Definition:Open Set

Topology
Let $$T = \left({X, \vartheta}\right)$$ be a topological space.

Then the elements of $$\vartheta$$ are called the open sets of $$T$$.

Thus "$$U \in \vartheta$$" and "$$U$$ is open in $$T$$" are equivalent statements.

Metric Space
Let $$M = \left({A, d}\right)$$ be a metric space.

Let $$U \subseteq M$$.

Then $$U$$ can be described as iff:
 * an open set in $$M$$;
 * open in $$M$$;
 * a $$d$$-open set;
 * $$d$$-open;

$$\forall y \in U: \exists \epsilon \left({y}\right) > 0: N_{\epsilon \left({y}\right)} \left({y}\right) \subseteq U$$.

That is, for every point $$y$$ in $$U$$, we can find an $$\epsilon > 0$$, dependent on that $$y$$, such that the $\epsilon$-neighborhood of that point lies entirely inside $$U$$.

Another way of saying the same thing is that one can not get out of $$U$$ by moving an arbitrarily small distance from any point in $$U$$.

It is important that the necessary value of $$\epsilon$$ is allowed to be different for each $$y$$.

Open Sets vs. Neighborhoods
It follows from the definition of $\epsilon$-neighborhood that every neighborhood is an open set.

However, not every open set is a neighborhood.

For example, let $$U \subseteq \reals^2: U = \left\{{\left({x_1, x_2}\right): a < x_1 < b, c < x_2 < d}\right\}$$.

Given $$x = \left({x_1, x_2}\right) \in U$$, it is easy to show that $$N_{\epsilon} \left({x}\right) \subseteq U$$ when $$\epsilon = \min \left\{{x_1 - a, b - x_1, x_2 - c, d - d_2}\right\}$$:



So $$U$$ is open in $$M$$, but it is not a neighborhood.

Complex Analysis
Let $$S \subseteq \C$$ be a subset of the set of complex numbers.

Suppose that $$\forall z_0 \in S: \exists \epsilon > 0: N_{\epsilon} \left({z_0}\right) \subseteq S$$

where $$N_{\epsilon} \left({z_0}\right)$$ is the $\epsilon$-neighborhood of $$z_0$$ for some real $$\epsilon > 0$$.

Then $$S$$ is described as open (in $$\C$$).

Note that $$\epsilon$$ may depend on $$z_0$$.

Real Analysis
Let $$I \subseteq \reals$$ be an open interval.

When $$\reals$$ is considered as a metric space, with the usual metric, then $$I$$ is an open set in $$\reals$$.