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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 $B(x,\epsilon )=\{x\}$.

For $X\subseteq {\mathbb{A}}^n$ and $P\in X$, the Zariski tangent space isThe dimension can be defined as ${\min }_{P\in X}\dim T_{P}X$. Equivalently, the krull dimension of an affine variety is the supremum of lengths of strict chains of irreducible closed subsets, or the Krull dimension of $k[X]$. A point is singular when $\dim T_{P}X>\dim X$.

Forthe differentials give the tangent spaces of a product of two nodal line pairsThe variety is a union of four two-dimensional linear spaces. If both pairs $(x_1,x_2)$ and $(x_3,x_4)$ are nonzero, the two displayed equations are independent and $\dim T_{P}X=2$. If exactly one pair is zero, the dimension is three; at the origin it is four. Thus the singular locus is the union of the loci where either coordinate pair vanishes.

For $Y$, rank at most one is equivalent to vanishing of all $2\times 2$ minors:These homogeneous equations prove that $Y$ is projectively Zariski closed.

The following affine charts show both its dimension and smoothness. On $y_0=1$,so $(y_1,y_2)$ are free. On $y_1=1$,so $(y_0,y_3)$ are free. On $y_3=1$,and on $y_4=1$,These four charts cover $Y$, since the point with only $y_2$ nonzero violates the second minor. Every chart is isomorphic to ${\mathbb{A}}^2$. Hence the smooth projective surface from overlapping rank-one coordinates satisfiesand is nonsingular everywhere.

The valency theorem states that for a nonconstant analytic map between compact connected Riemann surfaces,for every target point $w$.

For a rational map $f=p/q$ on the Riemann sphere, first cancel common factors; then the degree of a rational map of the Riemann sphere isThe analytic isomorphisms are exactly the degree-one maps, namely the Möbius transformations.

The required transformation isIt sends $\infty \leftrightarrow 1$ and $0\leftrightarrow -1$. Its fixed points satisfy

The octahedral rotation orbits on the Riemann sphere have possible sizescorresponding respectively to vertices, face centres, edge centres, and generic points.

In the displayed $F$, no numerator factor vanishes at $0$ or at a fourth root of one. Hence $0,\pm 1,\pm i$ are poles of order four. The numerator and denominator have degrees $24$ and $20$, so $F(z)\sim z^4$ at infinity; infinity is also a pole of order four. Thusand

Finally let $z$ have stabilizer of order $e$. A local coordinate turns its stabilizer action into rotation by $e$th roots of unity. Since $F$ is invariant, the first nonconstant term of its local expansion has exponent divisible by $e$, so $m_F(z)\ge e$. The orbit has $24/e$ points and therefore contributes at least $24$ to the fibre through $F(z)$. The valency theorem and $\mathrm{deg}F=24$ leave no room for any further point. The degree-sized invariant separates finite-group orbits, so $F(z)=F(w)$ implies that $z$ and $w$ lie in the same orbit.

The Hahn-Banach theorem says that if $Y$ is a linear subspace of a real normed space $X$ and $g:Y\to \mathbb{R}$ is bounded and linear, then there is a bounded linear extension $G:X\to \mathbb{R}$ with $\Vert G\Vert =\Vert g\Vert$.

For $x\in X$, the map $\hat{x}(f)=f(x)$ is linear on $X^{\prime }$ andso $\hat{x}\in X^{\prime \prime }$ and $\Vert \hat{x}\Vert \le \Vert x\Vert$. The assignment $Jx=\hat{x}$ is visibly linear. If $x\ne 0$, define $g$ on $\mathrm{span}\{x\}$ byIt has norm one, and Hahn--Banach extends it to an $f\in X^{\prime }$ of norm one with $f(x)=\Vert x\Vert$. ThereforeThus the canonical embedding into the bidual is an isometry and

On $C([0,1])$, point evaluation $\ell (f)=f(0)$ has norm one for the $L^{\infty }$ norm, because a continuous function's essential supremum equals its supremum. Hahn--Banach extends it to $\Lambda \in (L^{\infty }{)}^{\prime }$. If $\Lambda$ were represented by some $g\in L^1$, thenfor every continuous $f$. Choose continuous $0\le f_n\le 1$ with $f_n(0)=1$ and support in $[0,1/n]$. Absolute continuity of the integral makes the right side tend to zero, while the left side is always one, a contradiction. This singular functional on L infinity proves