## 1 Introduction

### 1.1 The classical set partition lattices

An equivalence relation on a set A uniquely determines a partition on this set. A partition of A is a collection $$A_i\subset A$$ ($$i\in I$$) of non-empty, pairwise disjoint sets (also called blocks) such that

\begin{aligned} \bigcup _{i\in I}A_i=A. \end{aligned}

The partition $$\pi$$ is a refinement of the partition $$\sigma$$ if the blocks in $$\sigma$$ arise as the union of some blocks in $$\pi$$. This is denoted by $$\pi \le \sigma$$. If $$\pi \ne \sigma$$ and $$\pi \le \sigma$$, we write $$\pi <\sigma$$. Thus, for example,

\begin{aligned} 1,4|2,3,7|5,6,8<1,2,3,4,7|5,6,8. \end{aligned}

Let $$\pi$$ and $$\sigma$$ be two partitions of A, and let $$\pi _e$$ and $$\sigma _e$$ be the corresponding equivalence relations. The equivalence relation $$\pi _e\vee \sigma _e$$ is defined to be the one for which $$a(\pi _e\vee \sigma _e)b$$ if and only if there is a sequence $$u_0=a,u_1,\dots ,u_k=b$$ of elements in A such that

\begin{aligned} u_0\rho _eu_1,u_1\rho _eu_2,\dots ,u_{k-1}\rho _eu_k. \end{aligned}

Here in every occurrence $$\rho _e$$ is either $$\pi _e$$ or $$\sigma _e$$. The corresponding partition is denoted by $$\pi \vee \sigma$$, and we say that $$\pi \vee \sigma$$ is the join of $$\pi$$ and $$\sigma$$. In other words, $$\pi \vee \sigma$$ is the smallest partition (with respect to the above-defined relation <) having both $$\pi$$ and $$\sigma$$ as its refinement.

The equivalence relation $$\pi _e\wedge \sigma _e$$ is the one for which

\begin{aligned} a(\pi _e\wedge \sigma _e)b\text { if and only if }a\pi _eb,\text { and }a\sigma _eb\quad (a,b\in A). \end{aligned}

The partition corresponding to $$\pi _e\wedge \sigma _e$$ is denoted by $$\pi \wedge \sigma$$, and called the meet of $$\pi$$ and $$\sigma$$. Clearly, $$\pi \wedge \sigma$$ is the greatest partition (with respect to <) which is the common refinement of $$\pi$$ and $$\sigma$$.

The collection of all partitions of A with respect to < is a partially ordered set (poset), and this poset with $$\vee$$ and $$\wedge$$ forms a lattice. This lattice is called partition lattice or equivalence lattice in the literature [1, 9]. The lattice structure depends only on the cardinality of A. For $$|A|=n$$, we denote the partition lattice by $${\mathcal {P}}(n)$$.

### 1.2 The restricted set partition lattices

#### 1.2.1 Definition

Our goal here is to define the modification of the partition lattice when the equivalence relations are not entirely arbitrary, but are subject of a restriction: Some fixed elements are not permitted to be equivalent. The enumerative problems of such partitions are well studied, see [4, 12]. But, as far as we know, lattice-theoretical questions of such equivalence relations have not been studied yet.

Let us introduce the notation $$[n]=\{1,2,\dots ,n\}$$. We will only consider finite lattices.Footnote 1 Without loss of generality, we can therefore suppose that A is some initial chain of the positive integers, i.e., $$A=[n]$$ for some positive integer n, and the elements which are not permitted to be equivalent belong to [r] for some $$1\le r\le n$$. These elements will be called special, as well as the blocks containing them. Partitions belonging to these restricted equivalence relations will be called restricted. The number of special elements is the degree of restriction, and we denote it by r.

The relation < and operation $$\wedge$$ can be defined in the same manner as before, but $$\vee$$ needs attention: The join of two restricted partitions may not satisfy the restriction. For example, if $$r\ge 2$$, then

\begin{aligned} 1,3|2,4|5,6|7\vee 1,4|2|3|5,7|6=1,2,3,4|5,6,7. \end{aligned}

In such cases, the join is defined to be a fictive element $$\hat{1}$$, and we fix it to be the greatest among all the elements in the restricted partition lattice. Clearly, $${\hat{1}}$$ is not a restricted partition, this is why we call it fictive. Now, it can be seen that $$(A,<,\vee ,\wedge )$$ forms a lattice. We call this lattice the restricted partition lattice and denote it by $${\mathcal {P}}_r(A)$$, or, when $$|A|=n$$, simply by $${\mathcal {P}}_r(n)$$. Its minimal element (the partition $$1|2|\cdots |n$$) will be denoted by $$\hat{0}$$

$${\mathcal {P}}_2(4)$$ is drawn in Fig. 1.

