# Complex analysis: How to show a set is open?

In complex analysis how do you show that a set is open or closed?

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A set...where, with what topology? Assuming you mean the complex plane $\,\Bbb C\,$ then it is exactly the same as showing a set is open in $\,\Bbb R^2\,$ , since these two topological spaces are homeomorphic under the usually given topologies. – DonAntonio Sep 27 '12 at 11:21
In practise, open sets are described by $<$ and $>$, whereas closed sets are described by $\leq$ and $\geq$. (Note that if there are any discontinuous functions around, this simplistic criteria might fail.) – Per Manne Sep 27 '12 at 11:29

In complex analysis, we (usually) consider $\mathbb{C}$ as being endowed with the Euclidean topology/metric/norm/etc.
So a set $U \subseteq \mathbb{C}$ is open if for each $z \in U$ there exists $\varepsilon >0$ such that $w \in U$ whenever $\left| z - w \right| < \varepsilon$.
Equivalently, $U$ is open if for each $z \in U$ there is an (open) disc of positive radius centred at $z$ which is contained in $U$.