# Obtaining the lattice of equivalence relations

Is there an easy method to obtain the lattice of all equivalence relations $L_n$ of a set with $n$ elements in Sage?

Obtaining the lattice of equivalence relations

2

I do not know whether there is such a builtin construction in Sage, so here is a possible construction.

Let `S`

be a set, e.g.:

```
sage: S = {'a','b','c','d'}
```

First, we define the list of partitions over `S`

:

```
sage: list(SetPartitions(S))
[{{'a', 'b', 'c', 'd'}},
{{'a', 'b', 'c'}, {'d'}},
{{'a', 'b', 'd'}, {'c'}},
{{'a', 'b'}, {'c', 'd'}},
{{'a', 'b'}, {'c'}, {'d'}},
{{'a', 'c', 'd'}, {'b'}},
{{'a', 'c'}, {'b', 'd'}},
{{'a', 'c'}, {'b'}, {'d'}},
{{'a', 'd'}, {'b', 'c'}},
{{'a'}, {'b', 'c', 'd'}},
{{'a'}, {'b', 'c'}, {'d'}},
{{'a', 'd'}, {'b'}, {'c'}},
{{'a'}, {'b', 'd'}, {'c'}},
{{'a'}, {'b'}, {'c', 'd'}},
{{'a'}, {'b'}, {'c'}, {'d'}}]
```

Second, we define a function that decides whether a partition refines another one:

```
sage: refine = lambda p,q : all(any(set(i).issubset(set(j)) for j in q) for i in p)
sage: refine(((1,), (2, 3)), ((1,), (2,), (3,)))
False
sage: refine(((1,), (2,), (3,)), ((1,), (2, 3)))
True
```

With both the list of partitions and the refinment order, we can construct the poset:

```
sage: P = Poset((list(SetPartitions(S)), refine))
sage: P
Finite poset containing 15 elements
sage: P.is_lattice()
True
sage: P.plot()
```

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Asked: ** 2021-05-24 13:18:33 +0200 **

Seen: **6,451 times**

Last updated: **May 24 '21**

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