#### 1.2.2 Some elementary properties of $${\mathcal {P}}_r(n)$$

Let us discuss first some elementary properties of the restricted partition lattice.

A lattice is atomistic, if every nonzero element can be represented as a join of atoms. $${\mathcal {P}}_r(n)$$ is atomistic, the atoms being the partitions

\begin{aligned} \pi _{a,b}=a,b|\text {singletons}\quad (a,b\in A). \end{aligned}

Obviously, the restricted partition lattice $${\mathcal {P}}_r(n)$$ is not a sublattice of $${\mathcal {P}}(n)$$ if $$r>1$$, because the join is defined differently in the two lattices. However, $${\mathcal {P}}(n-r)$$ is a sublattice of $${\mathcal {P}}_r(n)$$, because every element in $${\mathcal {P}}(n-r)$$ can be represented as a join of elements in $${\mathcal {P}}_r(n)$$ of the form

\begin{aligned} \rho _{a,b}=1|2|\cdots |r|a,b|\quad (a,b\in [n]{\setminus }[r]). \end{aligned}

$${\mathcal {P}}_r(n)$$ and $${\mathcal {P}}(n)$$ are similar; however, in that they are both ranked. To introduce the rank, we need the notion of covering. We say that $$\pi$$ is covered by $$\sigma$$ (or $$\sigma$$ covers $$\pi$$) if $$\pi <\sigma$$ and there is no element $$\gamma$$ satisfying $$\pi<\gamma <\sigma$$. We denote this property by $$\pi \prec \sigma$$.

The rank of the smallest element containing only singletons (the element $${\hat{0}}$$ that we introduced earlier) is zero: $$r({\hat{0}})=0$$. The rank of the atoms (the elements covering $$\hat{0}$$) is one, and, in general, if $$\pi \prec \sigma$$, then $$r(\sigma )=r(\pi )+1$$. The element of the highest rank is $$\hat{1}$$, and

\begin{aligned} r(\hat{1})=n-r+1. \end{aligned}
(1)

It can be seen that the rank is well defined: The rank of an element does not depend on the choice of its predecessor. The elements which are covered by $$\hat{1}$$ are called co-atoms. $$r(\pi )=n-r$$ for co-atoms, and in the co-atoms every block is special. More concretely, co-atoms are of the form

\begin{aligned} 1,\dots |2,\dots |\cdots |r,\dots . \end{aligned}

The number of elements of a given rank is well studied; these are the so-called r-Stirling numbers of the second kind [4]. They are denoted by $$\genfrac\rbrace \lbrace {0.0pt}{}{n}{k}_r$$, and they give the number of partitions on n elements into k-blocks such that the first r elements are in separate blocks. Therefore,

\begin{aligned} \left| \{\pi \in {\mathcal {P}}_r(n)\mid r(\pi )=i\}\right| =\genfrac\rbrace \lbrace {0.0pt}{}{n}{n-i}_r\quad (i=0,1,\dots ,n-r). \end{aligned}
(2)

These considerations also imply that

\begin{aligned} \left| {\mathcal {P}}_r(n)\right| =1+\sum _{i=0}^{n-r}\genfrac\rbrace \lbrace {0.0pt}{}{n}{n-i}_r=1+B_{n,r}. \end{aligned}
(3)

One must be added, because $$\hat{1}$$ is not a regular partition and it is not counted by the r-Stirling numbers. $$B_{n,r}$$ is the nth r-Bell number [12]:

\begin{aligned} B_{n,r}=\sum _{i=0}^{n-r}\genfrac\rbrace \lbrace {0.0pt}{}{n}{n-i}_r=\sum _{i=r}^{n}\genfrac\rbrace \lbrace {0.0pt}{}{n}{i}_r. \end{aligned}

