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\begin{document}

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\section*{Heyting algebra}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topos_theory}{}\paragraph*{{Topos Theory}}\label{topos_theory}

[[!include topos theory - contents]]

\hypertarget{category_theory}{}\paragraph*{{$(0,1)$-Category theory}}\label{category_theory}

[[!include (0,1)-category theory - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{relation_to_other_concepts}{Relation to other concepts}\dotfill \pageref*{relation_to_other_concepts} \linebreak
\noindent\hyperlink{to_logic}{To logic}\dotfill \pageref*{to_logic} \linebreak
\noindent\hyperlink{to_topology}{To topology}\dotfill \pageref*{to_topology} \linebreak
\noindent\hyperlink{ToBooleanAlgebras}{To Boolean algebras}\dotfill \pageref*{ToBooleanAlgebras} \linebreak
\noindent\hyperlink{proofs}{Proofs}\dotfill \pageref*{proofs} \linebreak
\noindent\hyperlink{to_toposes}{To toposes}\dotfill \pageref*{to_toposes} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

\begin{udefn}
A \textbf{Heyting algebra} is a [[lattice]] $L$ which as a [[poset]] admits an operation of [[implication]]

\begin{displaymath}
\Rightarrow: L^{op} \times L \to L
\end{displaymath}
satisfying the condition (really a [[universal property]])

\begin{displaymath}
(x \wedge a) \leq b \qquad if\;and\;only\;if \qquad x \leq (a \Rightarrow b)
\end{displaymath}
\end{udefn}
This is equivalent to the following definition.

\begin{udefn}
A Heyting algebra is a [[bicartesian closed category|bicartesian closed]] [[poset]], that is a poset which (when thought of as a [[thin category]]) is

\begin{itemize}%
\item [[finitely complete category|finitely complete]],


\item finitely cocomplete,


\item and [[cartesian closed category|cartesian closed]].



\end{itemize}
\end{udefn}
The implication $a\Rightarrow b$ is the [[exponential object]] $b^a$.

\begin{uremark}
Insofar as all these properties of a poset are described by [[universal properties]], being a Heyting algebra is a [[property-like structure]] on a poset; a poset can be a Heyting algebra in at most one way.

\end{uremark}
The definition of Heyting algebra may be recast into purely [[algebraic theory|equational form]]: add to the equational theory of lattices the inequalities $(x \Rightarrow y) \wedge x \leq y$ and $y \leq x \Rightarrow (y \wedge x)$, writing these inequalities in equational form via the equivalence $a \leq b$ iff $a = a \wedge b$. Hence we can speak of an \textbf{[[internalization|internal]] Heyting algebra} in any [[cartesian monoidal category|category with products]].

\begin{udefn}
A Heyting algebra [[homomorphism]] is a homomorphism of the underlying [[lattice]]s that preserve $\Rightarrow$. Heyting algebras and their homomorphisms form a [[concrete category]] [[HeytAlg]].

\end{udefn}
\hypertarget{relation_to_other_concepts}{}\subsection*{{Relation to other concepts}}\label{relation_to_other_concepts}

\hypertarget{to_logic}{}\subsubsection*{{To logic}}\label{to_logic}

In [[logic]], Heyting algebras are precisely algebraic models for [[intuitionistic logic|intuitionistic]] [[propositional calculus]], just as [[Boolean algebra|Boolean algebras]] model [[classical logic|classical]] propositional calculus. As one might guess from this description, the ``law of the [[excluded middle]]'' does not generally hold in a Heyting algebra; see the discussion below.

In a [[Heyting category]], every [[subobject poset]] $Sub(A)$ is a Heyting algebra. In particular, this holds for every [[topos]]. Furthermore, in a topos, the [[power object]] $\mathcal{P}(A)$ is an [[internalisation|internal]] Heyting algebra that corresponds to the external Heyting algebra $Sub(A)$. In a [[boolean topos]], the internal Heyting algebras are all internal [[boolean algebras]]. In general, however, the [[internal logic]] of a topos (or other Heyting category) is intuitionistic.

\hypertarget{to_topology}{}\subsubsection*{{To topology}}\label{to_topology}

One of the chief sources of Heyting algebras is given by [[topology|topologies]]. As a poset, the topology of a topological space $X$ is a [[complete lattice]] (it has arbitrary [[joins]] and [[meets]], although the infinitary meets are not in general given by [[intersection]]), and the implication operator is given by

\begin{displaymath}
(U \Rightarrow V) = int(U^c \vee V)
\end{displaymath}
where $U, V$ are open sets, $U^c$ is the set-theoretic complement of $U$, and $int(S)$ denotes the interior of a subset $S \subseteq X$.

