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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.

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  • $\begingroup$ s can't be in F - F is a collection of sets. $\endgroup$ – Zen Jun 1 '13 at 3:50
  • $\begingroup$ Zen, if S = {{1}, 2, 3}, s = {1}, and F = {{1}, {2}} then couldn't it be true that s is in F ? $\endgroup$ – Ozera Jun 1 '13 at 4:05
  • $\begingroup$ But {1} is not a subset of S, {{1}} is. But if you're allowing s to be a set as well as a number I guess you could construct S={1, {1}, 2, 3} and s={1} and F={{1}, {2}}. There are pitfalls to not distinguishing sets and collections though - what if s={S}? $\endgroup$ – Zen Jun 1 '13 at 4:12
  • $\begingroup$ If S = {{1}, 1, 2, 3}, s = {1}, and F = {{1}, {2}} then s $\in$ F isn't it? Now if we were to say s $\subseteq$ F that would be false since s is an element and not a subset. (I think) $\endgroup$ – Ozera Jun 1 '13 at 5:44
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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.

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