Before going to see the more involved properties of $${\mathcal {P}}_r(n)$$, still there is one easy structure attached to $${\mathcal {P}}_r(n)$$ we determine; this is the automorphism group, $$\mathrm {Aut}({\mathcal {P}}_r(n))$$. An automorphism $$\alpha$$ of a lattice is a bijection on the lattice as a set such that the meet and join are preserved by $$\alpha$$. It is well known—and easy to prove—that $$\mathrm {Aut}({\mathcal {P}}(n))\cong S_n$$ ($$S_n$$ being the symmetric group on [n]): The lattice structure is invariant under the “relabeling” of the elements of the set A. However, this is not true in the restricted partition lattice, when $$r>1$$: The special and non-special elements cannot be mapped into each other. When a permutation $$\alpha$$ interchanges, say, r and n, the result could lead out of $${\mathcal {P}}_r(n)$$:

\begin{aligned} \alpha (1,\dots |\cdots |s,n,\dots |\cdots |r,\dots |\cdots )=1,\dots |\cdots |s,r,\dots |\cdots |n,\dots |\cdots \not \in {\mathcal {P}}_r(n). \end{aligned}

Therefore, we infer that the special and non-special elements can be permuted only among themselves:

\begin{aligned} \mathrm {Aut}({\mathcal {P}}_r(n))\cong S_{n-r}\times S_r\quad (r>1). \end{aligned}

## 2 Properties of the restricted partition lattice $${\mathcal {P}}_r(n)$$

### 2.1 $${\mathcal {P}}_r(n)$$ is semimodular only when $$n=r+1$$

The $${\mathcal {P}}(n)$$ lattice is a semimodular lattice. This property, as we shall see soon, is not inherited by $${\mathcal {P}}_r(n)$$.

Semimodularity can be defined in many equivalent forms [9, Section V.2]; the definition that best fit our purpose is the following. A lattice is semimodular if (and only if) the following condition holds. If $$\pi$$ and $$\sigma$$ cover $$\pi \wedge \sigma$$, then $$\pi \vee \sigma$$ covers $$\pi$$ and $$\sigma$$, for all $$\pi$$ and $$\sigma$$.

### Proposition 1

Let $$2\le r<n$$ be integers. The restricted partition lattice $${\mathcal {P}}_r(n)$$ is semimodular only when $$n=r+1$$.

### Proof

To see why $${\mathcal {P}}_r(r+1)$$ is semimodular, choose $$\pi ,\sigma \in {\mathcal {P}}_r(r+1)$$ such that

\begin{aligned} \pi \wedge \sigma =1|2|\cdots |r|r+1. \end{aligned}

(Note that any other choice of $$\pi \wedge \sigma ={\hat{1}}$$ would lead to triviality.) Then, if we want $$\pi$$ and $$\sigma$$ to cover $$\pi \wedge \sigma$$ we have (without loss of generality) that

\begin{aligned} \pi&=1,r+1|2|\cdots |r,\\ \sigma&=1|2,r+1|3|\cdots |r. \end{aligned}

Now, $$\pi \vee \sigma ={\hat{1}}\gtrdot \pi ,\sigma$$, so semimodularity follows.

If $$n>r+1$$, the semimodularity does not hold. Indeed, if $$\pi \wedge \sigma$$ contains two non-special blocks, i.e.,

\begin{aligned} \pi \wedge \sigma =A_1|A_2|\cdots |A_r|A_{r+1}|A_{r+2}, \end{aligned}

then, among others, we have the below option to cover $$\pi \wedge \sigma$$ by $$\pi$$ and $$\sigma$$:

\begin{aligned} \pi&=A_1\cup A_{r+1}|A_2|\cdots |A_r|A_{r+2},\\ \sigma&=A_1|A_2\cup A_{r+1}|A_3|\cdots |A_r|A_{r+2}. \end{aligned}

Now,

\begin{aligned} \pi \vee \sigma =A_1\cup A_2\cup A_{r+1}|A_3|\cdots |A_r|A_{r+2}=\hat{1}. \end{aligned}

But $$\hat{1}$$ covers neither $$\pi$$ nor $$\sigma$$, since

\begin{aligned} \pi<A_1\cup A_{r+1}|A_2\cup A_{r+2}|\cdots |A_r<{\hat{1}}, \end{aligned}

and similarly for $$\sigma$$. $$\square$$

### 2.2 The complementation and relative complementation properties

Let P be an arbitrary lattice with 0 and 1. P is said to be complemented, if for any element a, there exists (a not necessarily unique) element b such that

\begin{aligned} a\wedge b=0,\quad \text{ and }\quad a\vee b=1. \end{aligned}

P is relatively complemented, if each of its intervals is complemented (relative to itself). Observe that a relatively complemented lattice is complemented.

