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19.1 Geometry and Arithmetic of Complex Numbers 675

EXAMPLE 19.4
Let M consist of all complex numbers with rational real and imaginary parts. M has no inte-
rior points, because every disk about any point must contain points with irrational real and/or
imaginary parts, hence it contains points not in M. This uses the fact that every real number has
irrational numbers arbitrarily close to it. No points of M are interior points, and M is not open.
Every complex number (whether in M or not) is a boundary point of M. M does not contain all
of its boundary points and is not closed.

EXAMPLE 19.5
Let K consist of all real numbers together with the number 5i (Figure 19.5). Because every open
disk about 5i contains points not in K and a point in K (namely 5i itself), 5i is a boundary point
in K, and K is not open. Every point of K is a boundary point, and these are its only boundary
points. Therefore, K contains all of its boundary points and is closed.

EXAMPLE 19.6
Any finite set of complex numbers is closed. Suppose S ={z 1 ,··· , z N } is a set of N numbers.
Each z j is a boundary point, and there are no other boundary points of S. Therefore, S contains
all of its boundary points and is closed.

THEOREM 19.1

Let S be a set of complex numbers. Then S is open if and only if S contains no boundary
points.

This theorem implies that a set having no boundary points must be open. It also means that
containing a single boundary point is enough to disqualify a set from being open. The reason the
theorem is true is that, if ζ is a boundary point of S that is in S, ζ is not an interior point, so not
every point of S is an interior point and S cannot be open.
An open set can have boundary points, but these cannot be in the set. We have seen this with
open disks. Every point on the bounding circle is a boundary point, but none of these are in the
open disk.

y

5i

x

FIGURE 19.5 K in Example 19.5.

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