The Cantor Set, Baire Category, and Measure Zero in $\mathbb{R}$

Let $A\subseteq \mathbb{R}$, with interior $A^{\circ }$ and closure $\bar{A}$ as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$.

• $A$ is nowhere dense when the interior of its closure is empty: $(\bar{A}{)}^{\circ }=\mathrm{\varnothing }\mathrm{.}$
• $A$ is meager, or of the first category, when there is a sequence $(A_n{)}_{n\in \mathbb{N}}$ of nowhere dense subsets of $\mathbb{R}$ with $A={\bigcup }_{n\in \mathbb{N}}{A}_n\mathrm{.}$
• $A$ is of the second category when it is not meager.
• $A$ is residual (also comeager) when $\mathbb{R}\setminus A$ is meager.

Why a sequence, and why that is the same as "an at most countable union". Sequences here are indexed by $\mathbb{N}$, which contains $0$. A finite family $A_0,\dots ,A_m$ of nowhere dense sets is turned into a sequence by setting $A_n:=\mathrm{\varnothing }$ for $n>m$, and $\mathrm{\varnothing }$ is nowhere dense because $\bar{\mathrm{\varnothing }}=\mathrm{\varnothing }$ has empty interior; the empty family is handled the same way and gives $A=\mathrm{\varnothing }$. So "a union of an at most countable family of nowhere dense sets" (Finite, countably infinite, countable, uncountable) and the displayed condition define the same class, and the sequence form is used below because it carries an explicit index and needs no case split.

Nowhere dense means exactly that the complement of the closure is dense. For $A\subseteq \mathbb{R}$,

$(\bar{A}{)}^{\circ }=\mathrm{\varnothing }⟺\mathbb{R}\setminus \bar{A}\text{ is dense in }\mathbb{R}\mathrm{.}$

Indeed, by the pointwise description of the interior (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$), $(\bar{A}{)}^{\circ }=\mathrm{\varnothing }$ says that no $x\in \mathbb{R}$ admits a real $\epsilon >0$ with $N_{\epsilon }(x)\subseteq \bar{A}$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$), that is, that every $N_{\epsilon }(x)$ meets $\mathbb{R}\setminus \bar{A}$. By claim 1 of The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points that says precisely that every $x\in \mathbb{R}$ is adherent to $\mathbb{R}\setminus \bar{A}$, that is, $\overline{\mathbb{R}\setminus \stackrel{‾}{A}}=\mathbb{R}$, which is density (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$).

A closed set is nowhere dense exactly when its interior is empty, since a closed set equals its own closure (claim 4 of The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). This is the form in which nowhere density is verified nearly every time below. (The phrase almost everywhere is avoided throughout this pair: it is a measure-theoretic term, and the only measure notion defined here is measure zero.)

Both classes are closed downwards. If $B\subseteq A$ then $\bar{B}\subseteq \bar{A}$ and hence $(\bar{B}{)}^{\circ }\subseteq (\bar{A}{)}^{\circ }$ (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$), so a subset of a nowhere dense set is nowhere dense. If $B\subseteq A={\bigcup }_n{A}_n$ with each $A_n$ nowhere dense, then $B={\bigcup }_n(A_n\cap B)$ and each $A_n\cap B$ is nowhere dense by the previous sentence, so a subset of a meager set is meager.

A union of two meager sets is meager. Let $M={\bigcup }_n{A}_n$ and $M^{\mathrm{\prime }}={\bigcup }_n{B}_n$ with all $A_n$ and all $B_n$ nowhere dense; fixing one witnessing sequence for $M$ and one for $M^{\mathrm{\prime }}$ is two instantiations of an existential statement, not a choice principle. Let $J:\mathbb{N}\times \mathbb{N}\to \mathbb{N}$ be a bijection ($\mathbb{N}\times \mathbb{N}\approx \mathbb{N}$) and define a sequence $(C_j{)}_{j\in \mathbb{N}}$ by

$C_{J(m,n)}:=\{\begin{array}{ll}{A}_n& m=0,\\ B_n& m\mathrm{\ne }0.\end{array}$

This is a total definition because $J$ is a bijection, every $C_j$ is nowhere dense, and ${\bigcup }_j{C}_j=M\cup M^{\mathrm{\prime }}$, since $A_n=C_{J(0,n)}$ and $B_n=C_{J(1,n)}$ and every $C_j$ is one of the $A_n$ or one of the $B_n$.