Finite, atomistic, semimodular lattices are relative complemented [1, Proposition 2.36]; and so, in particular, $${\mathcal {P}}(n)$$ is relatively complemented for all $$n=1,2,\dots$$. The restricted partition lattice fails to be semimodular (except when $$n=r+1$$), but we will show now that it is, nevertheless, relatively complemented (and so complemented).

To show that $${\mathcal {P}}_r(n)$$ is relatively complemented, we need to prove that any interval

\begin{aligned}{}[\pi ,\sigma ]=\{\rho \in {\mathcal {P}}_r(n)\mid \pi \le \rho \le \sigma \} \end{aligned}

is complemented (where the relative zero is $$\pi$$ and relative one is $$\sigma$$). We have to distinguish two cases: $$\sigma <\hat{1}$$ and $$\sigma =\hat{1}$$.

In the case when $$\sigma <\hat{1}$$, the interval $$[\pi ,\sigma ]$$ is isomorphic to an interval in a classical partition lattice, because on any chain from $$\pi$$ to $$\sigma$$, the special elements automatically remain in different blocks. Therefore, the relative complementation is inherited from the classical partition lattice.

When $$\sigma =\hat{1}$$, the interval $$[\pi ,\sigma ]$$ is of the following form:

\begin{aligned} \prod _{\rho \prec \hat{1}}[\pi ,\rho ]\oplus \hat{1}. \end{aligned}

Here, $$\oplus \hat{1}$$ is the operation of adding $$\hat{1}$$ to the top of the product of intervals, and connecting all the $$\rho$$s with it. The intervals in the product are relatively complemented by the previous point, and $$\hat{1}$$ is complemented with $$\pi$$ as its complement. We conclude that $${\mathcal {P}}_r(n)$$ is relatively complemented.

### 2.3 The Möbius function of $${\mathcal {P}}_r(n)$$

#### 2.3.1 The calculation of $$\mu ({\mathcal {P}}_r(n))$$

The Möbius function is an important map on partially ordered sets (posets); among others, it is useful in the development of the discrete version of integrals and differentials on posets [1, Section IV.2]. It is defined on a poset P as

\begin{aligned} \mu (\pi ,\pi )=1;\quad \mu (\pi ,\sigma )=-\sum _{\pi \le \rho <\sigma }\mu (\pi ,\rho )\quad (\pi ,\sigma \in P). \end{aligned}
(4)

The particular value $$\mu (\hat{0},\hat{1})$$ of the Möbius function is usually denoted by $$\mu (P)$$. It is well known [1, p. 154], [13, p. 128], that

\begin{aligned} \mu ({\mathcal {P}}(n))=(-1)^{n-1}(n-1)!\quad (n\ge 1). \end{aligned}

The following proposition determines the value of the Möbius function on the restricted partition lattice.

### Proposition 2

For all $$r\ge 2$$ and $$n\ge 0$$,

\begin{aligned} \mu ({\mathcal {P}}_r(n+r))=(-1)^{n-1}(r-1)^{{\overline{n}}}. \end{aligned}

Here, $$a^{{\overline{n}}}=a(a+1)\cdots (a+n-1)$$ is the rising factorial with $$a^{{\overline{0}}}=1$$.

### Proof

Since $${\mathcal {P}}_r(r)\cong {\mathcal {C}}_2$$, the chain of two elements, it follows that

\begin{aligned} \mu ({\mathcal {P}}_r(0+r))=-1=(-1)^{0-1}(r-1)^{{\overline{0}}}. \end{aligned}

For $$n=1$$, $${\mathcal {P}}_r(1+r)$$ contains $$\hat{0}$$, $$\hat{1}$$, and r number of atoms where the one non-singleton block is of the form $$\{s,r+1\}$$ ($$1\le s\le r$$). It thus follows from (4) that

\begin{aligned} \mu ({\mathcal {P}}_r(1+r))=-\sum _{\hat{0}\le \rho <\hat{1}}\mu (\hat{0},\rho )=-\left( \mu (\hat{0},\hat{0})+r(-1)\right) =r-1=(-1)^{1-1}(r-1)^{{\overline{1}}}. \end{aligned}

For $$n>1$$, instead of using the definition of the $$\mu$$ function, we recall Corollary 3.9.3 in [13]. This statement says that for any $$\pi \ne {\hat{1}}$$

\begin{aligned} \sum _{\sigma :\sigma \wedge \pi ={\hat{0}}}\mu (\sigma ,{\hat{1}})=0. \end{aligned}
(5)

