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For a set to be connected, it should have the property of not being able to be presented as a union of open disjoint sets(with respect to some topology- that I didn't quite get). I want to show that a ring $S=\{r<|z-z_0|<R\}$ is not simply connected by showing $\Bbb{C}\setminus S$ is not connected, but according to my notion of open sets, this set cannot be the union of two open sets as it is a closed set, but intuitively I have no doubt the is a "disconnection". Any clarification of the issue will help me.

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Crucial annoying topological fact: open and closed sets are not opposite to each other. A set can be both open and closed, or neither open nor closed. The union of two open sets is always open, but can be closed.

A topological space $X$ is called connected if the only sets which are both open and closed are $X$ and $\emptyset$. Otherwise, if $V$ is open and closed, then so is $X\setminus V$, so by your definition, $X$ can be expressed as a union of disjoint open sets.

In your definition, you are considering your set to be a subset of a topological space endowed with the subspace topology. So your open sets are those which are the intersection of an open set on the whole topology (i.e. $\mathbb C$) with your set.

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The key is that the definition speaks about relatively open sets, i.e., sets that are open in the topology that our set inherits from the original space (in this case the complex plane). The open sets in the relative topology are precisely the intersections of the original open sets (in the complex plane) with the space we are considering (the complement of the ring). It does not matter if that space is open in the complex plane.

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