MAY 16 SOL 1.
(a) A metric space (X, d) consists of a non-empty set X and a non-negative
real valued distance function d : X × X → R≥ which satisifes the axioms
(1) d(x, y) = 0 ⇐⇒ x = y for all x, y ∈ X
(2) d(x, y) = d(y, x) for all x, y ∈ X
(3) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.
A closed ball of radius r and centre x is a subspace
B r (x) := {y : d(y, x) ≤ r} ⊆ X,
for any r ≥ 0 and x ∈ X. If y ∈ B r (x), then d(y, x) ≤ r; so d(y, x) ≤ 2r, and
y ∈ B 2r (x). Thus B r (x) ⊆ B 2r (x).
[bookwork + unseen problem: 4+2]
(b) The taxicab metric d1 on Rn is given by
d1 (x, y) = |x1 − y1 | + · · · + |xn − yn |.
The Euclidean metric d2 on Rn is given by
d2 (x, y) =
(x1 − y1 )2 + · · · + (xn − yn )2
1/2
,
where the positive square root is understood.
So d1 satisfies the first axiom of (a) because |xi − yi | = 0 iff xi = yi , and the
second because |yi − xi | = |xi − yi | for any xi , yi ∈ R. For the third, observe that
n
n
n
n
X
X
X
X
|xi − zi | =
|(xi − yi ) + (yi − zi )| ≤
|xi − yi | +
|yi − zi |
i=1
i=1
i=1
i=1
for every 1 ≤ i ≤ n, because |r + s| ≤ |r| + |s| for any r, s ∈ R.
0
B 3 (0) is a closed disc of radius 3 centred at 0, and B 3 (0) is a closed diamond
with vertices (±3, 0) and (0, ±3).
[bookwork + variant of homework: 3+4]
0
(c) The ball B 1 (x) is a closed disc of radius 1 and centre x, whereas B p (x)
is a closed√diamond whose vertices are distant p√from x; its edges√are therefore
0
distant p/ 2 from x. So B 1 (x) ⊂ B p (x) iff p/ 2 ≥ 1, or p ≥ 2. Similarly,
0
B q (x) is a closed disc of radius q and centre x, and B 2 (x) is a closed diamond
√
√
0
whose edges are distant 2 from x. So B q (x) ⊂ B 2 (x) iff q ≤ 2.
0
Finally, B p (x) ⊂ B q (x) iff p ≤ q. So we must have
√
p ≤ q ≤ 2 ≤ p,
√
and the only possibility is that p = q = 2.
None of the above argument changes if all the balls are open.
[variant of homework: 5+2]
1
2
MAY 16 SOL 2.
(a) Let U ⊆ X in a metric space (X, d). The interior U ◦ ⊆ U consists of all
points x for which there exists > 0 such that the open ball B (x) ⊆ U . Then U
is open whenever U ◦ = U .
Let each of the sets Uj be open, for j = 1, 2, . . . , and consider a point u in
their union U = ∪j≥1 Uj . Then u lies in at least one of the Uj , say for j = k. But
Uk is open, so their exists > 0 such that B (u) ⊆ Uk . Hence B (u) ⊆ U , so U
is open.
If Vj is the open interval (−1/j, 1/j) in the Euclidean line, then V = ∩j≥1 Vj
consists of the single point 0; for any other real number y must satisfy |y| > 1/j
for sufficiently large j. So V is not open.
[bookwork: 3+2+2]
(b) In the Euclidean plane:
• The set R2 \ {(1, 1)} is a punctured plane. Given any non-zero x ∈ R2 ,
define r := d2 (x, (1, 1)) > 0. Then Br (x) is contained R2 \ {(1, 1)}, so x
must be interior. Therefore R2 \ {(1, 1)} is equal to its own interior, and
open.
• The set {x : 0 ≤ x1 ≤ 1, 0 < x2 < 1} is an open square Q with vertices
(0, 0), (1, 0), (0, 1), (1, 1), together with two edges. The points x ∈ Q are
interior, since B (x) ⊂ Q for < min{x1 , 1 − x1 , x2 , 1 − x2 }. The points
y on the edges are not interior, since y1 must be 0 or 1, and any B (y)
meets the complement of the set. So the interior is Q, and the set is not
open.
