Complex function open set, sequence, identically zero, proof

[binbagsss]

Homework Statement

Hi

I am looking at this proof that , if on an open connected set, U,there exists a convergent sequence of on this open set, and f(z_n) is zero for any such n, for a holomorphic function, then f(z) is identically zero everywhere.

##f: u \to C##Please see attachment

Homework Equations

The Attempt at a Solution

[/B]
I follow the proof on the open disc, but am just stuck on the ' by using connectedness we can extend this to ##U## .'

I have never taken a class in topology, nor, don't think I've ever had the definition of connectedness occur in a class. But from a quick google I see it is a set that cannot be partitioned into two nonempty subsets which are open in the relative topologyinduced on the set. Equivalently, it is a set which cannot be partitioned into two nonempty subsets such that each subset has no points in common with the set closure of the other.

I'm unsure how to,apply this to see that connnecteness implies that if the proof holds on an open disc, it,holds on ##U##

Many thanks

 

[mathwonk]

this is an interesting statement since it uses the imprecise word "any", in the phrase "for any n". If "any" means "some", as it probably should, then the statement is false. Of course here it is intended that "any" mean "all".

as to the argument, i suppose you could choose an arbitrary point w of the open set and connect w to z0 by a continuous path. Then you could ask where on that path f = 0. or just ask how far along the path, beginning at z0, f is identically zero. i.e. let z1 be the point such that f is identically zero for every point of the path between z0 and z1, but f is not identically zero on any longer portion of the path. If z1 ≠ w, this is a contradiction, by the argument just given, since the assumption that f is identically zero on one side of z1 along the path, implies that f is also identically zero on a disc centered at z1, hence also on some longer portion of the path. thus f is identically zero along the whole path, and hence also at w.

notice i am using a different definition of connectedness than you mentioned. i am using "path connectedness" instead of connectedness. a set is path connected iff for every pair of points p,q in the set, thre is a continuous path passing through both of them, or beginning at one and ending at the other. In an open subset of R^2, every point is contained in an open disc in which any two points are joined by a straight line path. Hence an open set in R^2 is at least locally path connected. Hence given point in such an open set, the subset that can be connected to it by a path is open, and also the subset that cannot be connected to it by a path is open. Hence if the subset is connected it is also path connected. I.e. a connected and locally path conncted set is path connected. hence an open connected set in R^2 is path connected. It is even easier to prove that a path connected set is connected.

 

[pasmith]

The proof relies on the identity theorem of holomorphic functions to conclude that if [itex]f = 0[/itex] on an open subset of [itex]U[/itex] then [itex]f = 0[/itex] on the whole of [itex]U[/itex]. The identity theorem requires that [itex]U[/itex] be connected.

Connectedness can be characterised as follows:

Let [itex]X[/itex] be a topological space. Then [itex]X[/itex] is connected if and only if every continuous [itex]f: X \mapsto \{0,1\}[/itex] in the discrete topology is constant.

This suggests the following example of why connectedness of [itex]U[/itex] is necessary for the identity theorem: Consider two open disjoint subsets [itex]V_1[/itex] and [itex]V_2[/itex] of [itex]\mathbb{C}[/itex] which are themselves connected, and let [itex]U = V_1 \cup V_2[/itex]. Define [tex]
f: U \to \mathbb{C} : z \mapsto \begin{cases}
0, & z \in V_1, \\
1, & z \in V_2.
\end{cases}[/tex] and [tex]
g: U \to \mathbb{C}: z \mapsto 0.
[/tex] Then [itex]f[/itex] and [itex]g[/itex] are holomorphic on [itex]U[/itex] and equal on the open subset [itex]V_1[/itex], but are not equal on [itex]U[/itex].

 

[FactChecker]

as to the argument, i suppose you could choose an arbitrary point w of the open set and connect w to z0 by a continuous path. Then you could ask where on that path f = 0. or just ask how far along the path, beginning at z0, f is identically zero. i.e. let z1 be the point such that f is identically zero for every point of the path between z0 and z1, but f is not identically zero on any longer portion of the path. If z1 ≠ w, this is a contradiction, by the argument just given, since the assumption that f is identically zero on one side of z1 along the path, implies that f is also identically zero on a disc centered at z1, hence also on some longer portion of the path. thus f is identically zero along the whole path, and hence also at w.

I think it would be better to cover the path with a finite set of open, overlapping disks and extend the zeros through all the disks. It seems like this way, you need to prove that there is a "first" non-zero point on the path.

 

[binbagsss]

The

The proof relies on the identity theorem of holomorphic functions to conclude that if [itex]f = 0[/itex] on an open subset of [itex]U[/itex] then [itex]f = 0[/itex] on the whole of [itex]U[/itex]. The identity theorem requires that [itex]U[/itex] be connected.

Connectedness can be characterised as follows:

Let [itex]X[/itex] be a topological space. Then [itex]X[/itex] is connected if and only if every continuous [itex]f: X \mapsto \{0,1\}[/itex] in the discrete topology is constant.

This suggests the following example of why connectedness of [itex]U[/itex] is necessary for the identity theorem: Consider two open disjoint subsets [itex]V_1[/itex] and [itex]V_2[/itex] of [itex]\mathbb{C}[/itex] which are themselves connected, and let [itex]U = V_1 \cup V_2[/itex]. Define [tex]
f: U \to \mathbb{C} : z \mapsto \begin{cases}
0, & z \in V_1, \\
1, & z \in V_2.
\end{cases}[/tex] and [tex]
g: U \to \mathbb{C}: z \mapsto 0.
[/tex] Then [itex]f[/itex] and [itex]g[/itex] are holomorphic on [itex]U[/itex] and equal on the open subset [itex]V_1[/itex], but are not equal on [itex]U[/itex].

I've never studied a topology class nor was it required, can I get by with another definition of connectedness for the identity theorem ? Thanks

 

[mathwonk]

the existence of such a first point is immediate from the least upper bound property of the real numbers. that property is also needed to prove the compactness of the closed bounded intervals, i.e. of the path, and hence to reduce an open covering by discs to a finite one.

and although i used a different definition of connectedness, i also gave an argument showing it follows from the other one. so yes you can use a different definition if you prove it holds.

 

[WWGD]

Re : connectedness somewhat fuzzy): Consider two open balls B1, B2 overlapping, where f(z) is identically zero in one of them, say B1. Can you see how you can apply the result to B2 on the overlap? In the case of a disconnection, the overlap is either inexistent, of minimal ( a finite collection of points).