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A sigma-algebra on a set $\Omega$ is a collection $\mathcal{F}$ of subsets such that $\Omega \in \mathcal{F}$, $A\in \mathcal{F}$ implies $\Omega \setminus A\in \mathcal{F}$, and $A_n\in \mathcal{F}$ for all $n$ implies ${\bigcup }_n{A}_n\in \mathcal{F}$. Thus it contains the empty set and is closed under countable intersections as well. This is precisely the closure needed to apply countable additivity without leaving the domain of a measure.

For $\Omega =\{a,b,c\}$,is a sigma-algebra. It describes observations that distinguish $a$ from the pair $\{b,c\}$, but cannot distinguish $b$ from $c$. The power set provides a finer sigma-algebra with all eight subsets. Both are valid; choosing a sigma-algebra records which questions the model can answer.

On the real numbers, the Borel sigma-algebra contains all open sets and everything generated from them by these operations. The Lebesgue measurable sets form a larger sigma-algebra, including all subsets of Borel null sets. Closure under countable unions accommodates constructions involving an entire sequence of observations or approximations.

Here simple random sampling means a random unordered subset $A$ of exactly $n$ distinct labels from $\{1,\dots ,N\}$, with every size-$n$ subset equally likely. The sample size is fixed in advance, $1\le n\le N$, and sampling without replacement prevents duplicate units. This is SRSWOR; some texts also use "simple random sampling" for independent uniform draws with replacement, which is a different design.^[ref]

Every label has inclusion probability $n/N$, but equal inclusion probabilities alone are insufficient. For example, from six labels choose either $\{1,2,3\}$ or $\{4,5,6\}$ with probability $1/2$. Each label is included with probability $1/2$, yet eighteen of the twenty size-three subsets are impossible: this is a two-group cluster sampling design. Unequal probability sampling deliberately changes inclusion probabilities, often to improve precision for a particular estimand, and generally requires an appropriately weighted estimator rather than the plain sample mean.

Consider the original mathematical population with labelled values $(2,4,6,8,10,12)$, and let $n=2$. The fifteen equiprobable subsets, followed by their sample means, areTheir sample means sum to $105$, giving expected value $105/15=7$. Their squared deviations from $7$ sum to $70$, giving variance $70/15=14/3$. This is an enumeration over hypothetical selections from six stipulated numbers, not measured survey data.

With independent uniform sampling with replacement, the $6^2=36$ ordered outcomes each have probability $1/36$. The two orders $(2,4)$ and $(4,2)$ are distinct outcomes, and $(2,2)$ is now possible. Forgetting order does not make the resulting unordered multisets equiprobable: $\{2,4\}$ has probability $2/36$ but $\{2,2\}$ has probability $1/36$. The thirty-six sample means still have expected value $7$, but their squared deviations sum to $210$, giving variance $210/36=35/6$. Avoiding duplicate labels improves precision for this fixed population.

For a group action of a group $G$ on a set $X$, the stabilizer subgroup of $x\in X$ consists of the transformations that leave $x$ fixed:It is a subgroup: the identity element fixes $x$, products of transformations fixing $x$ fix $x$, and the inverse element of a transformation fixing $x$ also fixes $x$. A stabilizer subgroup is a subset of $G$, whereas the orbit of a group action is a subset of $X$. See Ackerman's algebra notes, definition 0.4.1.2.

For the square group actions defined under group action, direct calculation givesThe reflection $s$ fixes $A$ and $C$ individually. The reflection $rs$ preserves the unoriented side of a polygon $AB$ by exchanging its two vertices of a polygon. The half-turn rotation $r^2$ exchanges the two ends of the unoriented diagonal of a polygon $AC$ while preserving that diagonal of a polygon. Thus fixing a side of a polygon or diagonal of a polygon as an object does not mean fixing every point on it; the latter condition uses a pointwise stabilizer. The four stabilizer subgroups have group orders $2,2,4,8$, matching group orbit sizes $4,4,2,1$ through the orbit-stabilizer theorem.