• The countably infinite version of the last observation is a different statement. To show that ${\bigcup }_i{M}_i$ is meager for a sequence $(M_i)$ of meager sets one must select a witnessing sequence of nowhere dense sets for every $i$ at once, which is an application of countable choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)); the two-set case above avoids it because two selections are two instantiations. Nothing on this page uses the countably infinite version, and every meager set met below is presented together with an explicit witnessing sequence.

• Nowhere dense is strictly stronger than having empty interior. $\mathbb{Q}$ has empty interior, since no neighbourhood consists of rationals alone, yet $\bar{\mathbb{Q}}=\mathbb{R}$ has interior $\mathbb{R}$, so $\mathbb{Q}$ is not nowhere dense. It is nevertheless meager, being a union of singletons; that computation is $\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$.

• First category, second category and residual are not a trichotomy. A set is meager or of the second category, and those two are exhaustive and exclusive by definition. Residual is a separate condition on the complement: a residual set is of the second category once $\mathbb{R}$ is known not to be meager in itself (Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets), but before that theorem nothing rules out a set that is both meager and residual.

• Category is a notion of topological smallness, and it is independent of smallness in the sense of measure. Neither of the two implications between "nowhere dense" and "measure zero" (Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover)) holds, and $\mathbb{R}$ itself splits into a meager set and a set of measure zero; the three items settling this are FALSE: every nowhere dense subset of $\mathbb{R}$ has measure zero, FALSE: every subset of $\mathbb{R}$ of measure zero is nowhere dense and $\mathbb{R}$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions ↗.

Let $A\subseteq \mathbb{R}$, with open and closed sets as in Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen.

• $A$ is an $F_{\sigma }$ set when there is a sequence $(F_n{)}_{n\in \mathbb{N}}$ of closed subsets of $\mathbb{R}$ with $A={\bigcup }_{n\in \mathbb{N}}{F}_n\mathrm{.}$
• $A$ is a $G_{\delta }$ set when there is a sequence $(V_n{)}_{n\in \mathbb{N}}$ of open subsets of $\mathbb{R}$ with $A={\bigcap }_{n\in \mathbb{N}}{V}_n\mathrm{.}$

The letters are the traditional ones: $F$ for fermé with $\sigma$ for somme, $G$ for Gebiet with $\delta$ for Durchschnitt.

The two classes are exchanged by complementation. $A$ is $F_{\sigma }$ if and only if $\mathbb{R}\setminus A$ is $G_{\delta }$. If $A={\bigcup }_n{F}_n$ with each $F_n$ closed, then $\mathbb{R}\setminus A={\bigcap }_n(\mathbb{R}\setminus F_n)$ by De Morgan, and each $\mathbb{R}\setminus F_n$ is open by the definition of closedness (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen); the converse is the same computation read backwards, using that the complement of an open set is closed, which is again Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen.

Every closed set is $F_{\sigma }$ and every open set is $G_{\delta }$, by the constant sequence $F_n:=A$, respectively $V_n:=A$. As with Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$, an at most countable family (Finite, countably infinite, countable, uncountable) may always be presented as a sequence: a finite list $F_0,\dots ,F_m$ of closed sets is extended by $F_n:=F_m$ for $n>m$, and a finite list of open sets likewise, so nothing is lost by indexing over $\mathbb{N}$.

• The classes are genuinely larger than the closed and the open sets. $\mathbb{Q}$ is $F_{\sigma }$ and is neither open nor closed, and the irrationals are $G_{\delta }$ and neither open nor closed; both computations are in $\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$. That $\mathbb{Q}$ is not also $G_{\delta }$ is the first genuinely hard fact about these classes and needs the Baire category theorem (Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets).

• Why the algebra of open sets is not enough. Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets gives that a finite intersection of open sets is open and a finite union of closed sets is closed. The definitions above are exactly what one gets by relaxing "finite" to "countable" once, and the point of the whole notion is that the relaxation is proper: a countable intersection of open sets need not be open, which is ${\bigcap }_k(-1\mathrm{/}k,1\mathrm{/}k)=\{0\}$ is not open.

