Open + Closed
Example 1. Let
be a metric space and
It may be easier to prove the complement is open than the prove
Sometimes it is hard to make a direct proof. Here we take an alternative route by proving that the complement is open. We pick a point not equal to and must show . This is done by picking a radius to be half the distance between and so that . We get the proof by rearranging the triangle inequality to show
We use the fact a set is open if and only if its complement is closed. We proceed by showing the complement set
is open. Recall
is open if and only if each element of
is an interior point. To this end, let
be given. It suffices to show
, and so
. Thus, setting
. For all
, the triangle inequality and choice of
is an interior point.
As noted previously, when proving something holds "for all elements,'' it is typically easiest to let an arbitrary element be given and prove the relation for that element. The result then will follow.
Example 2. Let
be a metric space. If
. Show both
Illustration of and a ball in . Note the ball is shown in red and is not the entire green circle. This follows from the fact a ball in can only contain elements in . Thus, in general, a ball might not be round.
be given. Set
to be the distance from
is closed and
, the distance
is positive.* Then consider the ball
. With the fact
, we deduce
is an interior point of
was arbitrarily chosen, every element of
is an interior point, i.e.
is open. An analogous argument applies to show
is open, completing the proof.
*This follows from a prior example.