For A={1,2,3,4,5}, P(A) is partially ordered by the set inclusion relation $ \subseteq$. For the set {2,3,5}, there are three immediate predecessors in P(A):{2,3}, {2,5}, and {3,5}. The empty set has no immediate predecessor. Also $\emptyset$, is the only immediate predecessor for {3}. We have {4} $ \subseteq$ {2, 4, 5} but {4} is not an immediate predecessor of {2, 4, 5} because {4} $\ne$ {4,5}, {4,5} $\ne$ {2,4,5} , {4} $\subseteq$ {4,5}, and {4,5} $\subseteq$ {2,4,5}
P(A) = { {}, {1}, {2}, {3}, {4}, {5}, {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5}, {1,2,3}, {1,2,4}, {1,2,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,3,4}, {2,3,5}, {2,4,5}, {3,4,5}, {1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, {2,3,4,5}, {1,2,3,4,5} }.
In the example above they are claiming
For the set {2,3,5}, there are three immediate predecessors in P(A):{2,3}, {2,5}, and {3,5}.
This is really confusing me. I placed a bold on {2,3,5} in the power set P(A) above. If you look at the immediate predecessor of {2,3,5} you will see it is {2,3,4}. So how are they getting the 3 immediate predecessors of {2,3,5} as {2,3}, {2,5}, and {3,5} ?
I am also confused about the reasoning of why they say {4} is not an immediate predecessor of {2,4,5}.
{4} is not an immediate predecessor of {2, 4, 5} because {4} $\ne$ {4,5}, {4,5} $\ne$ {2,4,5} , {4} ⊆ {4,5}, and {4,5} ⊆ {2,4,5}
I also placed in bold {4} and {2,4,5} in the power set P(A) above. If you look at the power set P(A) above you will see the immediate predecessor of {2,4,5} is {2,3,5}, so {4} cannot be the immediate predecessor of {2,4,5}. But then why are they trying to determine if {4} = {4,5}? Or whether {4,5} = {2,4,5}? Or whether {4} $\subseteq$ {4,5}? Or whether {4,5} $\subseteq$ {2,4,5} ?
The usual order $\le$ on the natural numbers is completely irrelevant here: the elements of $\wp(A)$ are sets of integers, not integers themselves, and they are ordered by $\subseteq$, subset inclusion. Thus, $\{2\}\subseteq\{2,3,5\}$ and $\{2,3\}\subseteq\{2,3,5\}$, so $\{2\}$ and $\{2,3\}$ are predecessors of $\{2,3,5\}$ in the order $\subseteq$, the only one that counts here. But $\{2,3,4\}\nsubseteq\{2,3,5\}$, so $\{2,3,4\}$ is not a predecessor of $\{2,3,5\}$ in the order $\subseteq$.
If $X$ and $Y$ are subsets of $A$, $X$ is a predecessor of $Y$ if and only if $X\subseteq Y$; that’s all that matters. The relative magnitudes of the members of $X$ and $Y$ are irrelevant: we care only that every member of $X$ is also a member of $Y$. If that’s not the case, then $X\nsubseteq Y$, and $X$ is not a predecessor of $Y$, even if $X=\{1\}$ and $Y=\{2\}$.
$X$ is an immediate predecessor of $Y$ if no subset of $A$ lies strictly between them. That is, $X$ is an immediate predecessor of $Y$ if there is no $Z\in\wp(A)$ such that $X\subsetneqq Z\subsetneqq Y$. This is why $\{2\}$ is not an immediate predecessor of $\{2,3,5\}$: it’s a predecessor of $\{2,3,5\}$, since $\{2\}\subseteq\{2,3,5\}$, but the set $\{2,3\}$ fits strictly between $\{2\}$ and $\{2,3,5\}$:
$$\{2\}\subsetneqq\{2,3\}\subsetneqq\{2,3,5\}\;.$$
There is no subset of $A$ that fits strictly between $\{2,3\}$ and $\{2,3,5\}$, however: if you add any element to $\{2,3\}$, either you get $\{2,3,5\}$ itself (if you add $5$), or you get something that isn’t a predecessor (subset) of $\{2,3,5\}$ at all (if you add $1$ or $4$). The sets $\{2,5\}$ and $\{3,5\}$ also have this property: you cannot add anything to either of them to get a proper subset of $\{2,3,5\}$: if, for instance, $\{2,5\}\subseteq Z\subseteq\{2,3,5\}$, then either $Z=\{2,5\}$, or $Z=\{2,3,5\}$.
Thus, the predecessors of $\{2,3,5\}$ in the order $\subseteq$ are the sets
$$\varnothing,\{2\},\{3\},\{5\},\{2,3\},\{2,5\},\{3,5\},\text{ and }\{2,3,5\}\;,$$
the subsets of $\{2,3,5\}$. (Terminology varies a little here: some people would not include $\{2,3,5\}$ as a predecessor of itself.)
No, $\{2,3,5\}$ is not even a predecessor of $\{2,4,5\}$, let alone an immediate predecessor: $\{2,3,5\}$ is not a subset of $\{2,4,5\}$. On the other hand, $\{4\}\subseteq\{2,4,5\}$, so $\{4\}$ is a predecessor of $\{2,4,5\}$. However, we can fit the set $\{4,5\}$ strictly between them:
$$\{4\}\subsetneqq\{4,5\}\subsetneqq\{2,4,5\}\;.$$
By definition, therefore, $\{4\}$ is not an immediate predecessor of ${2,4,5\$}.
If the familiar ordering $\le$ on the members of the set $A$ continues to get in the way of understanding the order $\subseteq$ on the subsets of $A$, replace $A$ by the set $\{\bullet,\heartsuit,\clubsuit,\diamondsuit,\square\}$ and look at its subsets; this is really the same problem with the names changed. Now $\{\heartsuit,\clubsuit,\square\}\nsubseteq\{\heartsuit,\diamondsuit,\square\}$, so $\{\heartsuit,\clubsuit,\square\}$ is not a predecessor of $\{\heartsuit,\diamondsuit,\square\}$, let alone an immediate predecessor. But $\{\diamondsuit\}\subseteq\{\heartsuit,\diamondsuit,\square\}$, so $\{\diamondsuit\}$ is a predecessor of $\{\heartsuit,\diamondsuit,\square\}$. It’s not an immediate predecessor, however, because
$$\{\diamondsuit\}\subsetneqq\{\diamondsuit,\square\}\subsetneqq\{\heartsuit,\diamondsuit,\square\}\;:$$
we can fit a subset of $\{\bullet,\heartsuit,\clubsuit,\diamondsuit,\square\}$ strictly between $\{\diamondsuit\}$ and $\{\heartsuit,\diamondsuit,\square\}$. We could also have demonstrated that $\{\diamondsuit\}$ is not an immediate predecessor of $\{\heartsuit,\diamondsuit,\square\}$ by observing that
$$\{\diamondsuit\}\subsetneqq\{\heartsuit,\diamondsuit\}\subsetneqq\{\heartsuit,\diamondsuit,\square\}\;.$$
The sets $\{\diamondsuit,\square\}$ and $\{\heartsuit,\diamondsuit\}$, on the other hand, are immediate predecessors of $\{\heartsuit,\diamondsuit,\square\}$ in this partial order, as is $\{\heartsuit,\square\}$: there is no subset of $\{\bullet,\heartsuit,\clubsuit,\diamondsuit,\square\}$ that fits strictly between any of these three sets and the set $\{\heartsuit,\diamondsuit,\square\}$.