Clarification of partial order set definition

Question

Consider the following definition of the set $\mathcal{X}$ where $\mathcal{S}$ is a partially ordered set. $$X\in\mathcal{X} = \{X\subseteq \mathcal{S} : x\in X\text{ and }y\succeq x\text{ imply }y\in X\}$$

I am a bit confused about what $\mathcal{X}$ looks like.

Consider the following lattice $\mathcal{S} = \{(1,1),(2,1),(1,2),(2,2)\}$ with the elementwise ordering. Note that $(1,1)$ is smaller than all other elements and $(2,2)$ is larger than all other elements.

What does the set $\mathcal{X}$ look like for this example?

Is it $$\mathcal{X} = \big\{\{(1,1),(1,2),(2,2)\},\{(1,1),(2,1),(2,2)\},\{(1,2),(2,2)\},\{(2,1),(2,2)\},\{(2,2)\}\big\}$$ or $$\mathcal{X} = \big\{\{(1,1),(1,2),(2,1),(2,2)\},\{(1,2),(2,2)\},\{(2,1),(2,2)\},\{(2,2)\}\big\}$$

Answer 1

I suspect that for partially ordered set $\langle \mathcal S,\succeq\rangle$, your definition should be $$\mathcal X=\{X\subseteq\mathcal S: \forall x\in X~\forall y\in\mathcal S~(y\succeq x\to y\in X)\}$$

Now, assuming the $\succeq$ ordering is a diamond lattice: $$\begin{array}{c}&&(2,2)\\&\nearrow&&\nwarrow\\(2,1)&&&&(1,2)\\&\nwarrow&&\nearrow\\&&(1,1)\end{array}$$

Then you would have:

$$\mathcal X=\big\{\{(1,1),(1,2),(2,1),(2,2)\}~,\{(1,2),(2,2)\}~,\{(2,1),(2,2)\}~,\{(2,2)\}~,\{\}\big\}$$

Answer 2

Thou doesth speak gibberish.
In consulation with Tweedledum and Tweedledee a translation:
X is the collection of upper subsets of S,
A in X iff A subset S and for x in A, y in S, (x <= y implies y in A).

In the example, the first X is the collection of all upper chains, the second X is the collection of all upper sets.

Who gave you that nonsense definition?