Subset
Task number: 3344
Determine which of the following conditions are not equivalent to the condition \(A \subseteq B\). When they are not, try to adjust them using the smallest possible change so that equivalence holds.
Variant
\(A\setminus B=\emptyset\)
Solution
Intuitively, if \(A\) is a subset of \(B\), then there exists no element which is in \(A\) but not in \(B\).
We can formalize this thought by transforming logical expressions, e.g. as follows:
\(A\subseteq B \quad \Longleftrightarrow \quad \forall x: x \in A \Rightarrow x \in B \quad \Longleftrightarrow \quad \forall x: \neg (x \in A \land x \notin B)\\ \quad \Longleftrightarrow \quad \neg (\exists x: x \in A \land x \notin B) \quad \Longleftrightarrow \quad A \setminus B = \emptyset \)
Answer
The condition \(A\setminus B=\emptyset\) is equivalent to \(A\subseteq B\).
Variant
\(A \cup B = B\)
Hint
When transforming expressions you may use the equivalence proven in the preceding variant.
Solution
The inclusion \(B\subseteq A\cup B\) is always true, so it suffices to prove the equivalence \(A \subseteq B \quad\Longleftrightarrow\quad A \cup B \subseteq B\).
Intuitively, if one side of the equivalence fails to hold because \(A\setminus B\) is non-empty, the second side fails to hold as well.
For a formal proof of equivalence it's convenient to use the equivalent condition \(A\setminus B=\emptyset\) from the previous variant in place of \(A\subseteq B\). After that it suffices to simplify set expressions, e.g. as follows.
We will prove that \(A\setminus B=\emptyset \quad\Longleftrightarrow\quad A \cup B = B\)
From left to right: \(A\cup B=(A\setminus B)\cup B =\emptyset \cup B = B\).
From right to left: \(A\setminus B=(A\cup B)\setminus B =B \setminus B = \emptyset\).
Answer
The condition \(A \cup B = B\) is equivalent to \(A\subseteq B\).
Variant
\(A \cap B = A\)
Solution
Intuitive argument: \(A \not\subseteq B\) exactly when \(A\cap B \subsetneq A\).
Analogously with the preceding variant we formally show that \(A\setminus B=\emptyset \quad\Longleftrightarrow\quad A \cap B = A\)
From left to right: \(A\cap B =A \setminus (A\setminus B) =A \setminus \emptyset = A\).
From right to left: \(A\setminus B=A \setminus (A\cap B) =A \setminus A = \emptyset\).
Answer
The condition \(A \cap B = A\) is equivalent to \(A\subseteq B\).
Variant
\(\overline A \setminus B \subseteq \overline B\)
Hint
By transforming the left side we determine that \(\overline A \setminus B = \overline (A \cup B) = \overline A \cap \overline B \subseteq \overline B\).
Answer
The condition \(\overline A \setminus B \subseteq \overline B\) is not equivalent to \(A\subseteq B\), because it is always true.
A possible adjustment is e.g. \(\overline A \cup B = \overline \emptyset\).
Variant
\(A \cap \overline B = \emptyset\)
Solution
We can easily see this is true, because \(A\setminus B= A \cap \overline B\).
Answer
The condition \(A \cap \overline B = \emptyset\) is equivalent to \(A\subseteq B\).
Variant
\(\overline A \subseteq \overline B\)
Answer
The condition \(\overline A \subseteq \overline B\) is not equivalent to \(A\subseteq B\), because it is equivalent to \(B \subseteq A\).
Possible adjustments are e.g. \(\overline B \subseteq \overline A\) or \(\overline A \supseteq \overline B\).