• Nothing here is a measure-theoretic notion. $F_{\sigma }$ and $G_{\delta }$ are defined from the topology of $\mathbb{R}$ alone and are used on this page to say precisely how far $\mathbb{Q}$ and its complement sit from being closed or open. They cut across Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover) completely: $\mathbb{Q}$ is $F_{\sigma }$ and has measure zero (Every at most countable subset of $\mathbb{R}$ has measure zero), while the Smith-Volterra-Cantor set is closed, hence $F_{\sigma }$, and does not (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero).

Let $(U_n{)}_{n\in \mathbb{N}}$ be a sequence of subsets of $\mathbb{R}$, each open (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen) and dense (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$). Then

${\bigcap }_{n\in \mathbb{N}}{U}_n\text{is dense in }\mathbb{R}\mathrm{.}$

Consequently, if $(A_n{)}_{n\in \mathbb{N}}$ is a sequence of nowhere dense subsets of $\mathbb{R}$ (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$), then ${\bigcup }_{n\in \mathbb{N}}{A}_n\mathrm{\ne }\mathbb{R}$: no meager subset of $\mathbb{R}$ exhausts $\mathbb{R}$, so $\mathbb{R}$ is of the second category in itself.

The selection is canonical, and the proof spends no choice principle. The textbook argument picks a nested interval at every stage in terms of the one before it, which is the axiom of dependent choice. The construction below instead fixes one enumeration $e$ of the rationals ($\mathbb{Q}$ is countably infinite, The rationals embed densely in the reals) and, at every stage, takes the interval whose two rational endpoints have least index among those meeting the requirements. The requirements are met by some rational-endpoint interval, which is what the refinement claim of the proof establishes, and the least such index is determined by The well-ordering principle; so the whole recursion is a single application of The recursion theorem to one total map. This is the same least-index device used later in the perfect-set development, transplanted here from perfect sets to dense open sets. What it does not settle is the strength of the theorem for general complete metric spaces; that metamathematical point is recorded separately on the choice ledger for this thread.

• What "dense" is doing at each end. Density of the $U_n$ is used exactly once, in the refinement claim, to find a point of $U_n$ inside a given open interval; openness is used exactly once, immediately after, to fit a whole closed interval with rational endpoints around that point. Neither hypothesis can be dropped. Without openness the conclusion fails: the family consisting of $\mathbb{Q}$ together with all the sets $\mathbb{R}\setminus \{q\}$ for $q\in \mathbb{Q}$ is an at most countable family of dense sets, all but the first of them open, and its intersection is empty. Without density it fails too, for the constant sequence $U_n:=(0,1)$ has intersection $(0,1)$, which is not dense in $\mathbb{R}$.

• Only nonemptiness of the nested intersection is used. The construction does not force the interval lengths to $0$ and does not need to: claim 1 of A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ already produces a point, and one point is all the argument wants. That is why no Archimedean step appears anywhere above.

• The consequence is the form used downstream. Applying it to the sequence of singletons $\{x_n\}$ of a supposed enumeration of $\mathbb{R}$ reproves that $\mathbb{R}$ is uncountable (Baire category gives a third proof that $\mathbb{R}$ is uncountable ↗); applying it to a supposed presentation of $\mathbb{Q}$ as a $G_{\delta }$ set is what shows that no such presentation exists ($\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$).

• Category is not measure. The intersection produced above is dense but may be very small in the sense of Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover); indeed $\mathbb{R}$ decomposes as a meager set together with a set of measure zero ($\mathbb{R}$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions ↗), so this theorem says nothing whatever about size in measure.

Write ${\mathbb{Q}}_{\mathbb{R}}$ for the image of $\mathbb{Q}$ in $\mathbb{R}$ under the canonical embedding $q\mapsto \widehat{q}$ (The rationals embed densely in the reals), the set usually written $\mathbb{Q}$ once the identification is made, and put $X:=\mathbb{R}\setminus {\mathbb{Q}}_{\mathbb{R}}$ for the irrationals. Then:

1. ${\mathbb{Q}}_{\mathbb{R}}$ is an $F_{\sigma }$ set ($F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$) and is meager (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$);
2. $X$ is a $G_{\delta }$ set and is residual;
3. ${\mathbb{Q}}_{\mathbb{R}}$ is not a $G_{\delta }$ set, and $X$ is not an $F_{\sigma }$ set.