A stabilizer subgroup records the transformations still available after imposing a constraint. Fixing a vertex of a polygon of the square leaves two choices; fixing its centroid leaves all eight. In a faithful group action, a stabilizer subgroup can still contain a nonidentity transformation, as $s\in (D_4{)}_A$ shows: a faithful group action requires that only the identity element fix every point of $X$ simultaneously, not that every individual stabilizer subgroup be trivial. The two-point cyclic group example under group action has both stabilizer subgroups equal to the group kernel $\{0,2\}$ of its associated group homomorphism.

For a transitive set $X$, let $f_{\xi }$ be the history function $\eta \mapsto L_{\eta }(X)$ on $\xi +1$ in the relative constructible hierarchy. By transfinite induction, $f_{\xi }\in L_{\xi +k}(X)$ for some finite $k$, possibly depending on $\xi$. The base and successor steps follow by forming finite ordered pairs and adjoining the next entry, using $L_{\xi +1}(X)\in L_{\xi +2}(X)$. These operations require finitely many subsequent definable power sets.

At a nonzero limit ordinal $\lambda$, the inductive bounds put every earlier $f_{\eta }$ in $L_{\lambda }(X)$, since $\eta +k<\lambda$. This level satisfies finite relation closure for set-theoretic coding, independently of the availability of history functions. Consequently its coded relative constructible stage predicate computes the endpoints correctly. Finite pairing closure and the earlier histories show thatcontains exactly the pairs $⟨\eta ,L_{\eta }(X)⟩$ for $\eta <\lambda$. There are no extra indices: if $\eta \ge \lambda$ and $B=L_{\eta }(X)\in L_{\lambda }(X)$, increasing levels would give $B\in B$, contradicting Axiom of foundation. Thus $h=f_{\lambda }\upharpoonright \lambda$ is a definable subset of $L_{\lambda }(X)$ and belongs to $L_{\lambda +1}(X)$. Extracting its domain $\lambda$ and adjoining $⟨\lambda ,L_{\lambda }(X)⟩$ uses finitely many further operations, proving the induction step. Every nonzero limit $\alpha$ therefore contains all $f_{\xi }$ with $\xi <\alpha$, as required by relative constructible level recognition.

For any nonzero limit ordinal $\delta$, a club set is an unbounded subset closed under limit points below $\delta$. A subset is stationary if it meets every such club set. Statements concerning the completeness of the club filter or Fodor lemma require regular uncountable height; they should not be transferred to arbitrary singular or countable limit ordinals without checking their hypotheses.

A sequence $D_{\alpha }\subseteq \alpha$, indexed by a stationary subset $S$ of a regular uncountable cardinal $\lambda$, guesses every $X\subseteq \lambda$ stationarily often: $\{\alpha \in S:D_{\alpha }=X\cap \alpha \}$ is stationary. This extends the usual stationary diamond principle at ${\omega }_1$.

For an uncountable regular cardinal $\kappa$, choose cofinal sequences $c_{\alpha }:\omega \to \alpha$ for the stationary set of ordinals of cofinality $\omega$. Above any bound, some fixed coordinate exceeds the bound on a stationary subset; Fodor lemma makes that coordinate constant on a stationary subset. Thus the stationary constant fibers, over all coordinates, have unboundedly many values. Regularity makes one coordinate have $\kappa$ such values. Its fibers are disjoint stationary sets; adding all leftover ordinals to one piece partitions $\kappa$.

For nonnegative integers $n,k$, the Stirling number of the second kindcounts the set partitions of an $n$-element finite set into exactly $k$ unlabelled blocks of a set partition. Set $S(0,0)=1$, $S(n,0)=0$ for $n>0$, and $S(n,k)=0$ for $k>n$. Also $S(n,1)=S(n,n)=1$ for $n\ge 1$. Grassl and Levin develop these counts through explicit set partitions and a recurrence relation. ^[ref]

