I'm currently going through Theories of Programming Languages (ToPL) and am struggling a bit with building up an intuition for posets in certain instances. While I can somewhat follow the math, I feel like I don't fully "get it", so hoping someone can help here.
Example
So, to take an example in something concrete, in ToPL page 32. We first get given that a function, $f$, is continuous if and only if, for all chains $x_0 \sqsubseteq x_1 \sqsubseteq x_2 ...$ in P,
$f(\bigsqcup\limits_{i = 0}^\infty x_i) = \bigsqcup\limits_{i = 0}^\infty{}' fx_i$
Now, suppose P is the vertical domain of the natural numbers, and P' is the two-element domain $\{ \bot', \top' \}$. Then the monotone function $fx=\text{if } x = \infty \text{ then } \top' \text{ else } \bot'$,
is not continuous, since the limit of $\{0,1,...\}$ is $\infty $ and $f \infty $ is $\top'$, but the limit of $\{f0, f1,...\} = \{\bot'\}$ is $\bot'$.
Question
Where I kinda get lost is, from the original given equation, I can certainly follow the LHS, where it goes $\bigsqcup\limits_{i = 0}^\infty x_i = \infty$ and then $f(\infty) = \top'$, but the RHS I don't know how they get $\{f0, f1,...\} = \{\bot'\}$. I would assume that in $x_i$ you could find $\infty$ and then you would end up with $\top'$, which means the constraints hold. Where does that $= \{\bot'\}$ come from?
Hope this made sense, think it got a bit messy :/
Pardon my abuse of notation here, I'm not sure what is used in the book. The chain $x = (x_0, x_1, x_2, \ldots) = (0, 1, 2, \ldots)$ does not include $\infty$, so the set $\{f(x_i) \mid i \in \{0, 1, 2, \ldots\}\}$ is just $\{\bot'\}$. Hence we find that
$$ \bigsqcup_{i=0}^\infty f(x_i) = \bigsqcup \{f(x_i) \mid i \in \{0, 1, 2, \ldots\}\} = \bigsqcup \{\bot'\} = \bot'$$
However, we have that no positive integer $n$ bounds the chain $x$ from above, the least element that does is $\infty$. So $\bigsqcup_{i = 0}^\infty x_i = \infty$. Then
$$ f\left(\bigsqcup_{i = 0}^\infty x_i\right) = f(\infty) = \top'$$