Somewhat more generally, a [[frame]] (a [[sup-lattice]] in which finite meets distribute over arbitrary sups) also carries a Heyting algebra structure. In a frame, we may define

\begin{displaymath}
(u \Rightarrow v) = \bigvee_{x \wedge u \leq v} x
\end{displaymath}
and the distributivity property guarantees that the universal property for implication holds. (The detailed proof is a ``baby'' application of an [[adjoint functor theorem]].)

Thus frames are extensionally the same thing as \emph{[[complete lattice|complete]] Heyting algebras}. However, \emph{intensionally} they are quite different; that is, a morphism of frames is not usually a morphism of complete Heyting algebras: they do not preserve the implication operator.

A [[locale]] is the same thing as a frame, but again the morphisms are different; they are reversed.

Topologies that are [[Boolean algebras]] are the exception rather than the rule; basic examples include topologies of [[Stone spaces]]; see [[Stone duality]]. Another example is the topology of a [[discrete space]] $X$.

\hypertarget{ToBooleanAlgebras}{}\subsubsection*{{To Boolean algebras}}\label{ToBooleanAlgebras}

In any Heyting algebra $L$, we may define a [[negation]] operator

\begin{displaymath}
\neg\colon L^{op} \to L
\end{displaymath}
by $\neg x = (x \Rightarrow 0)$, where $0$ is the bottom element of the lattice. A Heyting algebra is [[Boolean algebra|Boolean]] if the [[double negation]]

\begin{displaymath}
\neg \neg\colon L \to L
\end{displaymath}
coincides with the identity map; this gives one of many ways of defining a Boolean algebra.

In any Heyting algebra $L$, we have for all $a, b \in L$ the inequality

\begin{displaymath}
(\neg a \vee b) \leq (a \Rightarrow b) ,
\end{displaymath}
and another characterization of Boolean algebra is a Heyting algebra in which this is an equality for all $a, b$.

There are several ways of passing back and forth between Boolean algebras and Heyting algebras, having to do with the [[double negation]] operator. A useful lemma in this regard is

\begin{ulemma}
The double negation $\neg \neg\colon L \to L$ is a [[monad]] that preserves finite meets.

\end{ulemma}
The proof can be made purely equational, and is therefore internally valid in any category with products. Applied to the internal Heyting algebra $L = \Omega$ of a [[topos]], that is the [[subobject classifier]], this lemma says exactly that the double negation operator $\neg \neg\colon \Omega \to \Omega$ defines a [[Lawvere–Tierney topology]]. Similarly, we get the [[double negation sublocale]] of any [[locale]].

Now let $L_{\neg\neg}$ denote the poset of \emph{[[regular element]]s} of $L$, that is, those elements $x$ such that $\neg\neg x = x$. (When $L$ is the topology of a space, an open set $U$ is [[regular open subspace|regular]] if and only if it is the interior of its closure, that is if it is a regular element of the Heyting algebra of open sets described above.) With the help of the lemma above, we may prove

\begin{theorem}
\label{}\hypertarget{}{}
The poset $L_{\neg\neg}$ is a Boolean algebra. Moreover, the assignment $L \mapsto L_{\neg\neg}$ is the object part of a functor

\begin{displaymath}
F\colon Heyt \to Bool
\end{displaymath}
called \emph{Booleanization}, which is left adjoint to the full and faithful inclusion

\begin{displaymath}
i\colon Bool \hookrightarrow Heyt.
\end{displaymath}
The unit of the [[adjoint functor|adjunction]], applied to a Heyting algebra $L$, is the map $L \to L_{\neg\neg}$ which maps each element $x$ to its \emph{[[regular element|regularization]]} $\neg\neg x$.

\end{theorem}
Thus $\neg\neg\colon L \to L_{\neg\neg}$ preserves finite joins and finite meets and implication. In the other direction, we have an inclusion $i\colon L_{\neg\neg} \to L$, and this preserves meets but not joins. It also preserves negations; more generally and perhaps surprisingly, it preserves implications as well.

Regular elements are not to be confused with \emph{[[complemented element]]s}, i.e., elements $x$ in a Heyting algebra such that $x \vee \neg x = 1$, although it is true that every complemented element is regular. An example of a regular element which is not complemented is given by the unit interval $(0, 1)$ as an element of the topology of $\mathbb{R}$; a complemented element in a Heyting algebra given by a topology is the same as a [[clopen subset]].

Complemented elements furnish another universal relation between Boolean algebras and Heyting algebras: the set of complemented elements in a Heyting algebra $H$ is a Boolean algebra $Comp(H)$, and the inclusion $Comp(H) \to H$ is a Heyting algebra map which is universal among Heyting algebra maps $B \to H$ out of Boolean algebras $B$. In other words, we have the following result.

\begin{theorem}
\label{}\hypertarget{}{}
The assignment $H \mapsto Comp(H)$ is the object part of a right adjoint to the forgetful functor $Bool \to Heyt$.

