# Complex Analysis/Identity theorem, Liouville-type theorems, Riemann's theorem

After now having established the main tools of complex analysis, we may deduce the first corollaries from them, which are theorems about general holomorphic functions; for, it is the case that holomorphic functions follow a wide range of rules just by their holomorphy, and this and the following chapter cover some of them.

## Connectedness and the identity theoremEdit

For the famous identity theorem, we first need to frame the intuitively plausible notion of a connected set in mathematically precise terms.

**Definition 7.1**:

A topological space is called **connected** if and only if the only *nonempty* set which is simultaneously open and closed is .

It seems somewhat implausible that this is actually the "right" definition of connectedness. The following theorem shows that in fact, if the topological space equals (and thus in particular if ) and the subset is *open*, then there is a very intuitive characterisation of connectedness.

**Theorem 7.2**:

Let with the standard Euclidean topology. Then any nonempty open subset is connected (w.r.t. to the subspace topology induced by the standard Euclidean topology) if and only if for all there exists a continuous function such that and (such sets are also called **path-connected**, and the continuous map is called a **path**).

**Proof**:

Assume first that is path-connected and not connected. We first prove the following claim, which is actually an important alternative characterisation of connectedness:

**Claim**: A topological space is connected if and only if there do not exist open such that , , and (that is, we may not split up into two nonempty open disjoint subsets).

**Proof of claim**: Suppose first that is connected, and assume for a contradiction that we can write it as with nonempty, open, disjoint. Then is open, closed, nonempty and not the whole space, contradiction. Let now not be connected. We pick a nontrivial open and closed set , and contradicts the condition of the claim.

Further, we have the following:

**Claim**: If is connected and is a continuous function between topological spaces, then is connected.

**Proof of claim**: Assume otherwise. Then there exists open, closed, nontrivial. By the continuity of , is open, closed, nontrivial, contradiction.

**Claim**: The set is connected.

**Proof**:

Let now be connected, and let be an arbitrary point within ( was assumed to be nonempty). We define

- .

This set is certainly open, because if there exists a path from to and is any ball around contained in , then in fact since we may connect to using a straight line and to using the path we have. But it is also closed, since the complement is open for the same reason: Pick and such that ; if any point in is in , then, since is convex, the previous argument applies and , contradiction. Since is connected and nonempty ( ), and thus, by connecting any two points by first connecting and and then and , is path-connected.

Now we can state and prove the identity theorem.

**Theorem 7.3**:

An **accumulation point** of a sequence of complex numbers is a point such that for all , contains infinitely many distinct points

Let be a *connected* set, and let be holomorphic. If there exists a sequence that has an accumulation point such that , then .

**Proof**:

## Liouville-type theoremsEdit

## Riemann's theoremEdit

**Theorem 7.6**:

Let be a holomorphic function, where is an interior point of and assume that is bounded in a neighbourhood of . Then there exists a holomorphic function which *extends* , that is,

- .

Thus, is an extension of to .

**Proof**:

Consider the function

- .

We claim that is holomorphic in . Indeed, for it is differentiable by the product rule, and for we have

- , i.e. .

Hence, we may write as a Taylor series

where due to and . Thus,

- ,

and hence for , in the domain of convergence of the power series, we have

- ,

where the latter expression makes sense also at and is complex differentiable there. This means that

is the desired continuation of to .

## ExercisesEdit

- Prove that if is an entire function such that , then is constant.
- Prove that every uncountable subset of as an accumulation point. Conclude that if for uncountably many we have , then .