gives finitely many $V_{x,y_j}$ covering $Y$. Thenis an open neighbourhood of $x$, and $U_x\times Y$ is covered by the corresponding finitely many $W_{\lambda (x,y_j)}$.

The sets $U_x$ cover $X$. Compactness gives finitely many of them covering $X$, and the finitely many cover members associated to these choices now cover $X\times Y$. Hence the product is compact. This is the finite-product Tychonoff theorem proof, with the finite intersection creating a uniform neighbourhood along each fibre.

As printed, the assertion omits the nonempty hypothesis. If $X=\varnothing$ and $Y=\mathbb{R}$, the product is compact while $Y$ is not. The fully correct version without any nonemptiness assumption is

If both factors are Hausdorff, two distinct product points differ in at least one coordinate. Disjoint open neighbourhoods in that coordinate, multiplied by the other whole factor, give disjoint open neighbourhoods of the product points. Thus the product is Hausdorff.

Conversely, assume the product is Hausdorff. Fix $y_0\in Y$, using nonemptiness. For $x\ne x^{\prime }$, choose disjoint open neighbourhoods of $(x,y_0)$ and $(x^{\prime },y_0)$ and refine them to basis rectangles $U\times V$ and $U^{\prime }\times V^{\prime }$. Since $y_0\in V\cap V^{\prime }$, their disjointness forces $U\cap U^{\prime }=\varnothing$. These separate $x,x^{\prime }$, proving that $X$ is Hausdorff. Fixing a point of $X$ proves the same for $Y$.The nonempty hypothesis is essential: a product with the empty set is empty regardless of the other factor. This is the Hausdorff property of a nonempty product.

For every $k\in K$, the Hausdorff condition gives disjoint open neighbourhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ form an open cover of $K$, so choose a finite subcover $U_{k_1},\dots ,U_{k_m}$. PutThen $U,V$ are open sets, $K\subseteq U$ and $x\in V$. If a point belonged to $U\cap V$, it would belong to some $U_{k_j}$ and to $V_{k_j}$, contrary to their choice. ThusThe finiteness matters: a finite intersection of open sets is open. This is separation of a point and a compact set. If $K=\varnothing$, use $U=\varnothing$, $V=X$.

Every $x\in X\setminus K$ consequently has an open neighbourhood $V$ contained in $X\setminus K$. The complement is open, so a compact subset of a Hausdorff space is closed.

The conclusion can fail without the Hausdorff condition. Take the two-point set $X=\{0,1\}$ with the indiscrete topology $\{\varnothing ,X\}$. The singleton $K=\{0\}$ is compact, since one member of any open cover suffices. It is not closed, because its complement $\{1\}$ is not open. The two points cannot have disjoint open neighbourhoods, so this example is indeed not Hausdorff.

A subset $A$ is connected when it is a connected space in its subspace topology: there is no separation of a topological spaceinto two nonempty disjoint open subsets of $A$. Equivalently, the only clopen sets of $A$ are $\varnothing$ and $A$.

Since the given $A$ is not connected, choose such $P,Q$. Each is the complement of the other in $A$, so both are also closed subsets of $A$. They are compact by the closed subset of a compact space result. We now separate these two compact subsets by ambient open sets.

For each $p\in P$, part (b), applied to $Q$ and $p$, gives disjoint ambient open sets $G_p,H_p$ with $p\in G_p$ and $Q\subseteq H_p$. Choose $p_1,\dots ,p_m$ such that the $G_{p_j}$ cover $P$, using its compactness. DefineThen $P\subseteq U$, $Q\subseteq V$, and $U,V$ are disjoint open sets. Indeed a point of $U$ lies in some $G_{p_j}$, whereas every point of $V$ lies in the corresponding $H_{p_j}$. ConsequentlyThis proves the requested ambient separation, using the more general separation of disjoint compact subsets. The initially given relatively open sets $P,Q$ need not themselves be open in $X$; the finite construction supplies ambient open neighbourhoods.