Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $U$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.