Claims 1 and 2 are bookkeeping. Claim 3 is the substance and is exactly where Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets is spent: no argument from the algebra of open and closed sets alone can reach it, since ${\mathbb{Q}}_{\mathbb{R}}$ and $X$ are interchanged by complementation while $F_{\sigma }$ and $G_{\delta }$ are, so any such argument would prove the same thing about both sets and about neither.

• Where the two halves of the argument part company. Claim 1 is a listing argument: step 1.2 lists ${\mathbb{Q}}_{\mathbb{R}}$, and step 1.1 shows that each real singleton is nowhere dense; claim 3 uses the completeness of $\mathbb{R}$ through A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$, inside Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets. Indeed $\mathbb{Q}$ is a $G_{\delta }$ subset of itself, being the whole space, so no argument that ignores the ambient completeness can possibly give claim 3.

• The irrationals are also uncountable (The irrationals are uncountable), by a different and much cheaper argument that needs only the countability of $\mathbb{Q}$ and the uncountability of $\mathbb{R}$. Uncountability and being residual are independent properties: $\mathbb{Q}$ is meager and countable, the Cantor set is meager and uncountable (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points), and $X$ is residual and uncountable.

• The named witness for claim 3 is The irrationals form a residual $G_{\delta }$ set that is not $F_{\sigma }$ ↗, and the false statement it refutes is FALSE: $\mathbb{Q}$ is a $G_{\delta }$ subset of $\mathbb{R}$; the refutation is carried out here.

• Meagre and measure zero are not the same smallness. ${\mathbb{Q}}_{\mathbb{R}}$ is both, but the two notions diverge as soon as one leaves the countable case: $\mathbb{R}$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions ↗ writes $\mathbb{R}$ as a meager set together with a set of measure zero, and the set of measure zero there is residual. So being residual, which is what claim 2 gives for $X$, carries no information at all about size in measure.

Throughout, $\mathbb{R}$ is the complete ordered field (Complete ordered field (least-upper-bound property)), intervals and their lengths are as in Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, and a sequence is a function on $\mathbb{N}$, which contains $0$. Let $A\subseteq \mathbb{R}$.

• $A$ has measure zero, equivalently $A$ is null, when for every real $\epsilon >0$ there are sequences $(a_k{)}_{k\in \mathbb{N}}$ and $(b_k{)}_{k\in \mathbb{N}}$ of reals with $a_k\le b_k$ for every $k$, such that $A\subseteq {\bigcup }_{k\in \mathbb{N}}[a_k,b_k]\text{and}{\sum }_{k=0}^{\mathrm{\infty }}(b_k-a_k)\text{ converges with sum }\le \epsilon \mathrm{.}$
• $A$ has content zero when for every real $\epsilon >0$ there are $n\in \mathbb{N}$ and reals $a_0\le b_0,\dots ,a_n\le b_n$ with $A\subseteq {\bigcup }_{j\le n}[a_j,b_j]\text{and}{\sum }_{j=0}^n(b_j-a_j)\le \epsilon \mathrm{.}$

The number $b_k-a_k\ge 0$ is the length of $[a_k,b_k]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), and the sums are the series and the finite sums of Series, partial sums, convergence and the sum, divergence, and the tail series and Finite sums and finite products, by recursion.

Working form: only the partial sums have to be checked. All the terms $b_k-a_k$ are $\ge 0$, so by claim 2 of A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum the series converges exactly when its partial sums are bounded above, and its sum is then their supremum. Consequently, for a fixed $\epsilon >0$,

${\sum }_{k=0}^{\mathrm{\infty }}(b_k-a_k)\text{ converges with sum}\le \epsilon ⟺{\sum }_{k<n}(b_k-a_k)\le \epsilon \text{ for every }n\in \mathbb{N},$

since a supremum is $\le \epsilon$ exactly when $\epsilon$ is an upper bound of the set it is the supremum of (Complete ordered field (least-upper-bound property)). Every verification of nullity below checks the right-hand condition.

