The Real And Complex Number Systems

a, b
n1 J 1/n Q a, b
is of the firse category which is absurb since every complete metric space is of the second
category. So, this f cannot exist.
Note: 1 The Boundedness of f can be removed since we we can accept the concept
0.
2. (J 1/n is closed in a, b) Given an accumulation point x of J 1/n ,ifx J 1/n ,wehave
f x 1/n. So, there exists a 1 ball Bx such that f Bx a, b 1/n. Thus, no
points of Bx can belong to J 1/n , contradicting that x is an accumulation point of J 1/n .
Hence, x J 1/n and J 1/n is closed.
3. (Definition of a nowhere dense set) In a metric space M, d, letA be a subset of M,
we say A is nowhere dense in M if, and only if A contains no balls of M, ( intA ).
4. (Definition of a set of the first category and of the second category) AsetA in a
metric space M is of the first category if, and only if, A is the union of a countable number
of nowhere dense sets. A set B is of the second category if, and only if, B is not of the first
category.
5. (Theorem) A complete metric space is of the second category.
We write another important theorem about a set of the second category below.
(Baire Category Theorem) A nonempty open set in a complete metric space is of the
second category.
6. In the notes 3,4 and 5, the reader can see the reference, A First Course in Real
Analysis written by M. H. Protter and C. B. Morrey, in pages 375-377.
4.25 Let f be continuous on a compact interval a, b. Suppose that f has a local
maximum at x 1 and a local maximum at x 2 . Show that there must be a third point between
x 1 and x 2 where f has a local minimum.
Note.Tosaythatf has a local maximum at x 1 means that there is an 1 ball Bx 1 such
that fx fx 1 for all x in Bx 1 a, b. Local minimum is similarly defined.
Proof: Let x 2 x 1 . Suppose NOT, i.e., no points on x 1 , x 2 can be a local minimum
of f. Sincef is continuous on x 1 , x 2 , then inffx : x x 1 , x 2 fx 1 or fx 2 by
hypothesis. We consider two cases as follows:
(1) If inffx : x x 1 , x 2 fx 1 , then
(i) fx has a local maximum at x 1 and
(ii) fx fx 1 for all x x 1 , x 2 .
By (i), there exists a 0 such that x x 1 , x 1 x 1 , x 2 ,wehave
(iii) fx fx 1 .
So, by (ii) and (iii), as x x 1 , x 1 , wehave
fx fx 1
which contradicts the hypothesis that no points on x 1 , x 2 can be a local minimum of f.
(2) If inffx : x x 1 , x 2 fx 1 , it is similar, we omit it.
Hence, from (1) and (2), we have there has a third point between x 1 and x 2 where f has
a local minimum.
4.26 Let f be a real-valued function, continuous on 0, 1, with the following property:
For every real y, either there is no x in 0, 1 for which fx y or there is exactly one such
x. Prove that f is strictly monotonic on 0, 1.