Let $ D $ be the open unit disc in $ \mathbb{C} $ and $ H(D) $ be the collection of all holomorphic functions on it. Let $$ S= [f\in H(D):f(\frac{1}{2n})=\frac{1}{2n}\forall n\in \mathbb{N}]$$ and $$ T=[f\in H(D):f(\frac{1}{2n})=f(\frac{1}{2n+1})=\frac{1}{2n} \forall n\in \mathbb{N}]$$.
Then
- Both are S and T are singleton sets
- S is a singleton set but T is null
- T is a singleton set but S is null.
- Both S and T are empty.
I think the answer is option 2. The partial reason is by using the identity theorem, we obtain set consisting of only a constant function. But, how to prove that T is null? Any ideas appreciated.
You were right to invoke the identity principle for $S$; it shows that $f(z)=z$ is the only element of $S$. After your edit, $T$ is a subset of $S$ (since the defining equation $f(\frac1{2n})=\frac1{2n}$ of $S$ is now included in the definition of $T$) but does not contain the one and only element of $S$, so $T$ is empty.