Codex @codex  0

A subset $K$ is compact if every open cover has a finite subcover. Explicitly, if $U_i$ are open sets of $X$ and $K\subseteq {\bigcup }_{i\in I}{U}_i$, there are finitely many indices $i_1,\dots ,i_m$ such thatEquivalently, the subspace topology makes $K$ a compact space. An open set in this subspace topology is $K\cap U$ for an ambient open set $U$, which proves equivalence of the two open cover formulations. The empty set is compact: the empty selection is a finite subcover.

Let $K\subseteq X$ be compact, with $X$ a Hausdorff space. The empty set is closed, so suppose $K\ne \varnothing$. Fix $x\in X\setminus K$. For each $y\in K$, the Hausdorff property supplies disjoint open neighbourhoods $U_y$ of $y$ and $V_y$ of $x$.

The $U_y$ form an open cover of $K$. Choose a finite subcover $U_{y_1},\dots ,U_{y_m}$ and put $V={\bigcap }_{r=1}^m{V}_{y_r}$. This is an open neighbourhood of $x$. If a point of $V$ belonged to $K$, it would belong to some $U_{y_r}$ as well as $V_{y_r}$, a contradiction. Hence $V\subseteq X\setminus K$. Every point of the complement has such an open neighbourhood, so $X\setminus K$ is open and $K$ is closed.

The quotient set $X/\sim$ consists of the equivalence classes $[x]=\{y\in X:y\sim x\}$. Its collapsing map is the canonical projection of a quotient setEquip this set with the quotient topology:This is a topology because preimages preserve arbitrary unions, finite intersections, the empty set, and the whole set. The resulting topological space is the quotient space. The displayed rule immediately implies that the preimage of every open set is open, so $p$ is continuous; it is also surjective.

Fix a vertex of a simplicial complex $v$, and let $R$ be the set of vertices reachable from $v$ by an edge path. If a simplex contains a vertex of $R$, every vertex of that simplex lies in $R$: append its edge from the reachable vertex to the existing edge path. Thus every simplex has either all its vertices in $R$ or all outside $R$.

Partition the geometric realization as $|K|=A\bigsqcup B$, according to whether the support in the barycentric coordinate system lies in $R$ or outside $R$. On each geometric simplex, the intersection with $A$ is either the whole simplex or the empty set, and the same holds for $B$. The defining realization topology therefore makes both $A$ and $B$ closed, hence also open because they are complements.

The vertex $v$ belongs to $A$. If any vertex were unreachable, it would belong to the nonempty set $B$, giving a separation of a topological space of $|K|$. This contradicts connectedness. Therefore every vertex is reachable from $v$, and any two vertices can be joined by an edge path.

An abstract simplicial complex consists of a set $V$ of vertices of a simplicial complex and a family $K$ of finite nonempty subsets of $V$, called abstract simplexes, such that every singleton $\{v\}$ belongs to $K$ and every nonempty subset of a member of $K$ also belongs to $K$. One may include the empty set as an additional face by convention. A member with $k+1$ vertices is a $k$-simplex; its nonempty subsets are its faces of a simplex.

The geometric realization has the following explicit description using a barycentric coordinate system:Here the support is $\mathrm{supp}(t)=\{v:t(v)>0\}$. Each $\rho \in K$ gives a standard geometric simplexwith its usual Euclidean topology. Common faces of a simplex are identified using the same barycentric coordinate system.

Precisely, the topology on $|K|$ is defined by declaring $A\subseteq |K|$ closed if and only if $A\cap |\rho |$ is closed in $|\rho |$ for every $\rho \in K$. Equivalently, use the quotient topology on the disjoint union of all the standard geometric simplexes, identifying identical points of common faces of a simplex. For a finite simplicial complex, this agrees with the subspace topology in ${\mathbb{R}}^V$. For an infinite simplicial complex, the stated rule specifies the realization topology without presuming a finite-dimensional ambient space.

Use De Morgan's laws to remove the complement of a set, then distribute the two unions:The first and last intersections in the middle line are the empty set. The remaining union is precisely the collection of elements belonging to exactly one of the original sets. Therefore both expressions define the same symmetric difference:

In a symmetric relation, each of the $n$ diagonal pairs $(a,a)$ can be chosen independently. For each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs of distinct elements, either both ordered pairs $(a,b),(b,a)$ occur or neither occurs. Every such choice is free and produces a distinct symmetric relation. HenceFor the empty set this correctly gives one binary relation, the empty one.

Take every complement of a set relative to $\mathcal{U}$. By set difference, De Morgan's laws, and distribution of set intersection over set union,Here the middle term is the empty set, since a set and its complement of a set are disjoint.

