Key Concepts and Formulas

• Reflexive Relation: A relation $R$ on a set $A$ is reflexive if for every element $x \in A$, $(x, x) \in R$.
• Symmetric Relation: A relation $R$ on a set $A$ is symmetric if for every $x, y \in A$, if $(x, y) \in R$, then $(y, x) \in R$.
• Transitive Relation: A relation $R$ on a set $A$ is transitive if for every $x, y, z \in A$, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.

Step-by-Step Solution

Step 1: Understanding the Problem and the Relation We are given a set $W$, which represents all words in the English dictionary. A relation $R$ is defined on $W \times W$ such that $(x, y) \in R$ if and only if the words $x$ and $y$ have at least one letter in common. We need to determine if this relation $R$ is reflexive, symmetric, and transitive.

Step 2: Checking for Reflexivity

• Objective: To determine if for every word $x \in W$, the pair $(x, x)$ is in $R$.
• Condition for $(x, x) \in R$: According to the definition of $R$, $(x, x) \in R$ if word $x$ has at least one letter in common with itself.
• Reasoning: Consider any word $x$ from the English dictionary. A word is composed of letters. Any word $x$ will inherently share all its letters with itself. For instance, the word "hello" shares the letters 'h', 'e', 'l', 'l', 'o' with itself. Since a word always has at least one letter (in fact, many), it will always have at least one letter in common with itself.
• Conclusion: Therefore, for every $x \in W$, $(x, x) \in R$. The relation $R$ is reflexive.

Step 3: Checking for Symmetry

• Objective: To determine if for any two words $x, y \in W$, if $(x, y) \in R$, then $(y, x) \in R$.
• Condition for $(x, y) \in R$: This means word $x$ and word $y$ share at least one common letter.
• Reasoning: Suppose $(x, y) \in R$. This implies there exists a letter, say 'L', which is present in word $x$ and also present in word $y$. If 'L' is in $x$ and 'L' is in $y$, then it is also true that 'L' is in $y$ and 'L' is in $x$. The property of sharing a letter is mutual. If word $x$ has a letter in common with word $y$, then word $y$ must also have that same letter in common with word $x$.
• Conclusion: Therefore, if $(x, y) \in R$, then $(y, x) \in R$. The relation $R$ is symmetric.

Step 4: Checking for Transitivity

• Objective: To determine if for any three words $x, y, z \in W$, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.

• Condition for $(x, y) \in R$ and $(y, z) \in R$: This means word $x$ shares at least one letter with word $y$, and word $y$ shares at least one letter with word $z$.

• Reasoning: To check for transitivity, we need to see if the shared letter property "propagates" from $x$ to $y$ and then from $y$ to $z$, necessarily implying a shared letter between $x$ and $z$. We can test this with a counterexample.

  • Let $x = \text{'cat'}$. The letters in $x$ are $\{c, a, t\}$.
  • Let $y = \text{'bat'}$. The letters in $y$ are $\{b, a, t\}$.
  • Let $z = \text{'bee'}$. The letters in $z$ are $\{b, e\}$.

  Now, let's check the conditions:

  1. Is $(x, y) \in R$? Do 'cat' and 'bat' have at least one letter in common? Yes, they share 'a' and 't'. So, $(x, y) \in R$.
  2. Is $(y, z) \in R$? Do 'bat' and 'bee' have at least one letter in common? Yes, they share 'b'. So, $(y, z) \in R$.

  Now, we must check if $(x, z) \in R$ necessarily follows. Do 'cat' and 'bee' have at least one letter in common?

  • Letters in 'cat': $\{c, a, t\}$.
  • Letters in 'bee': $\{b, e\}$. The intersection of these sets is empty: $\{c, a, t\} \cap \{b, e\} = \emptyset$. Therefore, 'cat' and 'bee' do not share any letters. This means $(x, z) \notin R$.

  Since we found a specific instance where $(x, y) \in R$ and $(y, z) \in R$, but $(x, z) \notin R$, the relation $R$ is not transitive.

• Conclusion: The relation $R$ is not transitive.

Common Mistakes & Tips

• Transitivity Counterexamples: When checking for transitivity, always try to construct a scenario where the first two conditions hold but the third (the consequence) does not. This is often the most challenging property to prove or disprove.
• Word Properties: Remember that words are sets of letters. The order of letters or repetition of letters within a word does not affect whether they share a common letter for the purpose of this relation. For example, 'apple' and 'apply' share 'a', 'p', 'l'.
• Intersection of Sets: For transitivity, consider the sets of letters of the words. If $L(x)$ is the set of letters in word $x$, then $(x,y) \in R$ if $L(x) \cap L(y) \neq \emptyset$. Transitivity requires that if $L(x) \cap L(y) \neq \emptyset$ and $L(y) \cap L(z) \neq \emptyset$, then $L(x) \cap L(z) \neq \emptyset$. Our counterexample shows this is not always true.

Summary We analyzed the relation $R$ on the set of English words $W$, where $(x, y) \in R$ if words $x$ and $y$ share at least one letter. We found that the relation is reflexive because every word shares letters with itself. It is symmetric because if word $x$ shares a letter with word $y$, then word $y$ shares that same letter with word $x$. However, the relation is not transitive, as demonstrated by the counterexample of 'cat', 'bat', and 'bee', where 'cat' shares a letter with 'bat', and 'bat' shares a letter with 'bee', but 'cat' and 'bee' share no letters.

The final answer is $\boxed{A}$.