Is every subset of a lattice a lattice?

Lattice consists of a partially ordered set in which every two elements have to have unique supremum and infimum.

I'm confused about what the answer is. I considered a lattice $(L, \le)$ where $L$ is a set {1, 2, 3, 6} and $\le$ is relation of divisibility (a simplified version of this example) (e.g. 1 divides 2, 3 and 6, 2 divides 6, etc.).

Now if I take $\{1, 2, 3\}$ as a subset of this lattice, would this still be a lattice? Would $2$ and $3$ be missing a supremum? On one hand I think they would because $6$ is no longer in the set but on the other hand isn't supremum not required to be in the set in question? For example the set of real numbers on an open interval $(0, 1)$ has a supremum equal to 1, even though 1 is not in the set.

  • $\begingroup$ Your first interpretation is correct. The supremum needs to be an element of the lattice. The open unit interval isn't a lattice under the usual ordering. $\endgroup$ – Ethan Bolker Mar 8 '18 at 20:19

The example you gave is correct and also the smallest. Your poset is a rombus, where $6$ is the top vertex, $2$ and $3$ the mid ones, and $1$ the bottom vertex.

 . .

If you remove either the top vertex or the bottom you lose the existence of supremum or infimum respectively.

This is only a small correction to Ethan Bolker's comment. The open unit interval, as any totally ordered set is a lattice. The definition of lattice only requires the existence of join and meet for pairs of elements (and by induction for finite subsets). In a totally ordered set supremum of two elements is the maximum among them, and the infimum is the minimum among them.

In particular the totally ordered sets are lattices for which every subset is also a lattice.

  • $\begingroup$ Thank you for this clarification, I suspected the open unit interval is a lattice. I have one more question - I often read that $(0, 1)$ has a supremum $1$. But $1$ is not in $(0, 1)$. So why does not belonging to lattice disqualifies $6$ as being a supremum in my example? Is the definition of supremum different for lattice? $\endgroup$ – Matt Mar 8 '18 at 20:36
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    $\begingroup$ @Matt The question is which one is the total space, which one is the space where the supremum is being computed. $\{2,3\}$ has a supremum ... in $\{1,2,3,6\}$. Likewise $(0,1)$ has a supremum in $[0,1]$, or in $\mathbb{R}$. On the other hand $\{2,3\}$ doesn't have a supremum in $\{2,3\}$ (itself), and $(0,1)$ doesn't have a supremum in $(0,1)$. $\endgroup$ – crivair Mar 8 '18 at 20:40
  • $\begingroup$ Your last sentence is actually an equivalence: a lattice is totally ordered iff every subset is a lattice. For the converse, consider two-elements subsets. $\endgroup$ – Najib Idrissi Mar 8 '18 at 20:45

The space need to contain the supremum of every pair of objects in that subset. Althought, the statement is true for complete lattices.


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