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2014 Fall Mathematical Analysis III
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Solution 11
1. A dyadic rational number is a rational number of the form k/2n for some integers k and
n. Show that all dyadic rational numbers form a dense set in R.
Solution. Every number in [0, 1] can be expressed as binary number 0.a1 a2 a3 · · · with
aj ∈ {0, 1}. Those with finitely many non-zero terms belong to the set of all dyadic
rational numbers D. For each x and ε > 0, there exists some N such that 0 ≤ x − d ≡
0.a1 a2 · · · an aN −1 · · · − 0.a1 a2 · · · aN 0 · · · < ε. For other number the proof is similar.
S
2. Let {qj } be all rational numbers in [0, 1]. Is it true that [0, 1] ⊂ j B8−j (qj ) ? Recall that
Br (q) = (q − r, q + r).
P∞ −j
8−1
8
=
2
|B
< 1, it
−j (qj )| = 2
8
j=1
j=1
1 − 8−1
S
follows that [0, 1] is not contained in j B8−j (qj ). This gives you an example of a dense,
open set whose total “length” is less than one.
Solution. Since |
S
j B8−j (qj )| ≤
P∞
3. Let f : X → Y be continuous where X and Y are metric spaces. Let E be dense in X.
Prove that f (E) is dense in f (X).
Solution. Let y = f (x) ∈ f (X). Since E is dense in X, there exists a sequence {xn } ⊂ E
with xn → x. It follows from continuity that f (xn ) → y. Hence f (E) is dense in f (X).
4. Let D be a dense set in the complete metric space X. Show that every uniformly continuous
function defined in D can be extended to become a uniformly continuous function in X.
Solution. Let f be a uniformly continuous function defined in D. Suppose x ∈ X, there
exists xn → x with {xn } ⊂ D since D is dense in X. It follows from uniform continuity
that {f (xn )} forms a Cauchy sequence in R, hence f (xn ) tends to some limit y ∈ R. Also
another sequence {x0n } ⊂ D with x0n → x must have f (x0n ) → y due to uniform continuity.
Define f (x) = y. Given ε > 0, let δ > 0 be such that |f (z) − f (z 0 )| ≤ ε for any d(z, z 0 ) <
δ, z, z 0 ∈ D. For any x, x0 ∈ X, d(x, x0 ) < δ, let xn → x, x0n → x0 with {xn } ⊂ D, {x0n } ⊂ D.
It follows from the definition of f that |f (x) − f (x0 )| = limn |f (xn ) − f (x0n )| ≤ .
5. Here we present another proof of the separability of C[0, 1] without Weierstrass approximation theorem. For each n, divide [0, 1] into n many subintervals of length 1/n and
consider the collection Rn of all continuous functions which are linear on each subinterval.
Furthermore, they must be of the form bx + a, a, b ∈ Q, over each subinterval. Show that
∪n Rn forms a countable, dense subset in C[0, 1].
Solution The union of Rn forms a countable set in C[0, 1]. We show that it is dense.
Let f ∈ C[0, 1]. For ε > 0, there exists an n such that |f (x) − f (y)| < ε/2 whenever
|x − y| < 1/n. Divide the interval into n many subintervals of length less than 1/n and
consider the piecewise linear function g which is equal to f (xj ) at all endpoints of these
subintervals. For each x ∈ [0, 1], x ∈ [xj , xj+1 ] for some j, so
|f (x) − g(x)| = |f (x) − (1 − t)f (xj ) + tf (xj+1 ) |
≤ (1 − t)|f (x) − f (xj )| + t|f (x) − f (xj+1 )|
ε
ε
ε
≤ (1 − t) + t = .
2
2
2
Now, we can approximate g by some function h in Rn such that kg − hk∞ < ε/2 to finish
the job.
2014 Fall Mathematical Analysis III
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6. Show that the boundary of a nonempty open set in a metric space must be closed and
nowhere dense. Conversely, every closed, nowhere dense set is the boundary of some open
set.
Solution. Let U be a nonempty open set and let Γ be its boundary. Then U ∩ Γ = ∅,
since every point of U is an interior point. Γ is closed since the boundary of a set is always
a closed set. Let x ∈ Γ. Since x is a boundary point, any metric ball containing x must
contain some points in U . It follows that Γ is nowhere dense. Conversely, Let Γ be a closed
and nowhere dense set. Let U be the complement of Γ. Then U is open. Let x ∈ Γ. Since
Γ is nowhere dense, any metric ball containing x must contain some points in U . Hence
Γ ⊂ ∂U . Since every point of U is an interior point, Γ = ∂U .
7. Use Baire category theorem to show that transcendental numbers are dense in the set of
real numbers.
Solution. A number is called algebraic if it is a root of some polynomial with integer
coefficients and it is transcendental otherwise. Let A be all algebraic numbers and T be
all transcendental numbers so that R = A ∪ T . From MATH2050 or even earlier we know
that A is a countable set {aj }. Thus let An = {a1 , · · · , an } and we have T = ∩n R \ An .
As each R \ An is a dense, open set, T is a set of second category and therefore dense.
In case you don’t want to use the countability of algebraic numbers, you may let Bn =
{x : x is a root of some polynomial of degree not exceeding n, |x| ≤ n}. Then show that
each Bn is closed and nowhere dense. Therefore, A = ∪n Bn is of first category. To show
nowhere dense of Bn , you may assume the existence of at least one transcendental number
α, say. Then for every algebraic number a, show that a + n−1 α is a transcendental number
so you can always find a transcendental number no matter how close to a.
A final remark is, while it is easy to show transcendental numbers are dense, here we show
that it is of second category, a bit more information.
