METRIC SPACES EXAM QUESTIONS
AND CORRECT ANSWERS
Axiom of Choice - ANSWER Given a sequence of non-empty subsets S₁, S₂,
... of a set X, ∃ a sequence (xₙ) s.t. xₙ ∈ Sₙ ∀ n.
DEFN: A distance function's 3 properties - ANSWER (i) Positivity: d(x,y) ≥ 0,
and d(x,y) = 0 iff x = y
(ii) Symmetry: d(x,y) = d(y,x)
(iii) △ inequality: d(x,y) ≤ d(x,z) + d(y,z) (in other words, the distance between
two points via a third point is always at least the distance between just the two
points).
A metric space is a set and a distance function together.
Examples: d₁, d₂, d_∞, discrete metric - ANSWER d₁ (v,w) = Σ |v_i - w_i|
d₂ (v,w) = √[Σ (v_i - w_i)²]
d_∞ (v,w) = sup |v_i - w_i|
Referred to as l_# distances or norms.
Discrete metric: d(x,y) = 1 if x ≠ y
2-adic metric, ultrametric property - ANSWER Using the integers, d(x,y) =
2^(-m), where 2^m is the largest power of 2 dividing x-y.
Ultrametric property: d(x,z) ≤ max(d(x,y), d(y,z))
DEFN: Norm - ANSWER Let V be a v.s. over reals. ‖.‖ : V → [0, ∞) is a norm
if all of the following hold (x, y ∈ V):
‖x‖ = 0 iff x = 0
‖λx‖ = |λ|‖x‖ for all a in R
‖x + y‖ ≤ ‖x‖ + ‖y‖
A norm can form a metric with d(x,y) := ‖x - y‖ given x, y ∈ V
,"Normed space" a v.s. with a norm.
Properties of Metrics from Norms - ANSWER Translation Invariance, i.e.
d(x+z, y+z) = d(x, y)
Scalar Invariance, i.e. d(λx, λy) = |λ|d(x,y)
DEFN: Subspace - ANSWER Sps (X, d) is a m.s., let Y ⊆ X. Then the
restriction of d to Y × Y forms a metric for Y s.t. (Y, d|_Y×Y) is a m.s. Y is a
subspace.
Ex: Let X = Reals, Y can be any of [0,1], Q, Z.
DEFN: Product Space - ANSWER let x1, x2 ∈ X & y1, y2 ∈ Y.
d_X×Y ((x1, y1), (x2, y2)) = √(d_X (x1, x2)² + d_Y (y1, y2)²)
DEFN: Open and Closed Balls - ANSWER Open ball: B(a,r) = {x∈X : d(x, a)
< r}
Closed ball: B⁻(a,r) same thing but ≤.
Unit ball is B(0,1).
In the discrete metric, the open ball B(a, 1) is just a, the closed ball B⁻(a, 1) is
the entire space, since everything not a is exactly distance 1 from a.
B_Y (a, r) = Y ∩ B_X (a, r)
DEFN: Bounded, and its eqv defns. - ANSWER Let X be a m.s. and Y⊆X. Y
is bounded if Y is contained in some open ball.
EQVs:
Y is contained in some closed ball.
The set {d(y1, y2) : y1, y2 ∈ Y} is a bounded subset of Reals. (∆ ineq from
above)
From 3rd to 1st, sps d(y1, y2) ≤ K for all y1, y2 ∈ Y. If empty, bounded. O/W,
pick a in Y, then d(a, y) < K + 1 ∀ y ∈ Y, and so Y ⊆ B(a, K+1)
, DEFN: Limit, proof limits are unique - ANSWER Sps that (xₙ) is a sequence
of elements in a metric space (X, d). x∈X. Then xₙ → x or lim n→∞ xₙ = x if
for every ε>0, ∃N s.t. ∀ n≥N, d(xₙ, x) < ε.
Limits are unique. Sps a,b are both limits of xₙ. Let δ := d(a,b), and let ε = δ/2,
then d(xₙ, a), d(xₙ,b) < δ/2 for suff. large n, but δ = d(a,b) ≤ d(a, xₙ) + d(b,xₙ) < δ
by ∆ ineq. Contradiction.
DEFN: Continuity - ANSWER Let (X, d_X) and (Y, d_Y) be metric spaces. f :
X → Y is cts at a ∈ X if ∀ ε > 0, ∃ δ > 0 s.t. ∀ x ∈ X where d_X (a, x) < δ, d_Y
(f(x), f(a)) < ε.
f is continuous if it continuous at every a in X.
