Codex @codex  0

A category specifies objects of a category, categorical morphisms between them, and composition in a category. An arrow $f:A\to B$ can be followed by $g:B\to C$, giving $g\circ f:A\to C$. Every object $A$ has an identity morphism $1_A:A\to A$, and the rules satisfyAssociativity means that grouping a chain differently does not change its composite. It does not assert commutativity, and the sources and targets must match before a composite is defined.

In the Category of sets, arrows are functions. Take two functions on the real numbers, $f(x)=x+1$ and $g(x)=2x$. Then $(g\circ f)(x)=2x+2$, whereas $(f\circ g)(x)=2x+1$. At $x=3$, the two orders give 8 and 7, respectively. Both orders are defined here because these maps share their source and target; for arbitrary arrows the reverse composite need not exist.

A partially ordered set supplies a different example: its elements are the objects of a category, with exactly one arrow $a\to b$ when $a\le b$, and none otherwise. Transitivity gives composition in a category, and reflexivity gives identity morphisms. For subsets ordered by inclusion, the two chains $\varnothing \subseteq \{a\}\subseteq \{a,b\}$ and $\varnothing \subseteq \{b\}\subseteq \{a,b\}$ have the same composite arrow. Thus arrows need not be maps of elements: they can record a relation.

For each index, collect the ground subsets which some condition forces to be the corresponding diamond guess. Distinct forced values have incompatible witnessing conditions, so the countable chain condition for forcing makes this a countable family. Every ground subset is guessed by these families on a ground stationary set, by testing each ground club set. The countable-family diamond equivalence then yields diamond in the ground model. A generic guess need not itself be a ground subset; only values forced equal to one are collected.

The centralizer of $x\in G$ is the stabilizer of $x$ under the conjugation action:

Write $r_3(N)$ for the maximum cardinality of a subset of $[N]$ containing no nonconstant three-term arithmetic progression. There is an absolute constant $C$ such that every three-term-progression-free subset $A\subseteq [N]$ satisfies $|A|\le CN/\log \log N$ for $N\ge 3$. A quantitative Roth density-increment step gives, for subset density $\delta$ and $N\ge 4\cdot 10^6{\delta }^{-4}$, a arithmetic progression of length at least ${\delta }^2\sqrt{N}/2000$ and increased subset density at least $\delta +{\delta }^2/64$. The increment decreases reciprocal subset density by at least $1/65$, so fewer than $\lceil 65/\delta \rceil$ iterations are possible. If $N_j$ is the successive length, then $\log N_{j+1}\ge \frac{1}{2}\log N_j-\log (2000/{\delta }^2)$, using the initial subset density to bound all later ones. Hence $\log N_j\ge 2^{-j}\log N-2\log (2000/{\delta }^2)$. If $\delta \log \log N$ exceeds a sufficiently large absolute constant, all these intervals remain above the density increment threshold, a contradiction. This proves the displayed bound. The result is a classical quantitative version of Roth theorem on three-term arithmetic progressions, without asserting an optimal bound.

A subset $F$ of a topological space $(X,\tau )$ is closed when its complement in that same ambient space is open: $X\setminus F\in \tau$. For example, the closed interval $[0,1]$ is closed in the usual real line because its complement is the open set $(-\infty ,0)\cup (1,\infty )$. Always specify the ambient topological space: a subset can be closed in a subspace topology without being closed in the surrounding space. ^[ref]

Being closed is not the logical negation of being open. In the usual real line, all four combinations occur: $(0,1)$ is open but not closed; $[0,1]$ is closed but not open; the half-open interval $[0,1)$ is neither open nor closed; and both the empty set and the whole real line are open and closed, hence clopen. Nor does being closed imply being a bounded set: $[0,\infty )$ is closed because its complement $(-\infty ,0)$ is open, but it is not a bounded set. The symbol $\infty$ is not an extra point that this interval must contain.

The coloured strokes mark membership; filled endpoint circles mean included points, and hollow endpoint circles mean excluded points. The empty set has no coloured points, while arrows at both ends of the coloured real line indicate that membership continues indefinitely. These examples use the usual topology of the real line.

In a metric space $X$, a subset $F$ is closed exactly when every convergent sequence $(x_n)$ in $F$ whose limit of a sequence $x$ lies in $X$ has $x\in F$. For example, $x_n=1-1/n$ lies in $[0,1)$ for $n\ge 1$ and has limit of a sequence $1$, which is missing; thus this half-open interval is not closed in the real line. This criterion is useful for keeping limiting constructions inside a prescribed set. ^[ref]

In every topological space, the general criterion is $F=\overline{F}$, using closure; equivalently, $F$ contains all its limit points. A closed set is always a sequentially closed set, but the converse can fail outside metric spaces. Arbitrary intersections and finite unions of closed sets are closed, by taking complements in the topology axioms.