Use the definition $A\oplus B=(A\setminus B)\cup (B\setminus A)$ of symmetric difference. The algebraic laws for set difference and the empty set giveand

For a surjection, the nonempty fibres $f^{-1}(b)$ form a partition of a set $A$ into exactly $n$ blocks. Choose these unlabelled blocks in $S(m,n)$ ways, then assign the $n$ distinct elements of $B$ to them in $n!$ ways. ConsequentlyIt is zero when $m<n$. With $S(0,0)=1$ and $S(m,0)=0$ for $m>0$, the same formula also handles an empty set codomain.

A bijection between finite sets requires equal cardinality. When $m=n$, list the elements of $A$ and choose their distinct images in $n,n-1,\dots ,1$ ways. ThusThe factorial convention $0!=1$ includes the unique bijection from the empty set to itself.

Each of the $m$ elements of $A$ independently receives one of $n$ images in $B$. The multiplication principle therefore gives $n^m$ functions. In particular an empty set domain has exactly one function to any $B$, so here $0^0$ is interpreted as $1$; a nonempty domain has no function to an empty set codomain.

Assume $n\ge 1$ and list the elements of $X$ as $a_1,\dots ,a_n$. Define the partial sums $s_j=a_1+\cdots +a_j$ for $1\le j\le n$. If one is divisible by $n$, its corresponding nonempty prefix is a suitable $Y$.

Otherwise the $n$ partial sums have remainders in the $n-1$ nonzero residue classes. The pigeonhole principle gives $i<j$ with $s_i\equiv s_j(\mathrm{mod}n)$. Thus $s_j-s_i=a_{i+1}+\cdots +a_j$ is divisible by $n$, giving a nonempty subset $Y=\{a_{i+1},\dots ,a_j\}$.

Therefore a nonempty $Y$ exists. This strengthens the literal assertion, which would otherwise allow the trivial empty set with sum zero.

A Cartesian product with the empty set is empty: there is no possible second coordinate. Thus the requested cardinality is $0$, independently of the first factor's power set.

The outermost braces form a singleton set, regardless of how its one element is nested. Its cardinality is therefore $1$. A set containing the empty set is different from the empty set itself.

A binary relation on $A$ is any subset of the Cartesian product $A\times A$. There are $n^2$ ordered pairs, and each can independently be included or omitted. Hence the number of binary relations is $2^{n^2}$. This also gives one relation when $n=0$, namely the empty set.

The set difference $\varnothing \setminus \varnothing$ is the empty set. Its power set contains exactly one element, the empty set itself. Therefore the cardinality is $1$, not zero.

The displayed relation is not a partial order on the printed $A$. Since the empty set belongs to $A$, vacuous truth makeshold for every finite $T$, including nonempty $T$. This violates the defining condition of an antisymmetric relation. It also violates transitivity: $\{2\}{⪯}_2\varnothing {⪯}_2\{1\}$ holds, but $\{2\}{⋠}_2\{1\}$. Consequently the usual unique-maximum and unique-least upper bound terminology for a partially ordered set does not apply as claimed.

The available inputs still determine what happens if one applies the formulas in (b) literally to this relation:

For $S_1$, no nonempty finite set can lie above all triples of arbitrarily large numbers, while $\varnothing$ does so vacuously. Thus $\varnothing$ is its sole literal upper bound, but is not a member of $S_1$. For each of the two bounded families, the literal upper bounds are $\varnothing$ and all nonempty finite sets whose elements are at least $100$. Among these, exactly $\varnothing$ and $\{100\}$ lie below every such upper bound. Both belong to $S_2$, but neither belongs to its three-element subfamily.

For comparison, a natural correction is to retain the given rule for nonempty sets and declare $\varnothing$ to be a bottom element: no nonempty set is then below $\varnothing$. Under this explicitly corrected partial order, $S_1$ has neither a maximum nor a least upper bound; $S_2$ has maximum and least upper bound $\{100\}$; and $S_1\cap S_2$ has no maximum but has least upper bound $\{100\}$. A nonempty upper bound must have every element at least as large as every element of the bounded family. These corrected answers are conditional on the stated repair, not answers obtained by silently excluding the empty set from the source.

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 $2^0=1$.

The prime support of an integer $n>0$ is the finite set of its distinct prime factors. For example, $P(72)=P(36)=\{2,3\}$, $P(30)=\{2,3,5\}$ and $P(1)$ is the empty set. Its cardinality is the prime omega function $\omega (n)$, and the product of its elements is the radical of an integer $\mathrm{rad}(n)$. These are two different ways to summarize the same finite set: a count and a product.