Prove that the least upper bound of $\mathcal F$ is $\bigcup\mathcal F$ and the greatest lower bound of $\mathcal F$ is $\bigcap\mathcal F$.

Question

Not a duplicate of this or this.

This is exercise $4.4.23$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Prove theorem $4.4.11.$

Theorem $4.4.11.$ Suppose $A$ is a set, $\mathcal F\subseteq \mathscr P(A)$, and $\mathcal F\neq \emptyset$. Then the least upper bound of $\mathcal F$ $($in the subset partial order$)$ is $\bigcup\mathcal F$ and the greatest lower bound of $\mathcal F$ is $\bigcap\mathcal F$.

Here is my proof:

Let $F$ be an arbitrary element of $\mathcal F$. Let $x$ be an arbitrary element of $F$. Ergo clearly $x\in\bigcup\mathcal F$. Since $x$ is arbitrary, $F\subseteq\bigcup\mathcal F$. Therefore if $F\in\mathcal F$ then $F\subseteq\bigcup\mathcal F$. Since $F$ is arbitrary, $\bigcup\mathcal F$ is an upper bound for $\mathcal F$. Let $U$ be the set of all upper bounds for $\mathcal F$ and let $X$ be an arbitrary element of $U$. Let $y$ be an arbitrary element of $\bigcup\mathcal F$. So we can choose some $G_0\in\mathcal F$ such that $y\in G_0$. Since $X$ is an upper bound for $\mathcal F$ then $G_0\subseteq X$. Since $y\in G_0$, $y\in X$. Since $y$ is arbitrary, $\bigcup\mathcal F\subseteq X$. Thus if $X\in U$ then $\bigcup\mathcal F\subseteq X$. Since $X$ is arbitrary, $\bigcup\mathcal F$ is the smallest element of $U$ and hence the least upper bound for $\mathcal F$.

Let $F$ be an arbitrary element of $\mathcal F$. Let $x$ be an arbitrary element of $\bigcap\mathcal F$. Ergo clearly $x\in F$. Therefore if $F\in\mathcal F$ then $\bigcap\mathcal F\subseteq F$. Since $F$ is arbitrary, $\bigcap\mathcal F$ is a lower bound for $\mathcal F$. Let $L$ be the set of all lower bounds for $\mathcal F$ and let $Y$ be an arbitrary element of $L$. Let $y$ be an arbitrary element of $Y$. Since $Y$ is a lower bound for $\mathcal F$, $Y\subseteq F$. Since $y\in Y$, $y\in F$. Since $F$ is arbitrary, $y\in\bigcap\mathcal F$. Since $y$ is arbitrary, $Y\subseteq \bigcap\mathcal F$. Thus if $Y\in L$ then $Y\subseteq \bigcap\mathcal F$. Since $Y$ is arbitrary, $\bigcap\mathcal F$ is the biggest element of $L$ and hence the greatest lower bound for $\mathcal F$.

$Q.E.D.$

Is my proof valid$?$

Thanks for your attention.

Answer 1

Your proves look good. You could gain in clarity by explaining upfront what you're doing.

For example for least upper bond.

Let's first prove that $\bigcup \mathcal F$ is an upper bound. Proof...

And now let's prove that $\bigcup \mathcal F$ is less than any upper bound $U$. Proof...

This allows to conclude that $\bigcup \mathcal F$ is the least upper bound.