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A nonempty topological space is irreducible when it is not the union of two proper closed subsets, equivalently when any two nonempty open subsets intersect.

The Zariski topology on ${\mathbb{A}}^n$ is Noetherian because $k[x_1,\dots ,x_n]$ satisfies the ascending chain condition on ideals. Suppose a closed set $X$ that is not a finite union of irreducible closed sets were minimal among such counterexamples. It is reducible, say $X=X_1\cup X_2$ with both $X_i$ proper closed subsets. Minimality expresses each $X_i$ as a finite union of irreducible closed sets, and combining those decompositions contradicts the choice of $X$. Thus every closed $X$ has a finite irreducible decomposition.

Using $x_3=x_2^2$, the second defining polynomial reduces toConsequentlyEach component is the image of $t\mapsto (\pm t,t,t^2)$ and is isomorphic to ${\mathbb{A}}^1$, hence irreducible. In characteristic two the two displayed components coincide.

Suppose a polynomial $P(x,y)={\sum }_{j=0}^m{p}_j(x)y^j$ vanishes on every $(x,e^x)$. Thenas an entire function. These exponential-polynomial functions are linearly independent: applying $(D-m{)}^{\mathrm{deg}{p}_m+1}$ kills the term with largest exponent, while it acts injectively on $p_j(x)e^{jx}$ for $j<m$; induction on the number of terms finishes the proof. Hence every $p_j$ is zero, so no nonzero polynomial vanishes on the graph. Its Zariski closure is all of ${\mathbb{A}}^2$.

Finally refine an open cover of an affine variety $X$ by distinguished opens $D(f_j)$, choosing one inside an original member around each point. Since these distinguished opens cover,The Hilbert Nullstellensatz implies that the $f_j$ generate the unit ideal. Thusfor finitely many of them, so $D(f_{j_1}),\dots ,D(f_{j_r})$ cover $X$. The corresponding original open sets form a finite subcover. This is quasi-compactness of an affine variety.

The Seifert-van Kampen theorem says that if $X=U\cup V$, where $U,V,U\cap V$ are path connected open sets containing $x_0$, then

The cell attachment is the quotientwith the image of $x_0=f(\ast )$ as base point. For $n=2$, take one open set that deformation retracts onto $X$ together with a boundary collar, and another that consists of the interior of the disc with a collar and is contractible. Their intersection deformation retracts onto $S^1$. Its generator maps to $[f]$ in the first set and to the identity in the second. Van Kampen therefore proves the fundamental group after attaching a 2-cell formula

Applying van Kampen to the two circles givesAttach three discs along loops representing $a^2$, $b^3$, and $(ab{)}^2$. The resulting presentation complex for the symmetric group on three letters has groupSending $a$ to $(12)$ and $b$ to $(123)$ gives a surjection onto $S_3$. The relations imply $aba=b^{-1}$, so every word reduces to one ofThe presented group has at most six elements and surjects onto the six-element group $S_3$, so this map is an isomorphism.

The Baire category theorem states that a complete metric space is not a countable union of closed sets with empty interior. To prove it, let $G_1,G_2,\dots$ be open dense sets and start with any nonempty open set $V$. Inductively choose closed ballswith the first ball inside $V\cap G_1$. The centres are Cauchy. Completeness supplies a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense, which is the equivalent form of the theorem.

Now each $nS$ is closed and ${\bigcup }_{n\ge 1}nS=X$. Baire givesfor some $x,r,n$. Symmetry gives $B(-x,r)\subset nS$. If $\Vert z\Vert <r$, then $x+z$ and $-x+z$ lie in these two balls; convexity makes their midpoint $z$ belong to $nS$. Hencewhich proves the closed convex absorbing set has an origin neighbourhood result.

Convexity cannot be dropped. In $X=\mathbb{R}$, setThis is closed and symmetric but has gaps arbitrarily close to zero. For any $x\ne 0$, choose $k$ so that the intervalhas length at least one, and choose an integer $n$ in it. Then $x/n\in S$, so ${\bigcup }_{n}nS=\mathbb{R}$, while $S$ is not a neighbourhood of zero. This is a closed symmetric absorbing set without an origin neighbourhood.

The product $S(z)=f(z)f(1-z)$ is entire. From $f(z+1)=zf(z)$ and $f(1-z)=-zf(-z)$,Its modulus is therefore periodic with period one. The assumed bound for $f$ on $0\le \mathrm{Re}z\le 1$, together with the reflected bound for $f(1-z)$, bounds $S$ on that strip and hence on all of $\mathbb{C}$. By Liouville's theorem $S$ is constant, while $S(z+1)=-S(z)$ forces that constant to be zero.

