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I'm learning Complex Analysis, and we are given the following definitions:
Definition. Suppose that $\Omega \subseteq$ C and that $\Omega$ is open.
(1) The set $\Omega$ is connected if any two points of $\Omega$ can be joined by a polygonal path lying inside $\Omega.$

(2) The set $\Omega$ is simply connected if the interior of every simple closed polygonal path in $\Omega,$ lies in $\Omega$ that is, if “$\Omega$ has no holes”.
(3) The set is a domain if it is connected as well as open.

Later, my note makes a remark saying:
Note that connected is not defined for closed sets, but there are questions about closed sets being connected.

I don't understand this, it says that it's not defined for closed sets, yet it also says there are questions regarding closed sets being connected..?

Why is do closed sets not have connectedness defined?

loma
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Twenty-six colours
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    They do. I worry that your definitions really get at something called path-connectedness. I believe the two notions are the same for open sets in the plane, but I'm not sure they're the same for closed sets. I would re-ask your instructor about that last claim. – Randall Aug 22 '17 at 01:10
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    @Randall. Topological connectedness and topological path-connectedness are not the same for all closed subsets of $\mathbb C,$ or , equivalently, of $\mathbb R^2.$ The most commonly given example is called the topologists' sine curve. In $\mathbb R^2$ this is $A\cup B $ where $A={0}\times [-1,1]$ and $B={(x,\sin \frac {1}{x});:0<x<1}.$....... $A\cup B$ is closed, bounded, and connected, but not path-connected..... For an open set in the plane, connected, path-connected, and polygonally connected are equivalent. – DanielWainfleet Aug 22 '17 at 03:43

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The definition is in terms of polyonally path-connectedness, which is not the same as being topologically connected (or path-connected.)

For example, we would want to say something like $S^1 \subset \mathbb C$ is connected (or even path-connected,) but clearly it is not polygonally path connected. For this reason, it makes sense for the author to restrict attention to just open sets (for now.)

For open sets in the plane, the two notions of polyonally path-connected and connected agree

Andres Mejia
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Let $X$ be a topological space, e.g. a subset of $\mathbb{C}$. Your definition of what it means for $X$ to be connected is different from the usual definition of connectedness. However, the two definitions coincide when $X$ is open. That is why they didn't want to apply their definition for nonopen sets.

The usual definition of what it means for $X$ to be connected is that there are no proper nonempty subsets of $X$ which are both open and closed in $X$.

Your definition of being connected is what everyone else calls being polygonally path connected.

There is also the notion of just being path connected: for any $x, y \in X$, there exists a continuous map $\gamma: [0,1] \rightarrow X$ such that $\gamma(0) = x$ and $\gamma(1) = y$.

If $X$ is any topological space, then it is connected if it is path connected, but the converse is not true.

When $X$ is a subset of $\mathbb{C}$, it is path connected if it is polygonally path connected. But it is possible for $X$ to be path connected, but not polygonally path connected (the graph of $y = x^2$). And it is also possible for $X$ to be connected but not path connected (the union of the point $(1,0)$ with the closed line segments between $(0,0)$ and $(1, \frac{1}{n})$ for all $n \in \mathbb{N}$).

However, all three of these notions coincide when $X$ is open.

D_S
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