 Solution

Take the relational schema and functional dependencyThe attribute closure of $\{b\}$ is $\{b,a\}$, so $\{b\}$ is a candidate key. The attribute closure of $\{a\}$ is just $\{a\}$, so it is not a superkey. The set $\{a,b\}$ is a superkey but is not a candidate key, because removing $a$ leaves the superkey $\{b\}$. Thus the only candidate key is $\{b\}$, and $a$ occurs in none.

For example, records $(b,a)=(1,7),(2,7)$ illustrate that the same $a$ value may occur at different $b$ values. The conclusion about candidate keys follows from the declared functional dependencies, rather than from accidental uniqueness in any one sample.