20180525

1. To prove $A\subseteq A$, let $a\in A$. Then clearly $a\in A$ which means $A\subseteq A$ as desired.
2. We now prove $A\subseteq A\cup B$. Let $a\in A$. Then clearly $a\in A$ or $a\in B$. This means that $a\in A\cup B$, which proves that $A\subseteq A\cup B$.
3. We now prove $A\cap B\subseteq A$. Let $x\in A\cap B$. This means $x\in A$ and $x\in B$. In particular, notice that we know $x\in A$. This completes the proof that $A\cap B\subseteq A$.
4. Let $x$ be a real number. Let $P$ be the statement $x\in A$, and let $Q$ be the statement $x\in B$. Note that the statements $P\vee Q$ and $Q\vee P$ are logically equivalent from a truth table. The rule $A\cup B=B\cup A$ immediately follows. Alternately, we can prove this fact using the standard technique. Let $y\in A\cup B$. Then we know $y\in A$ or $y\in B$. This is equivalent to $y\in B$ or $y\in A$, which means $y\in B\cup A$. Hence we've shown $A\cup B\subseteq B\cup A$. The proof that $B\cup A\subseteq A\cup B$ is similar. Thus $A\cup B= B\cup A$.
5. The proof that $A\cap B=B\cap A$ is almost identical to the previous.
6. This is the next problem.
7. This is the next problem.
8. We now prove that $A\cup(B\cap C) = (A\cup B)\cap (A\cup C)$. We first show $A\cup(B\cap C) \subseteq (A\cup B)\cap (A\cup C)$. Let $x\in A\cup (B\cap C)$. This means that $x\in A$ or $x\in B\cap C$. We must show that $x\in A\cup B$ and that $x\in A\cup C$. There are two cases, namely $x\in A$ or $x\notin A$. Suppose first that $x\in A$. Then clearly $x\in A\cup B$ and $x\in A\cup C$ as $x$ is a member of the first set in each union. This shows that $x\in (A\cup B)\cap(A\cup C)$. Now suppose $x\notin A$. This means $x\in B\cap C$ (see the third sentences in this proof). Hence we know that $x\in B$ and $x\in C$. Since $x\in B$, we now $x\in A\cup B$. Since $x\in C$, we know $x\in A\cup C$. This shows that $x\in (A\cup B)\cap(A\cup C)$ as desired. This completes the proof that $A\cup(B\cap C) \subseteq (A\cup B)\cap (A\cup C)$. To finish, we must prove $(A\cup B)\cap (A\cup C) \subseteq A\cup(B\cap C)$. Let $y\in (A\cup B)\cap (A\cup C)$. We again use two cases. Suppose $y\in A$. Then clearly $y\in A\cup (B\cap C)$ by definition of union. The only other option is $y\notin A$. Recall we assumed that $y\in (A\cup B)\cap (A\cup C)$, which means $y\in (A\cup B)$ and $y\in (A\cup C)$. This means $y\in A$ or $y\in B$, and it means $y\in A$ or $y\in C$. Since we have assumed that $y\notin A$, this means that $y\in B$, and it means that $y\in C$. Together, this gives $y\in B\cap C$, which shows that $y\in A\cup (B\cap C)$.
9. Your proof should be very similar to the previous.

The answer here is complicated, though the best answer is probably, "NO." The answer depends on context, as described below.

If your initial answer was "yes", then good job; however, you made an assumption. As mathematicians, we have been trained from our past to assume that when a variable appears before us, then we will assume the unknown variable takes on any possible value that makes sense in the problem. Since we see $x\geq 4$, this means at some point we are comparing $x$ with the real number 4, so $x$ is possibly any real number. Since $x\in\mathbb{N}$ appears as well, we also assume that $x$ is a natural number. At this point, the second statement becomes

• For every $x\in \mathbb{R}\cap \mathbb{N}$, if $x\in \mathbb{N}$, then we have $x\geq 4$.

Since $\mathbb{R}\cap \mathbb{N}=\mathbb{N}$, we can rewrite this as

• For every $x\in\mathbb{N}$, if $x\in \mathbb{N}$, then we have $x\geq 4$.

Then notice that since $x\in \mathbb{N}$, the phrase after "if" is always true. This means that "If TRUE, then we have $x\geq 4$" is equivalent to "we have $x\geq 4$", which means we can rewrite the second statement as

• For every $x\in\mathbb{N}$, we have $x\geq 4$.

This now matches exactly the first statement. Notice that to obtain this conclusion, we had to make an assumption.

If your initial answer was "no", then great job. Note that $x\in\mathbb{N}$ is an open sentence, as we cannot determine the truth value of this sentence unless we know more information. The phrase "For every $x\in \mathbb{N}$, we have $x\geq 4$" fully describes which $x$ to consider - no assumptions must be made. The phrase "If $x\in \mathbb{N}$, then we have $x\geq 4$," does not provide any indication of what $x$ may or may not be, and hence the truth value cannot be determined without more information (possibly an assumption). The latter phrase can occur in many different contexts, and its truth value depends on the context. Consider each of the examples below.

• Suppose $x\in \mathbb{R}$. The sentence "If $x\in \mathbb{N}$, then we have $x\geq 4$," is false, because $x=1$ provides a counterexample.
• Suppose $x\in [3.5,\infty)$. The sentence "If $x\in \mathbb{N}$, then we have $x\geq 4$," is true because if we require both $x\in [3.5,\infty)$ and $x\in \mathbb{N}$, then we definitely have that $x\geq 4$.
• Suppose $x$ is irrational. The sentence "If $x\in \mathbb{N}$, then we have $x\geq 4$," is true because it is impossible for $x$ to be irrational and to have $x\in \mathbb{N}$ at the same time. Since the condition after the "if" part of the implication is false, then the entire implication is true.

Notice that the truth value of our implication depends entirely on the context. The truth value of, "For every $x\in \mathbb{N}$, we have $x\geq 4$," does not depend on any context, as the context is fully provided.

Consider as a last example the following paragraphs.

1. Suppose $x\geq 3.5$. For every $x\in \mathbb{N}$, we have $x\geq 4$.
2. Suppose $x\geq 3.5$. If $x\in \mathbb{N}$, then we have $x\geq 4$.

What is the difference? The first example tells me to let $x\geq 3.5$. Then the words "For every $x\in \mathbb{N}$" tell me to disregard, for a moment, the fact that we already assumed $x\geq 3.5$. Instead, we now let $x$ represent any natural number, and then claim that this always implies $x\geq 4$. The second sentence of this paragraph is false (as $x$ might equal 1), and is completely independent from the first sentence. Once the second sentence ends, we continue assuming $x\geq 3.5$. Now consider the second paragraph. Again we start by letting $x\geq 3.5$. However this time we don't forget this assumption when we continue reading. Instead, we use this choice of $x$ to determine the truth value of our implication, resulting in a true implication. The implication depends on the surrounding context, where as the fully quantified statement does not.

In conclusion, the two sentences are NOT logically equivalent.