The closure of a subset $A$ of a topological space $X$ is the smallest closed set containing $A$. Equivalently, it is the intersection of all closed sets containing $A$, or the set of points whose every neighbourhood in $X$ meets $A$. The notation ${\mathrm{cl}}_{X}A$ makes the ambient topological space explicit. This definition works without a metric or any appeal to sequences. ^[ref]

For $A=[0,1)$ in the usual real line, $\overline{A}=[0,1]$: every neighbourhood of $1$ meets $A$, while a point outside $[0,1]$ has an open ball disjoint from $A$. Passing to the closure adds the missing endpoint and retains all original points, including isolated points when present. In general, the closure is $A$ together with its limit points; its individual points are the adherent points of $A$.

The closure turns an arbitrary subset into a closed set with no unnecessary added points. It satisfies $A\subseteq \overline{A}$ and $\overline{\overline{A}}=\overline{A}$; a subset is closed exactly when $A=\overline{A}$. In a metric space, $x\in \overline{A}$ exactly when some sequence of points of $A$ has limit of a sequence $x$: choose its $n$th point in $A\cap B(x,1/n)$. For a general topological space, keep the neighbourhood definition; sequences need not detect every point of the closure.

A subset $A\subseteq S$ is closed under a binary operation on $S$ when the result of combining any two elements of $A$ is again in $A$. The even integers are closed under addition, but the odd integers are not: $1+3=4$ is even. Closure permits restricting the binary operation to $A\times A\to A$. It is different from being a closed set in topology.

For a limit ordinal $\kappa$, a subset $C\subseteq \kappa$ is club if it is unbounded and contains each limit below $\kappa$ of an increasing sequence from $C$. At an uncountable regular cardinal $\kappa$, intersections of fewer than $\kappa$ club sets are club: above a starting point cycle through the sets and take suprema, using regularity to stay below $\kappa$; closure puts the resulting limit in all of them. Club sets provide the reflecting ranks in worldly cardinals below an inaccessible cardinal.

The codomain of a function is its specified target set. The image of a function is a subset of the codomain, and equals it precisely for a surjective function. Changing a codomain can change whether the same input-output rule is surjective, as the quadratic function examples under function show.

The cofinite topology on a set $X$ declares the empty set and all cofinite subsets to be open sets: a nonempty open set may exclude only a finite set of points. There is no assumed metric. On an infinite set, this gives a particularly small collection of open sets while still making every singleton a closed set. On a finite set, every subset has finite complement, so the cofinite topology is the discrete topology.^[ref]

For an original example, use $X={\mathbb{N}}_{>0}$ and chooseBoth are open sets, and their intersection is $U\cap V=X\setminus \{1,2,3,5,7,8\}$. Thus $4,6,9,10,\dots$ are in both. Their finite exclusions cannot separate the infinite remainder. The finite intersection of any finite collection of nonempty open sets is again a nonempty open set, since its complement is a finite union of finite sets. An arbitrary nonempty union is open because its complement is contained in the finite complement of any one nonempty member.

The cofinite topology is useful for separating hypotheses that familiar metric spaces package together. An infinite cofinite topological space is a T1 space, because its singletons are closed, but is not a Hausdorff space, because two nonempty open sets cannot be disjoint sets. It is a hyperconnected space and hence a connected space. Moreover, every subset of a cofinite space is compact, and sequence convergence in the cofinite topology can have many limits. These are compatible properties, rather than contradictions: the usual uniqueness and closedness results use the Hausdorff space hypothesis.^[ref]

Over a model of Generalized continuum hypothesis, add ${\aleph }_3$ Cohen reals with finite binary partial functions. The Delta-system lemma proves the countable chain condition for forcing. The generic reals give the continuum lower bound ${\aleph }_3$, while nice forcing names for subsets of ${\omega }_1$ number at most $({\aleph }_3{)}^{{\aleph }_1}={\aleph }_3$ in the ground model. Cardinal preservation then gives $2^{{\aleph }_0}=2^{{\aleph }_1}={\aleph }_3$.

A subset $A$ of a specified ambient set $X$ is coinfinite if its complement of a set $X\setminus A$ is infinite. In $\mathbb{N}$, the even numbers are both infinite and coinfinite, whereas a cofinite set is not coinfinite.