Insert one distinguished element. For $n\ge 1$ and $1\le k\le n$, the Stirling numbers of the second kind obeyProof outline. Inspect the block of a set partition containing $n$. If it contains another element, removing $n$ leaves a set partition into $k$ blocks of a set partition; conversely, insert $n$ into any of those $k$ blocks of a set partition. If $n$ is alone, remove its singleton set and obtain a set partition into $k-1$ blocks of a set partition. The cases cannot overlap, and deletion reverses each construction. This explains the factor $k$ even though the blocks of a set partition have no names. ^[ref]

For $n=4,k=2$, start with the three two-block set partitions of $\{1,2,3\}$. Inserting $4$ into either block of a set partition gives six possibilities. The remaining possibility is $\{\{1,2,3\},\{4\}\}$, obtained by making $4$ a singleton set. Hence $S(4,2)=2\cdot 3+1=7$. The seven set partitions in the diagram comprise four with sizes $3+1$ and three with sizes $2+2$; this grouping by sizes independently gives $4+3=7$.

Three different counting questions. The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ selects one $k$-element subset. The Stirling number of the second kind $S(n,k)$ groups all $n$ distinct elements into $k$ nonempty unnamed blocks of a set partition. The power $k^n$ counts all functions to a specified $k$-element finite set, allowing some destination labels to go unused. At $n=4,k=2$, these counts are respectively $6$, $7$, and $16$. Choosing two people, making two unnamed nonempty teams, and assigning each person a red or blue label are different tasks.

When the destinations have labels. For fixed finite sets $X,Y$ of sizes $n,k$, respectively, the number of surjective functions $X\to Y$ isHere every element of $Y$ must be used. Each surjective function gives a set partition of $X$ into its nonempty preimages $f^{-1}(\{y\})$. Conversely, assign the $k$ distinct destination labels to the $k$ blocks of a set partition in $k!$ ways, where $k!$ is a factorial. For two labelled teams of four people this gives $2!\cdot 7=14$: of the $2^4=16$ assignments, exactly two use only one label. Dividing arbitrary function counts by $k!$ fails in general because unused labels can be permuted without changing a function. For example, $3^4/3!=13.5$ cannot count set partitions. The empty-to-empty function is a surjection, consistent with $0!S(0,0)=1$. Stanley gives the bijection underlying this qualification.^[ref]

For example, the inclusion-exclusion principle counts the surjections from four elements to three labelled destinations as $3^4-3\cdot 2^4+3\cdot 1^4=36$. Thus $S(4,3)=36/3!=6$. An independent check chooses the only two-element block of a set partition in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways; the other two elements form singleton sets.

Why these numbers also occur in algebra. Classifying all functions to an $m$-element finite set by the size $k$ of their image givesFor a fixed set partition, the falling factorial $(m{)}_k$ counts assignments of distinct labels to its $k$ blocks of a set partition, and is zero when $k>m$. At $n=3,m=4$, this reads $64=1\cdot 4+3\cdot 12+1\cdot 24$. Replacing $m$ by $x$ yields a polynomial identity: both sides agree at every nonnegative integer. Thus the Stirling numbers of the second kind translate powers into falling factorials. Grouping by equal values turns a counting rule into an algebraic change of coordinates. Summing $S(n,k)$ without destination labels instead gives the Bell number.

A subgroup $H\le G$ is a subset of a group $G$ that is itself a group under the same group operation. It must contain the identity element of $G$ and be closed under products and inverse elements. Associativity is inherited from $G$, so it needs no new proof. The whole group and the trivial group $\{e\}$ are always subgroups. A proper subgroup satisfies $H\ne G$.

A useful single-condition test isTo see why it works, choose $a\in H$; then $a{a}^{-1}=e\in H$. Taking $a=e$ gives every inverse element $b^{-1}\in H$. Applying the test to $a$ and $b^{-1}$ then gives $ab\in H$. Conversely, every subgroup satisfies the test by closure under products and inverse elements. This also makes the additive version memorable: a nonempty subset of an additive group is a subgroup exactly when it is closed under differences. See McKernan's MIT lecture 2, Proposition 2.3, for the equivalent closure criterion.

