The upper and lower closures of a set are the smallest upper and lower sets containing it

Question

Let $s$ be an element of a partially ordered set $(P,\le)$, the upper closure of $s$ denoted $\small{\uparrow} \large{s}$ is the set of all elements in $P$ greater than or equal to $s$,e.g.:

$$\left\{p \in\ P:\ s\le p\right\}$$

The lower closure can be defined analogously.

Given a subset $\mathbb S$ of a partially ordered set $(P,\le)$, the upper closure of $\mathbb S$ is the set of all closures of the elements contained in $\mathbb S$,e.g.: $$\uparrow \mathbb S := \bigcup_{s \in \mathbb S}\small{\uparrow} \large{s}$$

The lower closure can be defined analogously.

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Prove:

The upper and lower closures of a set are the smallest upper and lower sets containing it.

Updated.

My try:

Let $(S,≤)$ be a partially ordered , we want to show that if $U$ is an upper closure for $A⊆S$ , then U is also an upper set, so assume $u∈U$ and $s∈S:u≤s$, since U is an upper closure for $A$ so $∃a∈A:a≤u$ , by transitivity of $≤$ implies $a≤s$ , But I don't know how to show that $s∈U$ and finish the proof.

Answer 1

The upper set of $S$,
upper closure of $S = \left\{ x : \exists \;s\in S , \; s \le x \right\}$.
Let $U$ be an upper set with $S \subset U$.
Since $U$ is an upper set,
$u$ in $U$ and $u \le x$ implies $x$ in $U$.
As $S$ subset $U$, $s$ in $S$ and $s \le x$ implies $x$ in $U$.
Thus if in upper closure of $S$, exists $s$ in $S$ with $s \le x$.
Hence, from the previous, $x$ in $U$.
Consequently up $S$ subset $U$.
Now show upper closure of $S$ is an upper set and conclude
upper closure of $S$ is the smallest upper set containing S.

Answer 2

A detailed answer from a reputable source (that is me :-) ).

Let’s finish your proof.

Since $a\le s$, $s\in{\uparrow}a$ by the definition of ${\uparrow} a$. Since $a\in A$, $s\in{\uparrow} a$, and $U=\bigcup_{b\in A}{\uparrow} b$, we have $s\in U$. That is, we showed that for any $u\in U$ and any $s\in{\uparrow} U$ we have $s\in U$, that is, that $U$ is an upper set.

Now let $V\subset S$ be any upper set which contains $A$. We have to check that $V$ contains $U$. Indeed, let $u$ be any element of $U$. Since $U=\bigcup_{a\in A}{\uparrow} a$, there exists $a\in A$ such that $u\in{\uparrow} a$. Since $a\in A\subset V$, $a\le u$, and $V$ is an upper set, $u\in V$. Since this holds for any element $u\in U$, we have $U\subset V$. Since $U$ is an upper set containing $A$, and $U$ is contained in any upper set $V\supset A$, we see that $U$ is the smallest upper set containing $A$. The end of the proof.

The proof for the case of lower set and lower closure should be obtained from the above proof, by applying to it the respective changes. That is replacing each “$\le$” by “$\ge$”, “$\uparrow$” by “$\downarrow$”, “upper” by “lower” etc. Saying the truth, I did not check this, but I’ll be very surprised if this will not work. :-)