Since

\begin{aligned}{}[\sigma ,1]\cong {\mathcal {P}}_r(b(\sigma )), \end{aligned}
(6)

where $$b(\sigma )$$ is the number of blocks (special or not) of $$\sigma$$, a good choice of $$\pi$$ will lead to a recursive expression for $$\mu ({\mathcal {P}}_r(n+r))$$. Indeed, let us choose

\begin{aligned} \pi =1|\cdots |r|r+1,\dots ,n+r\in {\mathcal {P}}_r(n+r). \end{aligned}

Those $$\sigma$$’s which satisfy the equation $$\sigma \wedge \pi =0$$ are $$\sigma =0$$, or any non-singleton special block of $$\sigma$$ contains exactly two elements; and the non-special blocks are all singletons. Indeed, if a non-singleton special block contained at least two non-special elements, then these could be decomposed into atoms in which the special blocks are singletons; thus, $$\sigma$$ would have a non-$${\hat{0}}$$ meet with $$\pi$$. In addition, if $$\sigma$$ had a non-special block of at least two elements, then this could be decomposed into atoms of non-special blocks; but $$\pi$$ has a non-$$\hat{0}$$ meet with all of such atoms.

Let us suppose that there are k special blocks containing exactly two elements in $$\sigma$$ ($$0<k\le r$$). There are $$\left( {\begin{array}{c}n\\ k\end{array}}\right) r(r-1)\cdots (r-k+1)$$ such $$\sigma$$. Moreover, by (6) and by the properties of $$\sigma$$, $$[\sigma ,\hat{1}]\cong {\mathcal {P}}_r(n-k+r)$$. Therefore, (5) yields that

\begin{aligned} \mu ({\mathcal {P}}_r(n+r))+\sum _{k=1}^r\left( {\begin{array}{c}n\\ k\end{array}}\right) r(r-1)\cdots (r-k+1)\mu ({\mathcal {P}}_r(n-k+r))=0, \end{aligned}
(7)

the first term belonging to $$\sigma =\hat{0}$$.

We now use induction on n. It was seen above that $$\mu ({\mathcal {P}}_r(n+r))=(-1)^{n-1}(r-1)^{{\overline{n}}}$$ holds for $$n=0,1$$, so let $$n>1$$. Supposing that this formula holds up to $$n-1$$, (7) gives

\begin{aligned} \mu ({\mathcal {P}}_r(n+r))= & {} -\sum _{k=1}^r\left( {\begin{array}{c}n\\ k\end{array}}\right) r(r-1)\cdots (r-k+1)(-1)^{n-k-1}(r-1)^{\overline{n-k}}\\= & {} (-1)^n\sum _{k=1}^r\left( {\begin{array}{c}n\\ k\end{array}}\right) (-r)^{{\overline{k}}}(r-1)^{\overline{n-k}}. \end{aligned}

The last sum above is the binomial convolution of the rising factorial. In general [8, Chap. 5, Ex. 37]:

\begin{aligned} \sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) x^{{\overline{k}}}y^{\overline{n-k}}=(x+y)^{{\overline{n}}}, \end{aligned}
(8)

whence

\begin{aligned} \mu ({\mathcal {P}}_r(n+r))= & {} (-1)^n\sum _{k=1}^r\left( {\begin{array}{c}n\\ k\end{array}}\right) (-r)^{{\overline{k}}}(r-1)^{\overline{n-k}}\\= & {} (-1)^n\left( (-1)^{{\overline{n}}} -\left( {\begin{array}{c}n\\ 0\end{array}}\right) (-r)^{\overline{0}}(r-1)^{{\overline{n}}}\right) {\mathop {=}\limits ^{n>1}}(-1)^{n-1}(r-1)^{{\overline{n}}}. \end{aligned}

(Note that $$(-1)^{{\overline{n}}}$$=0 when $$n>1$$, and also that $$(-r)^{{\overline{k}}}=(-r)(-r+1)\cdots (-r+k-1)$$ is zero whenever $$k>r$$, so we can restrict the summation to the interval $$\{1,2,\dots ,r\}$$ when applying (8).) The formula for $$\mu ({\mathcal {P}}_r(n+r))$$ thus holds for arbitrary n. $$\square$$

