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U+D. By absoluteness of infinitude between transitive models, two transitive models of ZFC agree on the natural numbers and therefore on whether a shared set is a finite set or an infinite set.

True. If $x$ is a nonempty finite set, thenThis maximum is a successor ordinal, whereas the empty set has rank $0$. Hence a set of nonzero limit ordinal rank cannot be finite and must be an infinite set.

For every $g\in G$, left multiplication gives a bijectionwhose inverse is $x\mapsto g^{-1}x$. Hence every left coset has the same cardinality as $H$. If $H$ is an infinite set, every left coset is infinite; if $H$ is a finite set, every left coset is finite. These are the only possibilities.

Put $L={\bigcap }_{i=1}^k{H}_i$. Repeated application of part (iv), or mathematical induction on $k$, givesPart (ii) therefore implies that $L$ is an infinite set. If $G\setminus U$ is nonempty, choose $g\notin U$. We claim its left coset $gL$ is wholly outside $U$. If $g\ell \in H_i$ for some $\ell \in L$, then $\ell \in H_i$ and closure under products and inverses would give $g=(g\ell ){\ell }^{-1}\in H_i$, a contradiction.

Thus $gL\subseteq G\setminus U$. Part (i) makes $gL$ infinite, so the complement is infinite. If there is no such $g$, the complement is empty. This proves the complement of a finite union of finite-index subgroups property:

Take the first-order theory of an infinite set in the first-order language of logical equality alone, with an axiom asserting at least $n$ distinct elements for each positive integer $n$. It is complete and eliminates quantifiers: a new logical variable either equals a specified old logical variable or is required to avoid only finitely many existing values, which an infinite domain permits.

For every infinite cardinal $\kappa$, any two first-order models of size $\kappa$ are isomorphic because every bijection between their underlying sets preserves logical equality, the whole first-order language. ThusIts complete $n$-types record only the partition of coordinate indices according to logical equality. There are finitely many such partitions, and each is isolated by the corresponding logical equalities and inequalities. Ryll-Nardzewski therefore recovers countable categoricity; there are no nonprincipal complete finite-arity partial types to omit in this example.

The cardinality of a vector space is central: here “infinite” must continue to refer to the underlying set. Over a countably infinite field, every nonzero finite-dimensional space is already infinite. Thus the model class now consists of all nonzero $K$-vector spaces, not only infinite-dimensional ones.

Since independence of a pair is not definable over an infinite field, for (i), independence of a single vector is still $x\ne 0$, but independence of tuples of length at least two is not definable by one formula in this one-sorted language. Quantifier elimination follows from the linear-equation argument in the preceding part for all these infinite domains. Any candidate two-variable formula therefore involves only finitely many linear equations. Take an independent pair, and compare it with $(x,{\lambda }_{a}x)$, where $x\ne 0$ and $a\in K$ avoids the finitely many ratios making one of those nontrivial equations vanish. Both pairs satisfy the same listed atomic formulas, contradicting a proposed definition of independence. Longer tuple independence would define pair independence by existentially completing a pair inside an infinite-dimensional model, giving the same obstruction.

For (ii), the vector-space axiom schema together with $\exists x(x\ne 0)$ axiomatizes the infinite underlying spaces. For (iii), there is again no universal axiomatization, because the zero subspace is a substructure.

For finite axiomatizability with infinitely many scalar symbols, in (iv) there is no finite axiomatization in the full language: any finite axiom set mentions only finitely many scalar symbols. In a nonzero vector space alter an unused nonzero ${\lambda }_a$ to send zero to a nonzero vector, retaining all mentioned interpretations. The finite sentences keep their truth values, but a vector-space axiom is now false. Relative to the already imposed vector-space axioms, the extra condition is finitely axiomatized by the single nonzero-vector sentence. These absolute and relative answers must be distinguished.

For (v), the one-variable algebra still has two atoms and four Elements. In two variables the isolated atoms are $(0,0)$ and the countably many rank-one directions, represented by $x\ne 0,y={\lambda }_{a}x$ and by $x=0,y\ne 0$. There is also a nonisolated complete type of an independent pair. A formula using finitely many linear equations either picks finitely many isolated atoms, or contains that independent type and all but finitely many isolated atoms. Therefore $F_2/E_2(T)$ is the finite-cofinite Boolean algebra on a countably infinite set of atoms, with the independent type the extra nonisolated point in its Stone space. In particular this algebra is countably infinite, rather than the full power set of its atoms.

For (vi), dimension can be $1,2,3,\dots$ or ${\aleph }_0$ while the domain is countable, producing pairwise nonisomorphic countable models. Thus $T$ is not countably categorical. For an uncountable $\kappa$, a vector space over this countable field has cardinality $\kappa$ exactly when its dimension is $\kappa$, since in general a nonzero space has size $\max (|K|,\dim V)$ when that maximum is infinite. Basis bijections then give categoricity in every uncountable cardinal. Quantifier elimination also shows that these different positive dimensions are elementarily equivalent; independence of an arbitrarily long tuple is not being silently treated as definable over this infinite field.

Yes. The pure order on the reals is a dense linear order without endpoints and has quantifier elimination for dense linear orders without endpoints. Over a finite parameter set $a_1<\cdots <a_m$, a complete one-variable type either specifies equality to one of these points or specifies one of the open intervals and exterior rays they determine.

Each such interval or ray contains a real number. Hence every complete one-variable type over a finite parameter set is realized. The successive-coordinate characterization gives saturation for all finite tuples. This uses finite parameter sets; it does not say that arbitrary types over a countably infinite set of parameters are realized.

A pointwise periodic self-map is a self-map $f:X\to X$ for which every $x\in X$ satisfies $f^{n_x}(x)=x$ for some positive integer $n_x$ depending on $x$. Powers denote repeated function composition.

Such a self-map is a bijection. For surjectivity, $f^{n_x-1}(x)$ is a preimage of $x$. For injectivity, suppose $f(x)=f(y)$ and choose a common multiple $N$ of periods of $x$ and $y$; applying $f^{N-1}$ gives $x=y$. Thus $X$ decomposes into disjoint finite permutation cycles.

There need not be one positive $N$ with $f^N=\mathrm{id}$ on an infinite set. Take the disjoint union of one cycle of every positive length: any such $N$ would have to be divisible by every positive integer. On a finite set, in contrast, the least common multiple of the finitely many cycle lengths is a common period.