# Why is it meaningless for a closed set to be polygonal path connected?

My textbook (Complex Analysis by Saff & Snider) defines connectedness for open sets; the given definition of a connected open set is: a set in which every pair of points can be joined by a polygonal path that lies entirely in the set.

Using the given definition of connected set, I don't understand why it isn't similarly defined for closed sets too?

• The general definition of "connected" is more delicate, but for open subsets of ${\mathbb R}^n$ or ${\mathbb C}$ it is equivalent to the definition in your textbook. – Christian Blatter May 17 '13 at 8:04
• @ChristianBlatter Thanks Christian – Ryan May 17 '13 at 8:24

## 4 Answers

Polygonal connectedness (which is a stronger condition than topological connectedness) can be defined for any subset of the complex plane. The definition will be, as you'd expect, that a set $A\subseteq \mathbb C$ is polygonally connected if any two points in it can be connected by a polygonal path inside $A$.

The authors simply did not bother with the most general possibility. They will mostly care about subsets of the plane that are open and polygonally connected. The reason is that in complex analysis such subsets are very important. Openness is important because you want to be able to talk about a function being holomorphic in the given region, and if the region is not open this leads to annoying technical difficulties. Polygonal connectedness is important because it means any function defined on the region can be integrated between any two points in the region. These are things that happen often in complex analysis, and so you will see lots and lots of theorems that say something like "if $f$ is holomorphic on an open polygonally connected set, then ....".

• Thanks for the enlightenment. But I still don't understand, for sets in $\mathbb C$, why (polygonal-)connectedness isn't defined for closed sets, or at least be extended to apply to closed sets? – Ryan May 17 '13 at 8:25
• Because it's not a useful concept. You can define it, but it does not come up in any useful way. – Ittay Weiss May 17 '13 at 8:26
• Fair enough! Thank you. – Ryan May 17 '13 at 8:28

Because there is a difference between path connected and conected. And there is even a difference between polygonal connected and path connected. For example taking the set $\{ z \in \mathbb{C}: |z|=1\}$ has no polygonal paths connecting two points. But you intuitively says that this set should be connected. Connected in the pure topologic version sound this way:

A Topologic Space $(T,\tau)$ is called connected when there aren't any nonempty sets $O_1 \in \tau, O_2\in \tau$ with $O_1\cap O_2=\varnothing$ and $O_1 \cup O_2=T$. A subset of a Topological space is connected when it is connected as a topological space with the trace topology.

Every path connected space is connected but not every connected space is path connected. In $\mathbb{R}^n$ path connection, polygonal path connection and connection fall together for open sets, they are all equivalent when you know your set is already open.

This defintion is for sure much more complicated than your special case, and as you are interested in complex analysis about meromorphic functions most time you usually only use open sets anyway. so he wanted to avoid a complicated definition by using the special case of a more general one.

• Hey Dominic, is it really true though that in the complex plane, polygonally-connected and path-connected are equivalent? I am reading a lecture slide from uniMelb, and it says "A region is an open polygonally-connected set S together with all, some or none of its boundary points. We assume polygonal-connectedness to avoid inﬁnite length paths and fractal-like open sets." This seems to imply that "polygonally-connected" and "path-connected" are not equivalent (in the complex plane). – Ryan May 17 '13 at 12:40
• @Ryan as far as i can read, you take points of the boundary to the open set. Then the two things aren't equivalent anymore. Think of a spiral which contracts to the origin and is open. If you add the origin it will still be connected and path connected but not polygonally connected. – Dominic Michaelis May 17 '13 at 12:57
• I'm sorry, I don't understand what you are saying. I guess I don't get the significance of the last quoted line (my above comment)-- I think it is a mistake to suggest the need for specifically requiring polygonal-connectedness in the above definition of "region". – Ryan May 17 '13 at 17:39

The connectedness at here is path connectedness, which means any two points can be connected by a path traveling across the domain. For open sets it is not difficult to show polygon path connectedness and path connected is the same, because any polgyon path is a path, and any path can be approximated by a polygon path given the domain is open. Using some topology you can show path connected is the same as connected for open sets.

For closed sets the situation is different. What we normally regard as "connected" can differ from "path connected". For example the graph of $\sin[\frac{1}{x^{2}}]$ with $0$ when $x=0$ is connected, but not path connected. So complications arise this way. The book probably tried to sweep these issues under the rug when it introduce the concept to a first time learner. For more details on this you are welcome to read this:

http://en.wikipedia.org/wiki/Topologist%27s_sine_curve

• Thank you for your helpful response. – Ryan May 17 '13 at 8:29

Besides to @Dominic's neat answer, I am noting a good Theorem below. However it is about $\mathbb R$, you can get the points in another dramatized answer better:

Theorem: Let $E\subset\mathbb R$ including at least two points. Then $E$ is connected if and only if it is an interval.

In fact, if $E$ is not an interval,so $$\exists~p\notin E, a,b\in E~~ a<p<b$$ Now if you set $$G=(-\infty,p),~~H=(p,+\infty)$$ then $a\in G, b\in H$ and then $E\cap G, H\cap E$ are disjoint different non empty sets that can make $E$. This shows that $E$ is not connected!

• Good work, dear friend! $+\left( \lim_{x\to 0}\dfrac{\sin x}{x}\right)$ – amWhy May 18 '13 at 1:26
• @amWhy: Thanks dear friend. Daddy-Long-Legs – mrs May 18 '13 at 4:13