# upper set operation is a closure operator

In this entry, we shall prove the assertion made in
the main entry (http://planetmath.org/UpperSet) that $\uparrow $
is a closure operator^{}. This will be done by checking
that the defining properties are satisfied. To begin,
recall the definition of our operation^{}:

###### Definition 1.

Let $P$ be a poset and $A$ a subset of $P$. The *upper set*
of $A$ is defined to be the set

$$\uparrow A=\{b\in P\mid (\exists a\in A)a\le b\}$$ |

Now, we verify each of the properties which is required of a closure operator.

###### Theorem 1.

$\uparrow \mathrm{\varnothing}=\mathrm{\varnothing}$

###### Proof.

Any statement of the form “$(\exists a\in \mathrm{\varnothing})P(a)$”
is identically false no matter what the predicate^{} $P$ (i.e. it is
an antitautology) and the set of objects satsfying an identically
false condition is empty, so $\uparrow \mathrm{\varnothing}=\mathrm{\varnothing}$.
∎

###### Theorem 2.

$A\subseteq \uparrow A$

###### Proof.

This follows from reflexivity^{} — for every $a\in A$, one has
$a\le a$, hence $a\in \uparrow A$.
∎

###### Theorem 3.

$\uparrow \uparrow A=\uparrow A$

###### Proof.

By the previous result, $\uparrow A\subseteq \uparrow \uparrow A$. Hence, it only remains to show that $\uparrow \uparrow A\subseteq \uparrow A$. This follows from transitivity. In order for some $a$ to be an element of $\uparrow \uparrow A$, there must exist $b$ and $c$ such that $a\ge b\ge C$ and $C\in A$. By transitivity, $A\ge C$, so $a\in \uparrow A$, hence $\uparrow \uparrow A\subseteq \uparrow A$ as well. ∎

###### Theorem 4.

If $A$ and $B$ are subsets of a partially ordered set^{}, then

$$\uparrow (A\cup B)=(\uparrow A)\cup (\uparrow B)$$ |

###### Proof.

On the one hand, if $a\in \uparrow (A\cup B)$, then $a\ge b$ for some $b\in A\cup B$. It then follows that either $b\in A$ or $b\in B$. In the former case, $a\in \uparrow A$, in the latter case, $a\in \uparrow B$ so, either way $a\in (\uparrow A)\cup (\uparrow B)$. Hence $\uparrow (A\cup B)\subseteq (\uparrow A)\cup (\uparrow B)$.

On the other hand, if $a\in (\uparrow A)\cup (\uparrow B)$, then either $a\in (\uparrow A)$ or $a\in (\uparrow B)$. In the former case, there exists $b$ such that $a\ge b$ and $b\in A$. Since $A\subseteq A\cup B$, we also have $b\in A\cup B$, hence $a\in \uparrow (A\cup B)$. Likewise, in the second case, we also conclude that $a\in \uparrow (A\cup B)$. Therefore, we have $(\uparrow A)\cup (\uparrow B)\subseteq \uparrow (A\cup B)$. ∎

###### Theorem 5.

$\uparrow P=P$

###### Theorem 6.

$A\subseteq B$, $\mathrm{\uparrow}A\mathrm{\subseteq}\mathrm{\uparrow}B$

Title | upper set operation is a closure operator |
---|---|

Canonical name | UpperSetOperationIsAClosureOperator |

Date of creation | 2013-03-22 16:41:43 |

Last modified on | 2013-03-22 16:41:43 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 8 |

Author | rspuzio (6075) |

Entry type | Theorem^{} |

Classification | msc 06A06 |