MATH 2000 Review 2: Examples of Routine Arguments Fall 2006 1 ...

MATH 2000 Review 2: Examples of Routine Arguments Fall 2006
1. Prove that, if A is a nonempty bounded set A in R, and if b = sup A, then, for
every ε > 0, there exists x ∈ A such that x > b − ε.
2. Given a set A and a point a in R N , prove that a ∈ A if and only if there is a sequence
{an} in A approaching a, i.e. limn→ ∞ an = a.
3. Find the closure A of a given set A in R in each of the following cases: (a) A is an
open interval (a, b), where a, b ∈ R with a < b; (b) A = Q, the set of all rational
numbers; (c) A = {m −1 + 2 −n : m, n = 1, 2, 3, . . .}.
4. Prove that the closure of a convex set in R N is also convex.
5. Let K be a bounded closed subset of an open set U in R N . Prove that there exists
δ > 0 such that p − q ≥ δ for all p ∈ K and q ∈ U c ≡ R N \U.
6. Let F be a closed set in R N and let r be a positive number. Prove that the set
Fr = {x ∈ R N : x − y ≤ r for some y ∈ F } is closed.
7. Let f : X → Y be a continuous mapping, where X and Y are subsets of R N . (a)
Prove that, if X is open sets, then, for every open set U in R N , f −1 (U) is an
open set. (b) Prove that, if X is closed, for every closed set F in R N , f −1 (F )
is closed.
8. Prove that, if fn : X → Y are continuous mappings converging uniformly to f :
X → Y , then f is continuous. Furthermore, when uniform convergence is replaced
by pointwise convergence, this assertion no longer holds.
9. (a) Let f : R 2 → R be a function such that its partial derivatives D1f and D2f
exist and are nonnegative everywhere. Prove that, for all points (x1, y1) and (x2, y2)
in R 2 with x1 < x2 and y1 < y2, we have f(x1, y1) ≤ f(x2, y2). (b) Show that the
statement still holds if R 2 is replaced by any convex open set in R 2 .
10. Prove that if f : U → R is a function defined on an open set U in R N and if f

is differentiable at a point x0 in U, then the directional derivative
f(x0 + tv) − f(x0)
Dvf(x0) = lim
t→ 0 t
exists and is equal to Df(x0)v for each v ∈ R N .
11. Prove the product rule for functions of several variables: if f and g be differentiable
functions on an open set U in R N , then their product fg is also differentiable and
D(fg)(x) = f(x)Dg(x) + g(x)Df(x) for all x ∈ U. Recall: the differentiability of
f with the linear map Df(x) as its derivative means that for all x ∈ U we can
write f(x + h) − f(x) = Df(x)h + R(x, h) where limh→ 0 R(x, h)/h = 0.
Answers
1. By definition, b is the least upper bound of A. For any ε > 0, we have b − ε < b
and hence b − ε is not an upper bound of A, meaning that there is some x ∈ A
such that x > b − ε.
2. Suppose that a ∈ A. Then, for each positive integer n, we have B 1/n(a) ∩ A = ∅.
Take any an ∈ B 1/n(a) ∩ A. Then {an} is a sequence in A approaching a. Conversely,
suppose that there is a sequence {an} in A approaching a. Then, for any ε > 0,
there exists some n such that an − a < ε and hence an ∈ Bε(a) ∩ A, showing that
Bε(a) ∩ A is nonempty.
3. (a) (a, b) = [a, b]; (b) Q = R;
(c) A = {0} ∪ {m −1 : m = 1, 2, . . .} ∪ {m −1 + 2 −n : m, n = 1, 2, 3, . . .}.
4. Let A be a convex set in R N and let a, b ∈ A. Then there are sequences {an} and
{bn} in A converging to a and b respectively. We need to prove that, for each
t ∈ [0, 1], (1 − t)a + tb ∈ A. Since A is convex and since an and bn are in A, we
have (1 − t)an + tbn ∈ A. Now {(1 − t)an + tbn} is a sequence in A converging to
(1 − t)a + tb. Hence (1 − t)a + tb ∈ A.