• The set {x : x1 x2 ≥ 1} lies outside (and includes) the two arms of the
hyperbola x1 x2 = 1. The hyperbola provides all its non-interior points,
so its interior is {x : x1 x2 > 1}. The set is therefore not open.
The set Q2 is a 2–dimensional array of dust in R2 . No point q = (q1 , q2 ) is
interior, as any square open ball B (q) contains points with at least one irrational
coordinate. So (Q2 )◦ = ∅.
[variant of homework: 8]
(c) Choose f ∈ D, and let = |f (1)| > 0. Then for any g ∈ B (f ),
|g(1) − f (1)| ≤ sup |g(x) − f (x)| = dsup (g, f ) < |f (1)|;
0≤x≤1
so g(1) cannot be 0. Hence g ∈ D, and B (f ) ⊆ D. Therefore D is open, so its
complement A is closed. The function y = 1 − x is a non-constant element of A.
[variant of past exam question: 4+1]
3
MAY 16 SOL 3.
(a) An isometry f : X → Y is a distance preserving bijection; in other words,
dY (f (x), f (y)) = dX (x, y) for all x, y ∈ X.
Consider f : [0, 1] → [1, 2] given by f (x) = 1 + x, where the intervals are
subspaces of the Euclidean line. As 0 < x < 1 ⇒ 1 < 1 + x < 2, it follows that f
does indeed map [0, 1] to [1, 2]. Moreover, f has inverse given by f −1 (x) = x − 1,
because 1 + (x − 1) = (1 + x) − 1 = x; so f is a bijection. Furthermore,
d2 (f (x), f (y)) = |f (x) − f (y)| = |1 + x − (1 + y)| = |x − y| = d2 (x, y)
for any 0 < x, y < 1. So f is an isometry.
Let dX and dY be discrete, and consider any bijection g : X → Y . Then
g(x) = g(y) iff x = y, so g(x) 6= g(y) iff x 6= y. Thus dY (g(x), g(y)) = 0 iff
dX (x, y) = 0, and dY (g(x), g(y)) = 1 iff dX (x, y) = 1; since these are the only
possibilites, g is an isometry.
[unseen + variant of homework: 2+3+3]
(b) A continuous function g : X → Y , is one which satisfies
∀ > 0, ∃δ > 0 such that dX (x, x0 ) < δ ⇒ dY (g(x), g(x0 )) < for every x0 ∈ X.
If g is an isometry, take δ = . Then dX (x, x0 ) < δ ⇒ dY (g(x), g(x0 )) < δ = as required, for every x0 ∈ X.
[bookwork + homework: 2+2]
(c) The sequence (wn : n ≥ 1) converges to the limit w in (X, dX ) iff
∀ > 0, ∃N ∈ Z+ such that n ≥ N ⇒ dX (wn , w) < .
Suppose that g : X → Y is continuous at w, and assume that (wn ) converges
to w in X. For any > 0 there must therefore exist δ such that g(Bδ (w)) ⊆
B (g(w)), and N such that n ≥ N implies wn ∈ Bδ (w). Thus n ≥ N implies
g(wn ) ∈ B (g(w)), which means that g(wn ) converges to g(w).
Let h(x) = sin(1/x) for x 6= 0, and consider the sequence wn = 2/(4n + 1)π
for n ≥ 1. Then h(wn ) = sin(2nπ + π/2) = sin(π/2) = 1 for every n ≥ 1, which
converges to 1 ∈ R as n → ∞. But wn → 0, and h(0) = 0; so h(wn ) does not
tend to h(0). Hence h is not continuous at x = 0.
[bookwork + variant of homework: 2+3+3]
4
MAY 14 SOL 4.
(a) A subspace K of (X, d) is compact if and only if every open covering {Uα }
of K has a finite subcovering {Uαj : 1 ≤ j ≤ m}. The Heine-Borel Theorem
states that a closed and bounded subspace of n-dimensional Euclidean space is
necessarily compact.
(i) Since R1 with the discrete metric is not Euclidean, the Heine-Borel Theorem does not apply in this case. In fact [1, 2] is closed and bounded,
because these properties hold for any subset of a discrete metric space;
but it is not compact, because the open covering {{t} : 1 ≤ t ≤ 2} has no
finite subcovering.