8. Let F be a subset of C(X) where X is a complete metric space. Suppose that for each
x ∈ X, there exists a constant M depending on x such that |f (x)| ≤ M, ∀f ∈ F. Prove
that there exists an open set G in X and a constant C such that supx∈G |f (x)| ≤ C for
all f ∈ F. Suggestion: Consider the decomposition of X into the sets Xn = {x ∈ X :
|f (x)| ≤ n, ∀f ∈ F}.
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Solution. By assumption, X = n Xn . By the completeness of X we appeal to the Baire
category theorem to conclude that there is some n1 such that Xn1 has non-empty interior,
call it G. Then |f (x)| ≤ n1 , ∀x ∈ G for all f ∈ F.
9. A function is called non-monotonic if if is not monotonic on every subinterval. Show that
all non-monotonic functions form a dense set in C[a, b]. Hint: Consider the sets
En = {f ∈ C[a, b] : ∃x such that (f (y) − f (x))(y − x) ≥ 0, ∀y, |y − x| ≤ 1/n}.
Solution. We will show that each En is closed and nowhere dense. Let fk → f uniformly
and xk satisfy (fk (y) − fk (xk ))(y − xk ) ≥ 0 for y ∈ [xk − 1/n, xk + 1/n]. By passing to a
subsequence, one may assume xk → x0 . Then
fk (y) − fk (xk ) − (f (y) − f (xk )) ≤ |fk (y) − f (y)| + |f (xk ) − fk (xk )| ≤ 2kfk − f k∞ → 0,
which shows that
(f (y) − f (x0 ))(y − x0 ) = lim (f (y) − f (xk ))(y − xk ) = lim (fk (y) − fk (xk ))(y − xk ) ≥ 0,
k→∞
k→∞
2014 Fall Mathematical Analysis III
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hence En is closed. Next, if En has non-empty interior, we can find some f ∈ En such that
all functions in Bε (f ) are in En . Pick a polynomial p in Bε/2 (f ). We claim that there exists
some g, kp − gk∞ ≤ ε/2, does not belong to En . But kf − gk∞ < ε, contradiction holds.
Let ϕ be the jig-saw function that is described in our notes such that ϕ([a, b]) = [−1, 1]
and slope equal to a large number ±K and consider g = p + ε/2ϕ. Let x ∈ [a, b] and y > x
close to x, we have
ε
ε
(g(y)−g(x))(y−x) = (p(y)−p(x)+ (ϕ(y)−ϕ(x))(y−x) ≤ (L(y−x)+ (ϕ(y)−ϕ(x)))(y−x).
2
2
(L is a Lipschitz constant for p.) By the definition of ϕ, we can always choose some y close
to x from the right and K so large that L(y − x) + ε/2(ϕ(y) − ϕ(x)) < 0.
It shows that En is nowhere dense. Similarly, let
Fn = {f ∈ C[a, b] : ∃x such that (f (y) − f (x))(y − x) ≤ 0, ∀y, |y − x| ≤ 1/n}.
Then En is closed and nowhere dense for all n. Let the collection of all non-monotonic
functions be N . Since a function is non-monotonic if it is either increasing or decreasing
on some subinterval, we have
\
N =
C[a, b] \ En ∪ Fn .
n
By Baire’s theorem, N is of second category and hence dense. The proof here is similar but
simpler to the proof that nowhere differentiable functions form a set of second category.
Note: We construct the proof so that it looks like the proof of Proposition 3.16.
10. Optional. A basis of a vector space V is a set consisting of linearly independent vectors
satisfying, for each v ∈ V , there exist finitely vectors in this set such that v is a linear
combination of these vectors. Show that every basis of a Banach space must be an uncountable set. Recall that a Banach space is a vector space endowed with a norm whose
induced metric is a complete one. Hint: Try to decompose the Banach space into union
of finite dimensional subspaces. You may assume every finite dimensional subspace of a
Banach space is closed. The proof of this fact is not so easy.
Solution.
Assume on the contrary that there is a countable basis {zn }∞
n=1 . Let Wn
be the n-dimensional subspace spanned by the first
n
many
vectors.
For
each v ∈ V ,
S
v is a linear combination of some zj ’s, so V = n Wn . However, as each Wn is finite
dimensional and hence closed (you may assume this fact), we will have a contradiction
to Baire category theorem provided Wn is nowhere dense. Let us show that every proper
subspace in a normed space must be nowhere dense. For, let z ∈ V \ W where W is such
a subspace. For every ball Br (x) where x ∈ W and r > 0, the vector x + rz/2kzk lies in
this ball but does not belong to W , hence W is nowhere dense.
Here we outline how to show a finite dimensional subspace in a normed space is always a
closed set. We need to use the fact that all norms in the Euclidean space are equivalent.
Assuming this fact (I went over it in class), let k · k be the norm on X and Y be a finite
dim subspace.PLet {w1 , · · · , wn } bePa basis of Y . We introduce a new norm on Y by
setting |v| = ( j αj2 )1/2 where v = j αj wj . The map v 7→ (α1 , · · · , αn ) sets up a linear
bijection between Y and Rn . Let vn → v, vn ∈ Y in X. We want to show v ∈ Y . Since
{vn } is a Cauchy sequence in Y , by the equivalence of k · k and | · | it is also a Cauchy
sequence in | · |, which is simply the Euclidean norm in Rn . By the completeness of the
Euclidean space, it converges to some element of the P
form (β1 , · · · , βn ). Therefore, by the
equivalence of the two norms {vn } converges to w ≡ j βj wj which belongs to Y . By the
uniqueness of limit, we conclude that v = w ∈ Y .