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A subset $A\subseteq X$ is connected if it cannot be written as $A=U\cup V$, where $U,V$ are disjoint, nonempty sets open in the subspace topology on $A$.

If $f(X)$ were disconnected as $U\cup V$, continuity would make $f^{-1}(U)$ and $f^{-1}(V)$ a disconnection of $X$. Thus the continuous image of a connected space is connected.

For $Y=\{0,1\}$ with the discrete topology, a nonconstant continuous $h:X\to Y$ gives the disconnectionConversely, a disconnection $X=U\cup V$ defines a continuous nonconstant function by assigning $0$ on $U$ and $1$ on $V$. Hence $X$ is connected exactly when every such $h$ is constant.

Finally suppose $C$ is connected but $\mathrm{Cl}(C)=U\cup V$ is a disconnection. The intersections $C\cap U$ and $C\cap V$ cannot both be nonempty, so, after interchanging $U,V$, we have $C\subseteq U$. For any $v\in V$, the relatively open neighbourhood $V$ of $v$ in $\mathrm{Cl}(C)$ is disjoint from $C$, contradicting $v\in \mathrm{Cl}(C)$. Therefore the closure of a connected set is connected.

A function $f:U\to \mathbb{C}$ is holomorphic when it is complex differentiable at every point of the open connected set $U$. Morera's theorem states that a continuous function on a domain is holomorphic if its integral around every triangle whose interior lies in the domain is zero.

The integrand $e^{tz}/(1+t^2)$ is continuous jointly in $(t,z)$ on $[0,1]\times \mathbb{C}$, so the displayed integral defines a continuous function. For any triangle $T$, Fubini's theorem and the Cauchy integral theorem giveMorera's theorem therefore proves that $f$ is entire.

The function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ but has no antiderivative there, because an antiderivative would integrate to zero around every closed curve whereasThis is the standard period obstruction to a holomorphic antiderivative.

Two natural improvements are:

The first addresses discreteness and mean-dependent variance; the second addresses systematic lack of fit in the conditional mean.

Taking Laplace transforms and using $y(0)=0$ givesHence the transfer function isIts pole is at $-1/\alpha$, and its causal impulse response isIt is absolutely integrable exactly when

Start with the two families of uniqueness and injectivity clauses from part (i), and for every $x\in X$ add the finite clauseThis clause is a legitimate proposition because $A_x$ is finite and nonempty. It makes every $x$ have a value in $A_x$, while the earlier clauses make that value unique and make the resulting function injective. Conversely, every injection $g:X\to Y$ with $g(x)\in A_x$ satisfies all these clauses.

For distinct $x,x^{\prime }\in X$ and distinct $y,y^{\prime }\in Y$, take all clausesThe first family says that each $x$ is paired with at most one $y$; the second says that each $y$ is paired with at most one $x$. There is no existence clause, so the domain may be any subset of $X$. Thus the relations $f_v$ are exactly the injective partial functions from $X$ to $Y$.

The soundness theorem for propositional logic says that if $\Gamma ⊢t$, then every valuation satisfying every member of $\Gamma$ also satisfies $t$.

The proposed function need not respect the connectives. For example, if a primitive proposition $p$ is independent of $S$, then neither $p$ nor $\neg p$ is provable from $S$, so the definition gives

Order the consistent supersets of $S$ by inclusion. The union of a chain is consistent, since a finite proof of a contradiction would already use assumptions from one member of the chain. Zorn lemma therefore gives a maximal consistent extension $T$. It is deductively closed: if $T⊢t$, adjoining $t$ preserves consistency, so maximality forces $t\in T$. Moreover, for every $t$, exactly one of $t$ and $\neg t$ belongs to $T$. They cannot both belong by consistency; if neither belonged, the inconsistency of both proper extensions would give $T⊢\neg t$ and $T⊢\neg \neg t$, again a contradiction. The usual induction on formulae now shows thatdefines a valuation satisfying $T$, and hence $S$.

Now suppose every finite subset of $S$ has a model. By soundness every finite subset is consistent. Any proof of a contradiction from $S$ uses only finitely many assumptions, so $S$ itself is consistent. Applying the preceding maximal-consistent-extension construction gives a model of $S$. This proves the propositional compactness theorem.

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.

Let $F^{-1}(u)=\inf \{x:u\le F(x)\}$. The generalized-inverse identitygivesand similarly $\mathbb{P}({\tilde{X}}_n\le x)=F_{X_n}(x)$. Thus the coupled variables have the required marginal distributions.

At every continuity point $u$ of the nondecreasing function $F_X^{-1}$, convergence of the distribution functions at continuity points of $F_X$ impliesIndeed, values just below and above $F_X^{-1}(u)$ can be chosen at continuity points of $F_X$ and trap the two generalized inverses. A monotone function has at most countably many discontinuities, so this convergence holds for Lebesgue-almost every $u\in (0,1)$. Therefore ${\tilde{X}}_n\to \tilde{X}$ almost surely. This is the quantile coupling for convergence in distribution.

Lorentz invariance requires $A^{\mu }$ and $J^{\mu }$ to transform as four-vectors, $\chi$ as a scalar, $m$ and ${\mu }_0$ as scalars, and the integration domain and boundary conditions to respect the transformation. Then every contracted term and $d^{4}x$ are Lorentz invariant.

Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, the field strength is unchanged. The source term changes by a boundary term pluswhich vanishes for a conserved current and suitable boundary behaviour of $\chi$. The mass term is not gauge invariant. Thus the action has the usual local gauge invariance whenwith the gauge function or its boundary contribution suitably controlled. For $m>0$, the mass term breaks this gauge symmetry.

A generalized linear model has three components:

An elliptic function is a meromorphic $f:\mathbb{C}\to \hat{\mathbb{C}}$ with two real-linearly independent periods ${\omega }_1,{\omega }_2$. A fundamental cell may be taken as the half-open parallelogramwith $z_0$ chosen so that its boundary contains no zero or pole under discussion. The argument principle and cancellation on opposite sides show that a nonconstant elliptic function has the same number of zeros and poles in a cell, counting multiplicity.

If an elliptic function has no poles, it is entire. It is bounded on the closure of a fundamental parallelogram, and periodicity transfers that bound to the whole plane. Liouville's theorem therefore makes it constant.

Choose a fundamental parallelogram with no pole on its boundary. Opposite sides are translates traversed in opposite directions, so periodicity givesThe residue theorem then givesA lone simple pole has a nonzero residue by definition of its order, so an elliptic function cannot have exactly one simple pole in a fundamental cell.

For the fixed effective enumeration of unary partial computable functions, the infinite-domain index set is

The Rice theorem states that every nontrivial index set of partial computable functions is not computable.

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.