In Figure 1, we display the open unit disk, which is .

- is a neighborhood of each of its points.
- The subset
Im
is a neighborhhod of each of its points. If
, then for
min
,
(see Figure 2).

- The set Re is open. For any , we have Re (see Figure 3(a)).
- The unit open disk
is open: for any
, take
min
. Then
^{2.1}we have: (see Figure 3(b)).

- 1.
- Let . Suppose that ; as is open, there exists R>0 such that , whence . If , a similar argument works, whence the result.
- 2.
- Let . As and are open, there exist two positive numbers and such that and . Take . Thus and we are done.

- 1.
- Proposition 1.8 can be generalized to the union and the intersection of any finite number of open subsets of . The proof (by induction) is left to the reader.
- 2.
- For an infinite family of open subsets of , the intersection cannot be open. For example. take , for any complex number . Then , and it is easy to show that a set which contains only a single point is not open.

The following proposition will be very useful throughout our study of analytic functions (chapter chapter analytic functions) and further.

- 1.
- is closed.
- 2.
- is closed.

The proof is left to the reader, using De Morgan laws.

If , then but every open ball centered at contains points which do not belong to . Thus is not open.

Now consider the complementary subset of in the and take . Then , but every open ball centered at contains points which do not belong to . Thus is not open, whence is not closed.

- The punctured plane is connected and without a whole straight line is not connected.
- The subset
is not connected. See Figure 7: it is impossible to connect the points
and
with a finite sequence of segments totally included in
.

A **domain** is an open connected set.

- The unit disk is a domain (see Figure 8(a)).
- The whole of is a domain.
- The subset Re is not open, thus it is not a domain (see Figure 8(b)).

Let , where . Then, for any , the point is in the open ball , but not in . Thus is a boundary point of .

- If , take min ; then .
- If , take ; then .

A point such that is an interior point; take min , then .

- The annulus is bounded. For example, it is a subset of the disk, whose center is at the origin and whose radius is equal to (see Figure 11(a)).
- The whole of is not bounded.
- The set Re is not bounded. For every , there are complex numbers whose real part has a value between and and whose absolute value is greater than , e.g. (see Figure 11(b)).

- (i)
- The set is closed and bounded.
- (ii)
- The unit ball is bounded nut not closed.
- (iii)
- The set is closed but unbounded.
- (iv)
- The set is neither closed nor bounded.

Noah Dana-Picard 2007-12-24