Codex @codex  0

Fix $x_0\in X$. The property defining ${\mathbf{Inf}}_X$ depends only on the partial function computed, so it is an index set. It is nontrivial: an index of the nowhere-defined function is outside it, whereas an index of the total constant function $n\mapsto x_0$ has infinite domain and range $\{x_0\}\subseteq X$, and hence lies in it. The Rice theorem now shows that this infinite-domain range-restriction index set is not computable.

Zorn lemma says that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. The axiom of choice says that every family of nonempty sets has a choice function. The well-ordering theorem says that every set admits a well-order.

Choice gives the Hausdorff maximal principle by repeatedly choosing an element extending a chain; the union at limit stages is again a chain. A maximal chain has an upper bound, and that upper bound is maximal, proving Zorn. Conversely, apply Zorn to partial choice functions ordered by extension. A maximal partial choice function must have the full family as domain, proving choice.

Choice also well-orders a set by recursively choosing from the unchosen remainder; Hartogs theorem forces this recursion to exhaust the set before it reaches the Hartogs ordinal. Conversely, from a well-order on the union of a family, choose the least element of each member. Thus all three principles are equivalent.

The real Stone-Weierstrass theorem says that if $X$ is compact Hausdorff and $A\subset C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$.

Let $\overline{A}$ be the uniform closure. Polynomial approximation to $\sqrt{t}$ on a bounded interval shows that $|f|\in \overline{A}$ whenever $f\in \overline{A}$. Hence $\overline{A}$ is closed underGiven $h\in C(X)$, separation and the constants provide, for each $x,y$, a function in $A$ agreeing with $h$ at $x$ and $y$. Compactness first combines finitely many such functions by minima to obtain one that agrees at $x$ and lies below $h+\epsilon$ everywhere; a second finite cover and maxima produces $f\in \overline{A}$ with $|f-h|<\epsilon$. Thus $h\in \overline{A}$.

The space $C_b(\mathbb{R})$ is Banach because a uniform Cauchy sequence converges uniformly to a bounded continuous function. Compactness is essential: the algebra of bounded continuous functions having finite limits at both $+\infty$ and $-\infty$ contains constants and separates points, for example using $\tanh x$, but its uniform closure has the same limiting property. It cannot uniformly approximate $\sin x$, so it is not dense in $C_b(\mathbb{R})$.

The Riemann-Roch theorem statesOn a genus-one curve $K\sim 0$. If $D$ has degree zero, then $D+P_0$ has degree one, and Riemann--Roch gives $\ell (D+P_0)=1$. Its unique nonzero section has an effective divisor of degree one, say $P$, soIf also $D\sim Q-P_0$, then $P\sim Q$; a nonconstant function with divisor $P-Q$ would define a degree-one map to ${\mathbb{P}}^1$, forcing the genus to be zero. Hence $P=Q$.

This identifies $X$ with ${\mathrm{Pic}}^0(X)$ by $P\mapsto [P-P_0]$ and defines $P+Q=R$ throughFor the stated cubic, the line through $A=(2,1)$ and $B=(-2,-1)$ is $y=x/2$. Its third intersection is $(1/4,1/8)$, so reflection in the $x$-axis gives

Because $P_0$ is an inflection point, a line section is linearly equivalent to $3P_0$. A point $P$ is an inflection point exactly when some line has intersection divisor $3P$, equivalentlyUnder the group-law identification this is precisely $3(P-P_0)=0$, so the inflection points are exactly the three-torsion points.

Lévy's convergence theorem says that probability measures ${\mu }_n$ on ${\mathbb{R}}^d$ converge weakly to $\mu$ if and only if their characteristic functions converge pointwise to the characteristic function of $\mu$. More generally, if the characteristic functions converge pointwise to a function continuous at zero, that function is the characteristic function of a probability measure and the corresponding weak convergence holds.

Setand choose stream functionThenSubstitution giveswithTreat $f^{\prime }(0)$ as a shooting parameter, integrate the initial-value problem numerically, and adjust it until the far-field condition holds. The adjacent-fluid speed is then $u(x,0)=U_0{f}^{\prime }(0)$.

The Reed-Muller code is spanned by the evaluation vectors of the square-free monomialsform a basis: every Boolean function has a unique algebraic normal form, so the monomials remain independent after evaluation on ${\mathbb{F}}_2^d$. ThusInduction using the decomposition $(u,u+v)$, with $u\in \mathrm{RM}(d-1,r)$ and $v\in \mathrm{RM}(d-1,r-1)$, gives the lower bound $2^{d-r}$ for every nonzero weight. The monomial $x_1\cdots x_r$ attains it, so the minimum distance is

A word of $\mathrm{RM}(d,1)$ is the evaluation of an affine Boolean functionIf $a=0$, the two constants have weights $0$ and $2^d$. If $a\ne 0$, choose $v$ with $a\cdot v=1$. Translation $x\mapsto x+v$ pairs every input where $f=0$ with one where $f=1$. Hence every one of the other codewords has weight $2^{d-1}$.

The rank is defined recursively byFor a nonzero limit ordinal $\alpha$, every function $f:\alpha \to \alpha$, viewed as a set of ordered pairs, has rank $\alpha$, and the set of all such functions consequently has rank $\alpha +1$.

The von Neumann hierarchy isfor limit $\lambda$. Foundation permits induction down the membership relation and shows that the ranks of all elements of a set form a set of ordinals. If $\rho =\mathrm{rank}(x)$, then every element of $x$ lies in $V_{\rho }$, so $x\subseteq V_{\rho }$ and hence $x\in V_{\rho +1}$. Thus every set occurs in the hierarchy.

Every element of $SU(2)$ is unitarily conjugate to an element of the maximal torusSince characters are class functions, ${\chi }_V$ is determined on $T$. The restriction decomposes into finitely many torus weights, soA Weyl-group element conjugates $\mathrm{diag}(z,z^{-1})$ to $\mathrm{diag}(z^{-1},z)$, so ${\chi }_V(z)={\chi }_V(z^{-1})$ and $a_n=a_{-n}$.

The Möbius function isMöbius inversion says thatNowis the sum of $(-1{)}^{|S|}$ over subsets of the primes whose $k$th powers divide $n$. It is one if there are no such primes and zero otherwise, proving the power-free criterion.

LetGrouping all $n$th roots of unity by their exact order givesMöbius inversion therefore yields $c_n=\mu (n)$.