\end{theorem}
\hypertarget{proofs}{}\paragraph*{{Proofs}}\label{proofs}

We prove the lemma and theorems of the preceding section.

\begin{proof}
Since $\neg \neg$ preserves order, it is clear that $\neg \neg(x \wedge y) \leq \neg \neg x$ and $\neg \neg(x \wedge y) \leq \neg \neg y$, so

\begin{displaymath}
\neg \neg (x \wedge y) \leq (\neg \neg x) \wedge (\neg \neg y)
\end{displaymath}
follows. In the other direction, to show

\begin{displaymath}
(\neg \neg x) \wedge (\neg \neg y) \leq \neg \neg (x \wedge y),
\end{displaymath}
we show $(\neg \neg x) \wedge (\neg \neg y) \wedge \neg (x \wedge y) \leq 0$. But we have $\neg (x \wedge y) = (y \Rightarrow \neg x)$, and we also have the general result

\begin{displaymath}
(a \Rightarrow b) \wedge (b \Rightarrow c) \leq (a \Rightarrow c).
\end{displaymath}
Putting $a = y$, $b = \neg x$, $c = 0$, we obtain

\begin{displaymath}
\neg (x \wedge y) \wedge (\neg \neg x) \leq \neg y
\end{displaymath}
and so now

\begin{displaymath}
\neg (x \wedge y) \wedge (\neg \neg x) \wedge (\neg \neg y) \leq (\neg y) \wedge (\neg \neg y) \leq 0
\end{displaymath}
as required.

\end{proof}
\begin{proof}
Since $\neg \neg$ is a monad, and $L_{\neg \neg}$ is the corresponding category (poset) of $\neg \neg$-algebras, the left adjoint $\neg \neg \colon L \to L_{\neg \neg}$ preserves joins. Since this map is epic, this also gives the fact that $L_{\neg \neg}$ has joins. The map $L \to L_{\neg\neg}$ preserves meets by the preceding lemma, and $\neg \neg 1 = \neg 0 = 1$. Thus $L \to L_{\neg\neg}$ is a surjective lattice map, and it follows that $L_{\neg\neg}$ is distributive because $L$ is.

Working in $L_{\neg \neg}$ (where the join will be written $\vee_{\neg\neg}$ and the meet $\wedge_{\neg\neg}$), we have for any $x \in L_{\neg \neg}$ the equations

\begin{displaymath}
x \vee_{\neg \neg} \neg x = \neg \neg (x \vee \neg x) = \neg (\neg x \wedge \neg \neg x) = \neg 0 = 1
\end{displaymath}
\begin{displaymath}
\,
\end{displaymath}
\begin{displaymath}
x \wedge_{\neg\neg} \neg x = x \wedge \neg x = 0
\end{displaymath}
so that $\neg x$ is the complement of $x \in L_{\neg \neg}$. We have thus shown that $L_{\neg\neg}$ is a complemented distributive lattice, i.e., a Boolean algebra. This calculation also shows that $\neg\neg \colon L \to L_{\neg\neg}$ preserves negation.

To show $L \to L_{\neg\neg}$ preserves implication, we may start from the observation (see the following lemma) that in any Heyting algebra $L$, we have

\begin{displaymath}
\neg(a \Rightarrow b) = (\neg \neg a) \wedge (\neg b).
\end{displaymath}
Then

\begin{displaymath}
\itexarray{
\neg \neg(a \Rightarrow b) & = & \neg((\neg \neg a) \wedge (\neg b)) \\ 
 & = & \neg ((\neg \neg a) \wedge (\neg \neg \neg b)) \\
 & = & \neg \neg ((\neg a) \vee (\neg \neg b)) \\
 & = & \neg a \vee_{\neg \neg} (\neg \neg b) \\
 & = & \neg (\neg \neg a) \vee_{\neg \neg} (\neg \neg b)
}
\end{displaymath}
where the last expression is $(\neg \neg a) \Rightarrow (\neg \neg b)$ as computed in the Boolean algebra $L_{\neg \neg}$, since in a Boolean algebra we have $(x \Rightarrow y) = (\neg x \vee y)$.

Therefore $L \to L_{\neg\neg}$ is a Heyting algebra quotient which is the coequalizer of $1, \neg\neg \colon L \stackrel{\to}{\to} L$. It follows that a Heyting algebra map $L \to B$ to any Boolean algebra $B$ factors uniquely through this coequalizer, and the induced map $L_{\neg \neg} \to B$ is a Boolean algebra map. In other words, $L \to L_{\neg\neg}$ is the universal Heyting algebra map to a Boolean algebra, which establishes the adjunction.

\end{proof}
\begin{ulem}
In a Heyting algebra, $\neg(a \Rightarrow b) = (\neg \neg a) \wedge (\neg b)$.