(ii) The Heine Borel Theorem does apply to the subset {x : 1 ≤ x21 + x22 ≤ 9}
of the Euclidean plane. It is closed because its complement is the union
of the open ball B1 (0) and the complement of the closed ball B 3 (0), and
bounded because it lies in B 3 (0); thus it is compact.
[bookwork + variant of homework: 4+2+3]
(b) The Cantor set K ⊂ R is defined inductively. Let K0 = [0, 1], and define
K1 ⊂ K0 by deleting its open middle third (1/3, 2/3); so K1 = [1, 1/3] ∪ [2/3, 1]
is the union of two closed intervals. Assuming that Kn is the union of 2n closed
intervals, define Kn+1 inductively by deleting the open middle third of each; so
Kn+1 is the union of 2.2n = 2n+1 closed intervals. Then let
K := K1 ∩ · · · ∩ Kn ∩ . . . .
Since Kn is a finite union of closed intervals, it is closed for all n. Hence K is an
infinite intersection of closed sets, and is therefore closed. It is also bounded by
0 below and 1 above, so the Heine-Borel Theorem applies to confirm that K is
compact.
[bookwork: 8]
(c) In order for a real number a to lie in the Cantor set, it must satisfy 0 ≤ a ≤ 1
and have a ternary expansion containing only 0s and 2s.
Since
1/4 = 2/(32 − 1) = (2/32 ) (1 − 1/32 ),
we have
1/4 = 2/32 + 2/34 + 2/36 + · · · = 0 ·3 02020̇2̇,
which consists only of 0s and 2s. So 1/4 lies in the Cantor set.
[bookwork + version of homework: 3]
5
MAY 14 SOL 5.
(a) A Cauchy sequence in (X, d) is a sequence (xn ) that satisfies
∀ > 0, ∃N ∈ N such that m, n ≥ N ⇒ d(xm , xn ) < .
Then (X, d) is complete if every Cauchy sequence tends to a limit in X.
The subspace Q ⊂ R of the Euclidean line is not complete,
because the se√
quence 1, 1.4, 1.41, 1.414, of decimal approximations to 2 is Cauchy (being
convergent in R), but does not tend to a limit in Q.
[bookwork: 5]
(b) A map f : X → X is a contraction of (X, d) if there exists a constant K < 1
such that
d(f (x), f (y)) ≤ Kd(x, y) ∀x, y ∈ X .
To prove that such an f has at least one fixed point, choose x1 ∈ X at random,
and define xn = f (xn−1 ) for every n > 1. Then
d(xr , xr−1 ) ≤ Kd(xr−1 , xr−2 ) ≤ . . . ≤ K r−2 d(x2 , x1 )
for any r > 2, because f is a contraction. So the triangle inequality gives
d(xm , xn ) ≤ d(xm , xm−1 ) + · · · + d(xr , xr−1 ) + · · · + d(xn+1 , xn )
≤ (K m−2 + · · · + K r + · · · + K n−1 )d(x2 , x1 )
= K n−1 (K m−n−1 + · · · + K r−n+1 + · · · + 1)d(x2 , x1 )
= K n−1 (1 − K m−n )/(1 − K) d(x2 , x1 )
< K n−1 /(1 − K) d(x2 , x1 )
for any m ≥ n > 1. But K n−1 → 0 as n → ∞. So d(xm , xn ) < whenever
m ≥ n ≥ N for suitably large N , whence (xn ) is Cauchy. Since X is complete,
it follows that xn → p for some p ∈ X. Hence f (xn ) → f (p) (because f is
continuous) and f (xn ) = xn+1 → p as n → ∞; therefore f (p) = p.
Moreover, p is unique. For let f (q) = q; then d(p, q) = d(f (p), f (q)) ≤
Kd(p, q). Thus d(p, q) = 0, and p = q.
[bookwork: 10]
(c) If 0 ≤ x ≤ 2, then 1 ≥ 1 − x/2 ≥ 0; so f maps [0, 2] into [0, 2]. Since
|f (y) − f (x)| = |1 − y/2 − 1 + x/2| =
1
|y
2
− x|
for any 0 ≤ x, y ≤ 2, f is a contraction. Also, 1−x/2 = x has the unique solution
x = 2/3, which is the fixed point of f . It exemplifies (b) because [0, 2] is closed
in the complete space R, and hence is complete. [variant of example in class: 5]