If $f$ were nonzero at one point, it would be nonzero on a neighborhood, so $f(1-z)=0$ on an open set; the identity theorem would then give $f\equiv 0$, a contradiction to that choice. Therefore $f\equiv 0$ and $F=\Gamma$.

A neighbourhood of $x$ contains an open set containing $x$; $x_n\to x$ when every neighbourhood eventually contains every $x_n$. If two metrics have the same topology they plainly have the same convergent sequences. Conversely, if a $d$-open set $U$ were not $d\prime$-open, some $x\in U$ would have points $x_n\notin U$ with $d\prime (x_n,x)<1/n$; the common-sequence assumption would imply $d(x_n,x)\to 0$, contradicting $d$-openness. Symmetry finishes the proof. This fails generally: on an uncountable set the discrete and cocountable topologies differ, but in both every convergent sequence is eventually constant.

If $\Vert Tx\Vert \ge \delta \Vert x\Vert$ and $\Vert S-T\Vert <\delta$, then
$\Vert Sx\Vert \ge (\delta -\Vert S-T\Vert )\Vert x\Vert$. Thus bounded-below maps form an open set.

A subset is dense when every nonempty open set meets it, equivalently its closure is the whole space. Baire category theorem says a complete metric space has dense intersection for every countable family of open dense sets. Hereso Baire makes $S$ dense. It need not be open: enumerate the rationals $q_j$ and take $A_j=[0,1]\setminus \{q_1,\dots ,q_j\}$, giving the irrationals. Nor need ${\cap }_j{A}_j$ be nonempty: take $A_1=\varnothing$ and $A_j=[0,1]$ for $j\ge 2$.

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.

Let the common domain be an open set $U$. The uniform limit theorem makes the limit $f$ continuous. For every triangle whose closed interior lies in $U$, Cauchy's integral theorem gives ${\int }_{\partial T}{f}_n=0$. Part (ii) passes this identity to the limit, giving ${\int }_{\partial T}f=0$. Morera's theorem now proves that $f$ is holomorphic.

The same reasoning works for locally uniform convergence, since each triangle boundary is compact. No passage of derivatives through a limit has been assumed.

Morera's theorem says that a continuous complex-valued function $f$ on an open set $U\subseteq \mathbb{C}$ is holomorphic iffor every triangle $T$ whose closed interior is contained in $U$. Triangle boundaries are traversed with positive orientation. The continuity hypothesis matters: the zero-integral condition is combined with continuity, not used in isolation.

A metric space is connected when it has no separation of a topological space: it cannot be written as the union of two disjoint nonempty open subsets. Equivalently, the only subsets that are both open and closed are the empty set and the whole space.The open sets here are those of the metric topology.

Suppose a path-connected space had a separation $X=U\cup V$. Choose $u\in U$, $v\in V$, and a path $\gamma$ from $u$ to $v$. The inverse images ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$ are disjoint open sets in the subspace topology of $[0,1]$, cover it, and are nonempty because they contain its opposite endpoints. This contradicts the connectedness of an interval.

Therefore every path-connected space is connected. Notice that it is the subspace topology on $[0,1]$ that is used in the inverse-image argument.

One standard form of the Cauchy integral formula is this: if $f$ is holomorphic on an open set containing a contour that is a simple closed curve $\Gamma$ and its interior, and $\Gamma$ has positive orientation of a contour, then for every point $w$ inside it,The hypothesis includes the contour itself; singularities on the contour are excluded. In particular, the formula applies on a circle whose closed disk is contained in the domain of holomorphy.

One standard form of Cauchy's residue theorem is the following. Let $\Gamma$ be a positively oriented contour that is a simple closed curve, and let $f$ be a holomorphic function on an open set containing $\Gamma$ and its interior except for finitely many interior isolated singularities $z_1,\dots ,z_m$. There are no singularities on $\Gamma$. ThenThe residue $\mathrm{Res}(f,z_j)$ is the coefficient of $(z-z_j{)}^{-1}$ in the Laurent series at $z_j$. Each interior isolated singularity contributes once because the winding number of this positively oriented contour about each such point is one. Reversing the orientation of a contour reverses the sign. This form of Cauchy's residue theorem applies to the rectangle below.

The answers areDistinct natural numbers have $d(x,y)>1$. Taking $\epsilon =1$ in the Cauchy sequence condition forces every sufficiently late pair of terms to be equal. Every Cauchy sequence is therefore eventually constant and converges in $\mathbb{N}$, proving completeness.