#### 2.3.2 Another expression for $$\mu ({\mathcal {P}}_r(n))$$

There is another formula for the Möbius function which we use now. This expresses $$\mu ({\mathcal {P}}_r(n))$$ as an alternating sum of the number of chains of different lengths. More precisely, let $$c_i$$ be the number of chains of the form

\begin{aligned} \hat{0}=\pi _0<\pi _1<\cdots <\pi _i=\hat{1}. \end{aligned}

Thus, $$c_i$$ counts the chains of length i (and $$i+1$$ nodes). Clearly, $$c_0=0$$ and $$c_1=1$$ for every lattice having $$\hat{0}$$ and $$\hat{1}$$. Then, [13, Proposition 3.8.5]

\begin{aligned} \mu ({\mathcal {P}}_r(n))=c_0-c_1+c_2-c_3+\cdots . \end{aligned}
(9)

In $${\mathcal {P}}_r(n)$$, there are $$B_{n,r}-1$$ elements strictly between $$\hat{0}$$ and $$\hat{1}$$ (note that $$B_{n,r}$$ does not count $$\hat{1}$$ since $$\hat{1}$$ is not a partition). Therefore, in $${\mathcal {P}}_r(n)$$ the number of chains of length two is

\begin{aligned} c_2=c_2(n,r)=B_{n,r}-1. \end{aligned}

(To avoid confusion, we explicitly write out the n and r-dependence.) The number of chains of length i can be determined recursively based on the number of chains of length $$i-1$$. Indeed, let us choose a lattice element $$\pi \in {\mathcal {P}}_r(n)$$ of rank j. Since $$[\pi ,\hat{1}]\cong {\mathcal {P}}_r(n-j)$$, there are $$c_{i-1}(n-j,r)$$ chains in $${\mathcal {P}}_r(n)$$ with length $$i-1$$ starting from $$\pi$$ and ending in $$\hat{1}$$. (Equivalently, there are $$c_{i-1}(n-j,r)$$ chains in $${\mathcal {P}}_r(n-j)$$ of length $$i-1$$ starting from $$\hat{0}$$ and ending in $$\hat{1}$$, where $$\pi$$ plays the role of $$\hat{0}$$.) Note that, by (2), the total number of such $$\pi$$ partitions is $$\genfrac\rbrace \lbrace {0.0pt}{}{n}{n-j}_r$$. We now add $$\hat{0}$$ below $$\pi$$ so that our new chain will be of length i. Obviously, $$1\le j$$. On the other hand, from $$\hat{0}$$ we can go up until the rank $$j=r(\hat{1})-(i-1)=n-r+1-(i-1)=n-r-i+2$$ (see (1)) so that $$c_{i-1}(n-j,r)\ne 0$$.

Therefore, the total number of ways we can do the above construction to get the number of chains of length i is

\begin{aligned} c_i(n,r)=\sum _{j=1}^{n-r-i+2}\genfrac\rbrace \lbrace {0.0pt}{}{n}{n-j}_rc_{i-1}(n-j,r)=\sum _{j=r+i-2}^{n-1}\genfrac\rbrace \lbrace {0.0pt}{}{n}{j}_rc_{i-1}(j,r). \end{aligned}

According to (9), the alternating sum of these equals $$\mu ({\mathcal {P}}_r(n))$$.

## 3 Open questions

We would like to close the paper with some further questions and problems which could be interesting to study.

1. 1.

It is known that the classical partition lattice is simple. The proof (see [9]) uses perspectivity, direct indecomposability, and Dilworth’s theorem (it states that a geometric lattice is either simple, or decomposable). Geometricity does not hold in the restricted partition lattice in general (because of the lack of semimodularity), so simplicity does not follow. For which n and r is $${\mathcal {P}}_r(n)$$ simple?

2. 2.

Calculate or estimate the order dimension of the restricted partition lattice. This question was studied for $${\mathcal {P}}(n)$$ in [7].

3. 3.

How long is a maximal antichain in $${\mathcal {P}}_r(n)$$? There are tight bounds for the size of the longest antichains in the classical partition lattice. See the papers of E. R. Canfield [5], and Canfield, and L. H. Harper [6].

4. 4.

The number of not necessarily maximal chains starting from 0 and ending at 1 in $${\mathcal {P}}(n)$$ is denoted by $$Z_n$$. This sequence has some intriguing properties, and it was thoroughly studied by T. Lengyel [11]. See also [3, 10] for the number-theoretic properties of $$Z_n$$. What can we say about the number of such chains in $${\mathcal {P}}_r(n)$$?

5. 5.

Generalize $${\mathcal {P}}_r(n)$$ to infinite sets.