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Zorn lemma states that a partially ordered set in which every chain in a partially ordered set has an upper bound has a maximal element. Include the empty chain in a partially ordered set, so the hypotheses imply that the set $P$ is nonempty. Suppose there were no maximal element. Every chain in a partially ordered set $C$ has an upper bound $u$, and some $v>u$; therefore the set of elements strictly above every member of $C$ is nonempty. The axiom of choice selects one such element $f(C)$ for each chain in a partially ordered set (and an element of $P$ when $C$ is empty).

By Hartogs theorem, there is an ordinal $\kappa$ which does not inject into $P$. Transfinite recursion defines $x_{\alpha }=f(\{x_{\beta }:\beta <\alpha \})$ for all $\alpha <\kappa$: previous elements form a strictly increasing chain in a partially ordered set, including at limit ordinal stages. All $x_{\alpha }$ are distinct, giving the forbidden injection. This proves Zorn lemma; the choice of $f$ is the use of the axiom of choice.

Apply Zorn lemma to linearly independent subsets of $\mathbb{R}$ over $\mathbb{Q}$, ordered by inclusion. A union along a chain in a partially ordered set is linearly independent because each finite linear relation occurs in one member of that chain in a partially ordered set. A maximal independent set spans $\mathbb{R}$, since a vector outside its span could be adjoined. Thus $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$.

For a finite basis, the Steinitz exchange lemma proves equality of basis sizes. This also rules out an infinite basis when a finite spanning set exists: write each member of the finite spanning set using finitely many elements of another basis; their finite union spans, and independence leaves no other basis elements. If a basis $B$ is infinite with cardinality $\lambda$, the finite-support description of every vector and countability of $\mathbb{Q}$ giveHere infinite cardinal arithmetic uses the axiom of choice. Thus any infinite basis has cardinality $|V|$, proving all bases have the same cardinality. The zero space has its unique empty basis.

Use singular homology with integer coefficients. The Künneth theorem for triangulable spaces gives a split short exact sequence with tensor terms ${⨁}_{i+j=k}{H}_i(X)\otimes H_j(Y)$ and torsion terms ${⨁}_{i+j=k-1}{\mathrm{Tor}}_1(H_i(X),H_j(Y))$. When one factor's homology groups are free abelian, the torsion terms vanish and the tensor map is an isomorphism.

The circle has $H_0(S^1)=H_1(S^1)=\mathbb{Z}$ and all other groups zero. Inductively, the homology groups of a torus are free, andPascal's identity givesGeometrically the generators correspond to choosing $k$ of the $n$ circle factors and taking their product, consistent with the cellular model having one $k$-cell for each such subset and cancelling opposite boundary faces.

Let $F,G$ be disjoint closed subsets of $K$. They are compact. For each $x\in F,y\in G$, the Hausdorff space condition gives disjoint open neighbourhoods $U_{xy},V_{xy}$. For fixed $x$, finitely many $V_{xy}$ cover $G$; intersect the corresponding $U_{xy}$ and take the union of those $V_{xy}$ to obtain disjoint open sets $U_x\ni x$ and $V_x\supset G$. Finitely many $U_x$ cover $F$. Their union and the intersection of the corresponding $V_x$ are disjoint open neighbourhoods of $F,G$. The case of an empty set is immediate. Hence $K$ is a normal topological space.

Urysohn lemma states that in a normal topological space satisfying $T_1$, disjoint closed sets $F,G$ admit a continuous function $h:K\to [0,1]$ with $h=0$ on $F$ and $h=1$ on $G$.

For algebraic varieties that are irreducible topological spaces, a rational map $X⇢Y$ is an equivalence class of morphisms $U\to Y$ on nonempty open subsets of $X$, with two representatives equivalent if they agree on a nonempty open intersection. Such subsets are dense. A birational map has a rational inverse; equivalently it restricts to an isomorphism between dense open subsets. In function field language it induces an isomorphism $k(Y)\cong k(X)$ over the ground field.

U+D. The natural numbers are absolute between transitive models of ZFC, and the statement $x\subseteq \mathbb{N}$ is the bounded assertion $\forall y\in x(y\in \mathbb{N})$. Hence the property of being a subset of $\mathbb{N}$ is absolute.

The definable power set isThus it contains exactly the subsets of $A$ definable over the structure $(A,\in )$ using finitely many parameters from $A$.

Define the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ as the number of $r$-element subsets of an $n$-element set, equivalently as $n!/[r!(n-r)!]$. Partitioning the $r$-element subsets of $\{1,\dots ,n+1\}$ according to whether they contain $n+1$ proves Pascal's identityThus the forward difference operator satisfies

The restricted axiom of choice ${\mathsf{AC}}_X(Y)$ says that for every $X$-indexed family $(A_x{)}_{x\in X}$ of nonempty subsets of $Y$, there is a choice function $c:X\to Y$ such that $c(x)\in A_x$ for every $x\in X$.

A path component of a topological space $X$ is a nonempty subset maximal among its path-connected subsets. The path component through $x$ can be described asEvery point along such a continuous path also belongs to $P_x$: the initial part of the path reaches that point. Thus the paths witnessing membership stay inside $P_x$. This makes path components useful for deciding which allowed positions can be reached from one another by continuous motion.