The sequence $x_n=n$ has no convergent subsequence, since convergence would force its sufficiently late terms to lie within distance $1$ of its limit of a sequence, hence to equal that limit. Thus sequential compactness fails.

Every singleton is an open ball of radius $1$, so the metric topology is the discrete topology. For example, $\{1\}$ and $\mathbb{N}\setminus \{1\}$ form a separation into two nonempty open sets, so $X$ is not connected. It is consequently not path-connected either.

The inverse tangent $T(x)=\arctan x$ is a homeomorphism from the real line with its Euclidean metric onto $I=(-\pi /2,\pi /2)$ with its subspace topology: $T$ and its inverse $\tan$ are continuous. It is also an isometry from $(\mathbb{R},\rho )$ onto $I$, because $\rho (x,y)=|T(x)-T(y)|$. Consequently, for $U\subseteq \mathbb{R}$,For a direct open ball interpretation, $B_{\rho }(x,r)=T^{-1}((T(x)-r,T(x)+r)\cap I)$ is $d$-open by continuity of $T$. Conversely, continuity of $\tan$ at $T(x)$ supplies a sufficiently small $\rho$-ball inside any prescribed $d$-ball about $x$. Thus both metrics generate exactly the same open sets.

Both properties are preserved by a continuous map.

For sequential compactness, take any sequence $(y_n)$ in $f(X)$ and choose $x_n\in X$ with $f(x_n)=y_n$. If $X$ is sequentially compact, some subsequence satisfies $x_{n_k}\to x\in X$. By continuity, $y_{n_k}=f(x_{n_k})\to f(x)\in f(X)$. Hence $f(X)$ is a sequentially compact space.

For connectedness, suppose instead that $f(X)=U\cup V$ is a separation into disjoint nonempty relatively open sets. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint nonempty open sets of $X$ whose union is $X$, contradicting connectedness. Therefore $f(X)$ is a connected space.

Assume condition (1). For fixed $a$, the supremumis nondecreasing as $r$ increases. Therefore its limit as $r\downarrow 0$ exists in the extended real numbers and equals ${\inf }_{r>0}{M}_a(r)$. Because $a\in B(a,r)$, it is at least $f(a)$. Condition (1) says precisely that this limit is $f(a)$.

It follows that for every $\epsilon >0$ there is $r>0$ with $M_a(r)<f(a)+\epsilon$. Fix $C\in \mathbb{R}$ and $a$ with $f(a)<C$. Taking $\epsilon =C-f(a)$ produces an open ball on which $f(x)<C$. Thus $\{x:f(x)<C\}$ is an open set, and its complementThis proves (1) implies (2). The property being characterized is upper semicontinuity.

Let $F(z)=\overline{\sinh z}$. More generally, for a holomorphic function $h$, the increment expansion at $z_0$ givesAlong real increments $w$ the first factor $\overline{w}/w$ is $1$, while along purely imaginary increments it is $-1$. Thus complex differentiability of $\overline{h}$ requires $h^{\prime }(z_0)=0$. Conversely, if $h^{\prime }(z_0)=0$, the displayed expression tends to zero for arbitrary complex increments, so the complex derivative exists and equals zero.

For $h=\sinh$, this condition is $\cosh z_0=0$, equivalent to $e^{2z_0}=-1$. ConsequentlyThese points form a discrete subset of the complex plane and contain no nonempty open set. Every neighbourhood of each one contains points where $F$ is not complex differentiable. Therefore $F$ is holomorphic on no nonempty open set, even though it has complex derivatives at the specified points.

The identity theorem states that if two holomorphic functions on a connected open set agree at a sequence of distinct points with an accumulation point inside that set, then they agree everywhere on it. Equivalently, the zeros of a holomorphic function which is not identically zero cannot accumulate inside its domain.

Because $f,g$ never vanish, their logarithmic derivatives are holomorphic. The holomorphic functionvanishes at every $a_n$. Since the distinct $a_n$ converge to $a\in U$, the identity theorem gives $H\equiv 0$. Applying the quotient rule now givesA holomorphic function with zero derivative on a domain is constant. One can see this locally from its Taylor series, then propagate the constant using connectedness. Hence $f/g=c$ on $U$, with $c\ne 0$, andThe conclusion can fail if the accumulation occurs outside $U$. Take the punctured complex plane $U=\mathbb{C}\setminus \{0\}$ andBoth functions are nowhere vanishing and holomorphic on $U$, and $a_n\to 0\notin U$. The chain rule givesso these logarithmic derivatives agree at every $a_n$. However, $f$ is not constant: for example $f(1/\pi )=e^{-1}$ and $f(1/(2\pi ))=e$. Therefore no constant $c$ gives $f=cg$.