Power set of A is a complete lattice

Question

I am currently trying to proof that the power set of A is a complete lattice.

Since $\mathcal{P}(A),\subset$ is a partially ordered set, we still have to proof that $\sup(X)$ and $\inf(X)$ exist, for every not empty subset of $\mathcal{P}(A)$.

One can see, making a sketch that: $$ \sup(X)=\cup_{C \in X} C $$

$$ \inf(X)=\cap_{C \in X} C $$

It is easy to proof that $\cup_{C \in X} C$ is a lower bound and that $\cap_{C \in X} C$ is an upper bound. I don't get how to proof that they are the largest lower bound and the smallest upper bound.

Answer 1

Let $\mathcal X\subseteq\wp(A)$ and let $B$ be an upperbound of $\mathcal X$ w.r.t. the inclusion.

That means that $C\subseteq B$ for each $C\in\mathcal X$.

Then also $\bigcup\{C\mid C\in\mathcal X\}\subseteq B$ (do you see why?).

Next to that $\bigcup\{C\mid C\in\mathcal X\}$ itself is also an upper bound of $\mathcal X$.

(In your question you call it incorrectly a lower bound)

(To avoid confusion: $\bigcup\{C\mid C\in\mathcal X\}=\bigcup_{C\in\mathcal X}C$)

These two facts allow the conclusion that $\bigcup\{C\mid C\in\mathcal X\}$ is the least upper bound of $\mathcal X$.

Likewise it can be shown that $\bigcap\{C\mid C\in\mathcal X\}$ serves as greatest lower bound of $\mathcal X$.

Answer 2

Suppose that $B$ is another upper bound of $X$. So for all $C \in X$, $C \subseteq B$. But then $\cup_{C \in X} C \subseteq B$ as well ($x$ in the union, so $x$ is some $C'$ for $C' \in X$. And then $C' \subseteq B$, so $x \in B$). So the union is smaller than then any other upper bound, hence it by definition the smallest upper bound.

A dual argument can be held for the intersection: suppose $B \subseteq C$ for all $C \in X$ (so $B$ is a lower bound for $X$). Then $B \subseteq \cap_{C \in X} C$ (if $x \in B$, then for any $C \in X$, $B \subseteq C$, so $x \in C$. As $C$ was arbitratry, we have proved the inclusion). So $B$ is smaller than the intersection of $X$, so the latter is the maximal lower bound, as required.