\end{ulem}
\begin{proof}
Since $\neg$ is contravariant and $a \Rightarrow -$ is covariant, we have

\begin{displaymath}
\neg(a \Rightarrow b) \leq \neg(a \Rightarrow 0) = (\neg \neg a).
\end{displaymath}
Since $- \Rightarrow b$ is contravariant, we have

\begin{displaymath}
\neg(a \Rightarrow b) \leq \neg(1 \Rightarrow b) = (\neg b).
\end{displaymath}
We conclude that $\neg(a \Rightarrow b) \leq (\neg \neg a) \wedge (\neg b).$ On the other hand, we have

\begin{displaymath}
(\neg \neg a) \wedge (\neg b) \wedge (a \Rightarrow b) \leq (\neg \neg a) \wedge (\neg a) \leq 0
\end{displaymath}
whence $(\neg \neg a) \wedge (\neg b) \leq \neg (a \Rightarrow b)$, which completes the proof.

\end{proof}
\begin{uremark}
It follows from this lemma that double negation on a Heyting algebra $\neg \neg \colon L \to L$ preserves implication, since

\begin{displaymath}
\neg \neg(a \Rightarrow b) = \neg ((\neg \neg a) \wedge (\neg b)) = 0^{(\neg \neg a) \wedge (\neg b)} = (\neg \neg b)^{(\neg \neg a)} = (\neg \neg a) \Rightarrow (\neg \neg b).
\end{displaymath}
This is important for the [[double negation translation]].

\end{uremark}
\begin{proof}
In a Heyting algebra $H$, the elements $0$ and $1$ are clearly complemented. If $x$ and $y$ are complemented, then so is $x \wedge y$ since

\begin{displaymath}
1 = 1 \wedge 1 = (x \vee \neg x \vee \neg y) \wedge (y \vee \neg x \vee \neg y) = (x \wedge y) \vee (\neg x \vee \neg y)
\end{displaymath}
\begin{displaymath}
\,
\end{displaymath}
\begin{displaymath}
(x \wedge y) \wedge (\neg x \vee \neg y) = (x \wedge y \wedge \neg x) \vee (x \wedge y \wedge \neg y) = 0 \vee 0 = 0.
\end{displaymath}
By a similar proof, $x \vee y$ is complemented. Finally, $x \Rightarrow y$ has complement $x \wedge \neg y$: writing $x \Rightarrow y = y^x$ for typographical clarity, we have

\begin{displaymath}
1 = 1 \wedge 1 = (\neg x \vee x) \wedge (y \vee \neg y) \leq (y^x \vee x) \wedge (y^x \vee \neg y) = y^x \vee (x \wedge \neg y),
\end{displaymath}
\begin{displaymath}
\,
\end{displaymath}
\begin{displaymath}
y^x \wedge x \wedge \neg y \leq y \wedge \neg y = 0.
\end{displaymath}
Thus the complemented elements form a Heyting subalgebra $Comp(H) \hookrightarrow H$. Clearly $Comp(H)$ is a Boolean algebra, and clearly if $B$ is Boolean, then any Heyting algebra map $B \to H$ factors uniquely through $Comp(H) \hookrightarrow H$. This proves the theorem.

\end{proof}
\hypertarget{to_toposes}{}\subsubsection*{{To toposes}}\label{to_toposes}

An [[elementary topos]] is a [[vertical categorification]] of a Heyting algebra: the notion of Heyting algebra is essentially equivalent to that of [[(0,1)-topos]]. Note that a [[Grothendieck topos|Grothendieck]] $(0,1)$-topos is a [[frame]] or [[locale]].

\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\begin{prop}
\label{}\hypertarget{}{}
For $\mathcal{T}$ a [[topos]] and $X \in \mathcal{T}$ any [[object]], the poset $Sub(X)$ of [[subobject]]s of $X$ is a Heyting algebra.

In other words, every topos is a [[Heyting category]].

\end{prop}
In particular for $X = \Omega$ the [[subobject classifier]], $Sub(\Omega)$ is a Heyting algebra.

In $\mathcal{T} =$ [[Set]] for every set $S$ we have that $Sub(S)$ is the [[Boolean algebra]] of subset of $S$.

More details and examples are spelled out at [[internal logic]].

\begin{prop}
\label{}\hypertarget{}{}
A [[frame]] $L$ is a Heyting algebra.

\end{prop}
\begin{proof}
By the [[adjoint functor theorem]], a right adjoint $x \Rightarrow -$ to the map $x \wedge -: L \to L$ exists since this map preserves arbitrary joins.

\end{proof}
\hypertarget{references}{}\subsection*{{References}}\label{references}

Named after [[Arend Heyting]].

[[!redirects Heyting algebra]] [[!redirects Heyting algebras]]



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