Every path component lies inside one connected component, because every path-connected space is connected. Thus each connected component is a union of path components. In a locally path-connected space the two classifications agree, by connected locally path-connected spaces are path-connected. For a subset of the real line with its usual subspace topology, the classifications also agree: every connected component is a real interval, and interpolation supplies a continuous path within it.

The topologist's sine curve shows why the distinction matters in general. Use specifically the closed versionwith the subspace topology from the usual Euclidean plane. It has one connected component, $T$, but two path components, $G$ and $L$. The graph of a function $G$ is a continuous image of a connected space, and its closure is $T$, so closure of a connected set gives connectedness. Each of $G,L$ is path-connected, but a continuous path crossing from $L$ into $G$ would force its second coordinate to oscillate between $1$ and $-1$ arbitrarily near its departure time, contradicting continuity; the topologist's sine curve article gives the short argument. Also $G$ is not closed in $T$, so path components need not be closed, unlike connected components.^[ref]

Being joined by a continuous path is an equivalence relation. Proof outline: the constant path joins a point to itself; $t\mapsto \gamma (1-t)$ reverses a path; and a path from $x$ to $y$ followed by one from $y$ to $z$ gives a path from $x$ to $z$. For the last construction, run each path on half of $[0,1]$; their values agree at $1/2$, so the pasting lemma for closed subspaces gives continuity. The equivalence classes are path-connected, are maximal, and form a partition of a set of $X$ into path components.^[ref]

Suppose nontrivial groups $G_1,G_2$ act on a set and there are disjoint nonempty subsets $X_1,X_2$ such that every nonidentity element of $G_1$ maps $X_2$ into $X_1$, while every nonidentity element of $G_2$ maps $X_1$ into $X_2$. Assume at least one factor has at least three elements. Then the subgroup generated by $G_1$ and $G_2$ is their free product. The cardinality hypothesis excludes the counterexample in which $G_1=G_2=C_2$ acts by swapping $X_1$ and $X_2$. Versions allowing both factors to have order two need an additional separation hypothesis.

The power set $\mathcal{P}(X)$ is the set of all subsets of $X$, including the empty set and $X$ itself. It is useful whenever an object is a selection rather than an individual member: a possible collection of active vertices in a graph, for example, belongs to the power set of the vertex set.

For $X=\{a,b,c\}$, the power set has eight members:Each of the three set elements is either included or excluded. More generally, if $|X|=n<\infty$, these $n$ independent binary choices give $|\mathcal{P}(X)|=2^n$. This is a bijection between subsets and binary strings of length $n$, not a probabilistic independence assumption.

Ordering the subsets by inclusion gives a Boolean lattice. In the picture an upward edge adds one member. Complementation reverses inclusion, and union and intersection combine selections. The empty-input case is worth checking: $\mathcal{P}(\varnothing )=\{\varnothing \}$ has one member, not zero, consistently with $2^0=1$.

For a forcing name $\tau$, let $U$ consist of pairs $(\rho ,p)$ where $(\rho ,r)\in \tau$ for some $r\ge p$. Every ground-model subset $\nu$ of $U$ is a name whose value is contained in ${\tau }^G$. Every subset ${\sigma }^G\subseteq {\tau }^G$ has an equivalent such name: retain those $(\rho ,p)\in U$ with $p⊩\rho \in \sigma$. The atomic membership truth lemma for forcing proves equality of values. Collect all these names from the ground-model power set ${\mathcal{P}}^M(U)$ into a single outer name, pairing each with every condition. Its value is the full internal power set, without presupposing that power set in the extension.

For a function $f:X\to Y$ and a subset $U\subseteq Y$, its preimage isIt is a subset of the original domain of a function. The notation does not require an inverse function: the preimage of a singleton set can have no members, one member, or many members. In the finite example, $f^{-1}(\{0\})=\{a,b\}$, $f^{-1}(\{1\})=\{c\}$ and $f^{-1}(\{3\})=\varnothing$. The preimage of a singleton set is also its fiber of a function.

For subsets $U,V\subseteq Y$, preimages preserve both unions and intersections:For the intersection, the same input $x$ belongs precisely when its single output $f(x)$ belongs to both $U$ and $V$. There is no opportunity to choose different witnesses, which is what made the image counterexample fail. The identities also hold for arbitrary families of output subsets.

In the finite example, choose $U=\{0,1\}$ and $V=\{0,2\}$. ThenThe intersection of the two preimages is exactly the same set $\{a,b\}$. Complements are preserved with respect to the appropriate source and target sets:For example, the unused output three contributes no extra input: with $U=\{0,2\}$, both sides of the complement identity are $\{c\}$. See Lebl, Proposition 0.3.15: preimages of unions, intersections and complements.

This makes preimages useful for transporting conditions. The solution set of $g(x)=0$ is $g^{-1}(\{0\})$, and a zero set is a preimage of zero. Requiring two output conditions takes the intersection of their preimages. In topology, a continuous function pulls every open set back to an open set; a measurable function pulls the specified measurable sets in its codomain back to measurable sets in its domain of a function. These definitions use preimages because their set operations remain consistent even when the function is noninjective.