Let $$W$$ denote the words in the English dictionary. Define the relation $$R$$ by : $$R = \{(x, y) \in W \times W \mid$$ the words $$x$$ and $$y$$ have at least one letter in common $$\}$$. Then $$R$$ is

For a relation $$R$$ on a set to be

• reflexive ⇔ $$(x,x)\in R$$ for every element $$x$$ of the set.
• symmetric ⇔ $$(x,y)\in R \Rightarrow (y,x)\in R$$.
• transitive ⇔ $$(x,y)\in R \text{ and } (y,z)\in R \Rightarrow (x,z)\in R$$.

The set under consideration is $$W$$ = “all words that appear in an English dictionary”. The relation is

$$R=\{(x,y)\in W\times W \mid x \text{ and } y \text{ have at least one common letter}\}.$$

Reflexive Every word shares all its own letters with itself, so for every $$x\in W$$ the ordered pair $$(x,x)$$ belongs to $$R$$. Hence $$R$$ is reflexive.

Symmetric Suppose $$(x,y)\in R$$. Then $$x$$ and $$y$$ contain at least one common letter, say “$$\ell$$”. Because $$\ell$$ is also in $$y$$, the pair $$(y,x)$$ shares the same letter $$\ell$$, giving $$(y,x)\in R$$. Therefore $$R$$ is symmetric.

Not transitive To test transitivity we need an example with words $$x,y,z$$ such that $$(x,y)\in R$$ and $$(y,z)\in R$$ but $$(x,z)\notin R$$.

Take the dictionary words
$$x=\text{“BALL”},\qquad y=\text{“LAMP”},\qquad z=\text{“MOCK”}.$$

• “BALL” and “LAMP” share the letters $$A$$ and $$L$$, so $$(x,y)\in R$$.
• “LAMP” and “MOCK” share the letter $$M$$, so $$(y,z)\in R$$.
• “BALL” and “MOCK” have no common letter, so $$(x,z)\notin R$$.

Since a single counter-example is enough to break transitivity, $$R$$ is not transitive.

Thus $$R$$ is reflexive and symmetric, but not transitive.

Option B which is: reflexive, symmetric and not transitive