Why could not we choose the first element as bottom?

Question

I am trying a solve the Exercise 4 of Section 2.4 in the book Category Theory for Computing Science.

The exercise is "Show that for every set $S$, the poset $(\mathcal{P}(S),\subseteq)$ is a strict $\omega$-CPO."

The definition of $\omega$-CPO is "A poset $(\mathcal{P}(S),\leq)$ is an $\omega$-complete partial order, or $\omega$-CPO, if every chain has a supremum. If the poset also has a minimum element, it is called a strict $\omega$-CPO."

The answer of the book says "Let $\mathcal{L}=(C_0,C_1,C_2...)$ be a chain in $(\mathcal{P}(S),\subseteq)$, the bottom element is $\emptyset$".

However, I think the bottom element is $C_0$. What is wrong with my understanding?

Answer 1

To prove that $(\mathcal P(S), \subseteq)$ is a strict $\omega$-CPO, there are two parts:

1. Every chain in $(\mathcal P(S), \subseteq)$ has a supremum
2. $(\mathcal P(S), \subseteq)$ has a bottom element.

The proof first considers a chain $\mathcal L$ to show that it has a supremum. Having done this, it moves on to part 2, the bottom element, and observes that the bottom element of $(\mathcal P(S), \subseteq)$ is $\emptyset$.

It does not say that the bottom element of $\mathcal L$ is $\emptyset$.