5. Suppose the contrary that such δ > 0 does not exist. Then, for each positive integer
n, there exists pn ∈ K and qn ∈ U c such that pn − qn < 1/n. Since {pn}
is a sequence in a bounded closed set K, by Bolzano-Weierstrass theorem, it has a
subsequence, say {pnk}, converging to some p in K. From pn − qn < 1/n we
see that {qnk } also converges to p as well. Since qnk are in the closed set U c , we
deduce that the limit p is also in U c . Thus p is in both K and U c , contradicting
the fact that K and U c do not intersect.
6. Let {xn} be a sequence in Fr converges to some point x. It is enough to show
that x ∈ Fr. By assumption, for each n, there exists some yn ∈ F such that
xn − yn ≤ r. Let zn = xn − yn. Then zn ≤ r for all n and hence {zn} is
a bounded sequence. By Bolzano–Weierstrass theorem, we can extract a convergent
subsequence of {zn}, say {znk } with limk→ ∞ znk = z. Since znk ≤ r for all k,
we have z ≤ r as well. Now ynk = xnk − znk converges to x − z ≡ y as k → ∞.
Since ynk ∈ F and since F is closed, we have y ∈ F . Thus we have y ∈ F and
x − y = z ≤ r, showing that x ∈ Fr.
7. (a) Take any point x0 ∈ f −1 (U). We are going to show that there exists δ > 0 such
that Bδ(x0) ⊆ f −1 (U). Now x0 ∈ f −1 (U) means f(x0) ∈ U. Since U is open, there
exists ε > 0 such that Bε(f(x0)) ⊆ U. Since f is continuous at x0, there exists
δ0 > 0 such that f(Bδ0 (x0) ∩ X) ⊆ Bε(f(x0)). In particular, f(Bδ0 (x0) ∩ X) ⊆ U.
Since X is open, there exists δ1 > 0 such that Bδ1 (x0) ⊆ X. Let δ = min{δ0, δ1}.
Then Bδ(x0) ⊆ Bδ0 (x0) ∩ X and hence f(Bδ(x0)) ⊆ U. The last inclusion can be
rewritten as Bδ(x0) ⊆ f −1 (U). This shows that U is open.
(b) It is enough to show that a point x0 in f −1 (F ) is in fact in f −1 (F ). Notice
that since X is closed and since f −1 (F ) ⊆ X, we have f −1 (F ) ⊆ X. Hence
x0 ∈ X and thus f(x0) is defined. Take any sequence {xn} in f −1 (F ) converging
to x0; (this is possible since x0 is in the closure of f −1 (F )). Then f(xn) ∈ F for
all n. By the continuity of f, {f(xn)} converges to f(x0). Since F is closed, we
have f(x0) ∈ F and consequently x0 ∈ f −1 (F ).
8. Take any point x0 ∈ K. We are going to prove that f is continuous at x0. Let ε > 0
be given. Since {fn} converges to f on K uniformly, there exists some N such that
|fN(x) − f(x)| < ε/3 for all x ∈ K. Since fN is continuous, there exists δ > 0 such

that |fN(x) − f(x0)| < ε/3 for all x ∈ K ∩ Bδ(x0). Hence, for all x ∈ K ∩ Bδ(x0),
|f(x) − f(x0)| = |f(x) − fN (x) + fN(x) − fN (x0) + fN (x0) − f(x0)|
≤ |f(x) − fN (x)| + |fN (x) − fN (x0)| + |fN (x0) − f(x0)|
≤ ε/3 + ε/3 + ε/3 = ε.
Hence f is continuous at x0. To show that the assumption of uniform convergence is
essential, let us entertain the following example. Define a sequence of functions {fn}
on [0, 1] by putting fn(x) = x n . Then limn→ ∞ fn(x) = f(x) for all x ∈ [0, 1],
where
f(x) =
Clearly f is discontinuous at 1.
0 if 0 ≤ x < 1;
1 if x = 1.
9. (a) Given points (x1, y1) and (x2, y2) in R 2 with x1 < x2 and y1 < y2, consider
the functions f1 and f2 on R given by f1(x) = f(x, y1) and f2(y) = f(x2, y).
The mean value theorem tells us that f1(x2) − f1(x1) = f ′ 1(ξ)(x2 − x1) for some ξ ∈
[x1, x2]. The last identity can be rewritten as f(x2, y1) − f(x1, y1) = D1f(ξ, y1)(x2 −
x1). Since D1f ≥ 0 and x2 > x1, we have f(x2, y1) ≥ f(x1, y1). Similarly we have
f2(y2) − f2(y1) = f ′ 2 (η)(y2 − y1) for some η ∈ [y1, y2], or f(x2, y2) − f(x2, y1) =
D2f(x2, η)(y2−y1), from which we obtain f(x2, y2) ≥ f(x2, y1) by noticing D2f ≥ 0.
Thus we have f(x2, y2) ≥ f(x2, y1) ≥ f(x1, y1).
10. Write f(x0 + h) = f(x0) + Df(x0)h + R(x0; h). Let ε > 0 be given. By assumption,
there exists δ > 0 such that |R(x0; h)|/h ≤ ε (or |R(x0, h)| ≤ εh) for all
h with h < δ. Let η = δ/v if v = 0 and η = 1 otherwise. Then, when
0 < |t| < η, we have |tv| < δ and hence
f(x0
+ tv) − f(x0)
t
− Df(x0)v
=
f(x0
+ tv) − f(x0) − Df(x0)tv
t
=
R(x0; tv)
t ≤ εtv = εv.
|t|
This shows that the directional derivative Dvf(x0) exists and is equal to Df(x0)v.
11. By assumption, at each point x ∈ U, we have
f(x + h) − f(x) = Df(x)h + Rf(x, h), g(x + h) − g(x) = Dg(x)h + Rg(x, h)

with limh→ 0 Rf(x, h)/h = 0 and limh→ 0 Rg(x, h)/h = 0. Hence
f(x + h)g(x + h) − f(x)g(x)
= f(x + h)g(x + h) − f(x)g(x + h) + f(x)g(x + h) − f(x)g(x)
= g(x + h)(Df(x)h + Rf (x, h)) + f(x)(Dg(x)h + Rg(x, h))
= g(x)Df(x)h + (g(x + h) − g(x))Df(x)h + Rf (x, h)g(x + h)
+ f(x)Dg(x)h + f(x)Rg(x, h)
= (f(x)Dg(x) + g(x)Df(x))h + R(x, h),
where R(x, h) = (g(x + h) − g(x))Df(x)h + Rf (x, h)g(x + h) + f(x)Rg(x, h) satisfies
|R(x, h)|
h ≤ |g(x + h) − g(x)| Df(x) + |Rf(x, h)|
g(x + h) + |f(x)|
h
|Rg(x, h)|
h
which converges to 0 as h → 0.