Closed intervals lose nothing. A bounded interval with endpoints $a\le b$ is contained in $[a,b]$ and has the same length (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), so a cover by intervals of any of the four bounded forms yields a cover by closed intervals with the same lengths. The definition is therefore stated with closed intervals once and for all. Covers by open intervals are a genuinely different demand, and passing to one costs a little extra length: the enlargement $[a_k,b_k]\subseteq (a_k-{\delta }_k,b_k+{\delta }_k)$ is carried out where it is needed, in A sequence of intervals covering $[a,b]$ has total length at least $b-a$, so no interval of positive length has measure zero and in For a compact subset of $\mathbb{R}$, measure zero and content zero coincide.

Both notions are inherited by subsets. If $B\subseteq A$ and $A$ is null, then any cover of $A$ covers $B$, so $B$ is null; the same sentence with finite covers shows a subset of a set of content zero has content zero.

A finite cover is a countable cover, so content zero implies measure zero. Padding the list $[a_0,b_0],\dots ,[a_n,b_n]$ with the degenerate intervals $[0,0]$ for $k>n$ leaves the total length unchanged, by the splitting law for finite sums (Laws of finite sums and finite products). This is recorded as a lemma with its proof, A set of content zero has measure zero, because it is cited on its own.

• The two notions genuinely differ. $\mathbb{Q}\cap [0,1]$ is null and does not have content zero ($\mathbb{Q}\cap [0,1]$ has measure zero and not content zero, although it is bounded ↗), so the two quantifier patterns, "a sequence of intervals" and "a finite list of intervals", are not interchangeable. They do agree for compact sets (For a compact subset of $\mathbb{R}$, measure zero and content zero coincide), and the compact case is the only one in which content zero is used anywhere on this pair of pages. Nothing is claimed about what later pages will do with it.

• Why "content" and not "measure" for the finite version. The finite-cover notion is the vanishing of the Jordan outer content, and the countable-cover notion is the vanishing of the Lebesgue outer measure. Neither outer quantity is available at this point in the reading order. Jordan outer content is defined later in Jordan inner and outer content and Jordan measurable bounded sets in ${\mathbb{R}}^m$ ↗; Lebesgue outer measure is still not defined. No item on this page assigns a nonzero size to any set. Every statement is of the shape "can, or cannot, be covered by intervals of total length below such and such a bound". That is a deliberate restriction of scope at this point in the reading order, not a claim that the general notions are unavailable in mathematics.

• Measure zero is not vacuous and not universal. No interval with two distinct endpoints is null (A sequence of intervals covering $[a,b]$ has total length at least $b-a$, so no interval of positive length has measure zero), while every at most countable set is (Every at most countable subset of $\mathbb{R}$ has measure zero) and so is the uncountable Cantor set (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points). The three facts together are what make the notion worth having.

Let $a,b\in \mathbb{R}$ with $a\le b$, let $n\in \mathbb{N}$, and let $c_0\le d_0,\dots ,c_n\le d_n$ be reals such that

$[a,b]\subseteq {\bigcup }_{j\le n}[c_j,d_j],$

the intervals being those of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length. Then

${\sum }_{j=0}^n(d_j-c_j)\ge b-a\mathrm{.}$

The same bound holds for a cover by bounded intervals of any of the four bounded forms, since an interval with endpoints $c\le d$ is contained in $[c,d]$ and has the same length $d-c$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length); replacing each covering interval by the closed interval on its endpoints changes no length and only enlarges the union. In particular a finite family of intervals of total length strictly below $b-a$ cannot cover $[a,b]$, which is the form in which this lemma is used throughout the page.

This is the one quantitative fact underlying everything about measure zero here. Without it nothing forbids a set of measure zero from being all of $[0,1]$. Four items on this page rest on it: A sequence of intervals covering $[a,b]$ has total length at least $b-a$, so no interval of positive length has measure zero directly, and through that lemma The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero and FALSE: every set of measure zero has content zero. Two of the worked items on the companion page rest on it as well.