Let $S$ be a set. Suppose that $s$ is an element of $S$, $T$ is a subset of $S$, and $F$ is a set of subsets of $S$. How many statements of the form $X \mathbin{R} Y$ are possible, where $X$ and $Y$ are each taken from $\{S,s,T,F\}$ and $R$ is taken from $\{\in,\subseteq\}$? Classify each statement as always true, possibly true, or always false.
$$\begin{align*}
S \in S &\text{ is }\text{ Always False}\\
S \in s &\text{ is }\text{ Always False}\\
S \in T &\text{ is }\text{ Always False}\\
S \in F &\text{ is }\text{ Possibly True }(?)\\[0.2in]
s \in S &\text{ is }\text{ Always True}\\
s \in s &\text{ is }\text{ Always False}\\
s \in T &\text{ is }\text{ Possibly True}\\
s \in F &\text{ is }\text{ Possibly True}\\[0.2in]
T \in S &\text{ is }\text{ Possibly True}\\
T \in s &\text{ is }\text{ Always False}\\
T \in T &\text{ is }\text{ Always False}\\
T \in F &\text{ is }\text{ Possibly True}\\[0.2in]
F \in S &\text{ is }\text{ Always False }(?)\\
F \in s &\text{ is }\text{ Always False}\\
F \in T &\text{ is }\text{ Always False }(?)\\
F \in F &\text{ is }\text{ Always False}\\[0.2in]
S \subseteq S &\text{ is }\text{ Always True}\\
S \subseteq s &\text{ is }\text{ Always False}\\
S \subseteq T &\text{ is }\text{ Possibly True}\\
S \subseteq F &\text{ is }\text{ Possibly True}\\[0.2in]
s \subseteq S &\text{ is }\text{ Always False }(?)\\
s \subseteq s &\text{ is }\text{ Always False }(?)\\
s \subseteq T &\text{ is }\text{ Possibly True }(?) \\
s \subseteq F &\text{ is }\text{ Possibly True }(?)\\[0.2in]
T \subseteq S &\text{ is }\text{ Possibly True}\\
T \subseteq s &\text{ is }\text{ Always False}\\
T \subseteq T &\text{ is }\text{ Always True}\\
T \subseteq F &\text{ is }\text{ Possibly True}\\[0.2in]
F \subseteq S &\text{ is }\text{ Possibly True, }\\
F \subseteq T &\text{ is }\text{ Possibly True} \\
F \subseteq s &\text{ is }\text{ Always False, }\\
F \subseteq F &\text{ is }\text{ Always True}
\end{align*}
$$
I was wondering if there were any errors in my answers.
Best Answer
I'll just focus on the question marked statements that happen to be wrong.
Note that $F$ is not necessarily the power set of $S$; it could be the case that $F=\{\emptyset\}$, for example, since $\emptyset \subseteq S$. Thus, if $T=S=\{\{\emptyset\}, 3\}$, then we have $F\in T$ and thus $F \in S$.
It's possible for sets to themselves be elements. Thus, if $s=\emptyset$ and $S=\{\emptyset, 3\}$, then we have $s \subseteq S$ and $s \subseteq s$ (since the empty set is a subset of any other set).
Also, note that by the definition of $T$, $T \subseteq S$ is always true.