For the square dihedral group $D_4$ above, $R=\{e,r,r^2,r^3\}$ is a cyclic subgroup: products add exponents modulo four, and $(r^k{)}^{-1}=r^{-k}$. The smaller collections $\{e,r^2\}$ and $\{e,s\}$ are subgroups of group order two. But $\{e,r\}$ is not a subgroup, since $r\cdot r=r^2$ lies outside it. A plausible-looking list of symmetries therefore need not constitute a complete collection of reversible operations.

Inside $(\mathbb{Z},+)$, the even numbers $2\mathbb{Z}$ form a subgroup: $2m-2n=2(m-n)$ is an even number. Its two cosets consist of the even numbers and the odd numbers, so its index of a subgroup is two. The parity group homomorphism $k\mapsto k\mathrm{mod}2$ has $2\mathbb{Z}$ as its kernel of a group homomorphism. A subgroup can thus identify exactly which information a composition-preserving map forgets.

Subgroups also describe residual symmetry. In the group action of $D_4$ on the four vertices, the stabilizer subgroup of $(1,1)$ is $\{e,sr\}$: the diagonal reflection $sr$ fixes that vertex, and the other six motions move it. Requiring a feature to stay fixed still leaves a group, because both composition and reversal preserve that requirement. This is why subgroups are useful for studying a group action one feature at a time.

A system of distinct representatives for a finite indexed family of finite sets $(A_i{)}_{i\in I}$ chooses one element $x_i\in A_i$ per index, with all choices distinct. Form a bipartite graph with left vertex set $I$, right vertex set ${\bigcup }_{i\in I}{A}_i$, and edges $ix$ whenever $x\in A_i$. Matching saturation of $I$ is precisely a system of distinct representatives. Hence Hall's marriage theorem says that such choices exist exactly when $|{\bigcup }_{i\in J}{A}_i|\ge |J|$ for every subset $J\subseteq I$. Repeated equal sets still have separate indices and separate demands.

For $n\ge 1$, a topological $n$-manifold with boundary is a Hausdorff second-countable space locally homeomorphic to relatively open sets in the half-space ${\mathbb{H}}^n$. The openness here is in the subspace topology of ${\mathbb{H}}^n$, not necessarily in ${\mathbb{R}}^n$. Its intrinsic boundary $\partial M$ consists of points sent to $x_n=0$; the remaining points have ordinary coordinate charts with images open in Euclidean space. The intrinsic boundary is independent of the chosen chart.^[ref]

The closed interval $[0,1]$ illustrates the distinction: its interior points have open interval charts, while $0$ and $1$ have charts modeled on $[0,\epsilon )\subseteq {\mathbb{H}}^1$, using $x$ and $1-x$. Its intrinsic boundary is $\{0,1\}$. The circle has empty intrinsic boundary even though its boundary of a set as a subset of the Euclidean plane is the whole circle. A topological manifold without boundary is therefore a special case with $\partial M=\varnothing$, and a smooth atlas on the half-space charts gives a manifold with boundary in the smooth sense.

A topological space is a set $X$ together with a topology $\tau \subseteq \mathcal{P}(X)$, a family of subsets satisfyingHere $\mathcal{P}(X)$ is the power set; the union may have arbitrarily many members, while the intersection must be finite. The members of $\tau$ are the open sets. These are axioms about subsets, with no assumed metric or surrounding Euclidean space. ^[ref]

For $X=\{a,b\}$, the discrete topology is ${\tau }_d=\{\varnothing ,\{a\},\{b\},X\}$, while the indiscrete topology is ${\tau }_i=\{\varnothing ,X\}$. Both satisfy the axioms. They describe different topological spaces on the same set: the first is Hausdorff and disconnected, while the second is connected and not Hausdorff. Each is compact because $X$ is a finite set.

The identity function $i:(X,{\tau }_d)\to (X,{\tau }_i)$ is a continuous map: the only target open sets have inverse images $\varnothing$ and $X$. Its inverse is not a continuous map, since the target open set $\{a\}$ has inverse image $\{a\}$, which is not an open set in ${\tau }_i$. Thus even a continuous bijection need not be a homeomorphism.