Proof
We consider the situation
,
and show the existence of such an
with the bisection method. For that, we put
and
,
we consider the arithmetic mean
,
and we compute
-
If
,
then we put
-
and if
,
then we put
-
In each case, the new interval
is lying inside the initial interval and has half of its length. It fulfills again the condition
,
therefore we can apply the same defining method again and get recursively a family of
nested intervals.
Let
denote the real number which is determined by these nested intervals. For the lower bounds of the intervals, we have
,
and this carries over to the limit
, due to the
criterion for continuity in terms of sequences.
Hence,
.
For the upper bounds, we have
,
and this again carries over to
, so
. Therefore,
.