DEFN: Uniform Continuity - ANSWER f : X → Y is unif cts if for any ε > 0,
∃ δ > 0 s.t. ∀ x₁, x₂ ∈ X where d_X (x₁, x₂) < δ, d_Y (f(x₁), f(x₂)) < ε
LEMMA: Continuity written with Limits and proof - ANSWER f : X → Y. f is
cts at a iff for any sequence (xₙ) with xₙ → a, we have f(xₙ) → f(a)
pf: 1st, Sps f is cts at a. ∴∀ε>0, ∃δ>0 s.t ∀x∈X where d(x,a)<δ, d(f(x),f(a))<ε.
Now if (xₙ) is a seq with limit a, then ∃N s.t. d(a,xₙ)<δ ∀n≥N.
∴ d(f(a), f(xₙ))<ε, ∴ f(xₙ)→f(a)
2nd, Sps f not cts at a. ∴∃ ε>0 s.t. ∀ δ > 0 ∃ x ∈ X where d(x,a) < δ but d(f(x),
f(a)) ≥ ε.
Taking δ = 1/n, for each n ∃ xₙ ∈ X where d(xₙ,a) < 1/n but d(f(xₙ), f(a)) ≥ ε.
∴ xₙ → a but f(xₙ) does not limit to f(a).
LEMMA: Let f : V → W be a linear map between normed v.s. f is cts iff {‖f(x)‖
: ‖x‖ ≤ 1} is bounded. Proof - ANSWER (→) Sps f is cts, and so is cts at 0.
Taking ε = 1, ∃ δ > 0 s.t. d(f(x), 0) < 1 when ‖x‖ < δ. (f(0) = 0).
∴ ‖f(x)‖ ≤ 1 for ‖x‖ < δ.
Take ‖v‖ = 1, so ‖δv/2‖ < δ, and so ‖f(δv/2)‖ ≤ 1.
∴ ‖f(v)‖ ≤ 2/δ, and hence is bounded.
(←) Sps ‖f(v)‖ < M ∀ v s.t. ‖v‖ ≤ 1.
Let ε > 0 and set δ = ε/M.
AND CORRECT ANSWERS
Axiom of Choice - ANSWER Given a sequence of non-empty subsets S₁, S₂,
... of a set X, ∃ a sequence (xₙ) s.t. xₙ ∈ Sₙ ∀ n.
DEFN: A distance function's 3 properties - ANSWER (i) Positivity: d(x,y) ≥ 0,
and d(x,y) = 0 iff x = y
(ii) Symmetry: d(x,y) = d(y,x)
(iii) △ inequality: d(x,y) ≤ d(x,z) + d(y,z) (in other words, the distance between
two points via a third point is always at least the distance between just the two
points).
A metric space is a set and a distance function together.
Examples: d₁, d₂, d_∞, discrete metric - ANSWER d₁ (v,w) = Σ |v_i - w_i|
d₂ (v,w) = √[Σ (v_i - w_i)²]
d_∞ (v,w) = sup |v_i - w_i|
Referred to as l_# distances or norms.
Discrete metric: d(x,y) = 1 if x ≠ y
2-adic metric, ultrametric property - ANSWER Using the integers, d(x,y) =
2^(-m), where 2^m is the largest power of 2 dividing x-y.
Ultrametric property: d(x,z) ≤ max(d(x,y), d(y,z))
DEFN: Norm - ANSWER Let V be a v.s. over reals. ‖.‖ : V → [0, ∞) is a norm
if all of the following hold (x, y ∈ V):
‖x‖ = 0 iff x = 0
‖λx‖ = |λ|‖x‖ for all a in R
‖x + y‖ ≤ ‖x‖ + ‖y‖
A norm can form a metric with d(x,y) := ‖x - y‖ given x, y ∈ V
,"Normed space" a v.s. with a norm.
Properties of Metrics from Norms - ANSWER Translation Invariance, i.e.
d(x+z, y+z) = d(x, y)
Scalar Invariance, i.e. d(λx, λy) = |λ|d(x,y)
DEFN: Subspace - ANSWER Sps (X, d) is a m.s., let Y ⊆ X. Then the
restriction of d to Y × Y forms a metric for Y s.t. (Y, d|_Y×Y) is a m.s. Y is a
subspace.
Ex: Let X = Reals, Y can be any of [0,1], Q, Z.
DEFN: Product Space - ANSWER let x1, x2 ∈ X & y1, y2 ∈ Y.
d_X×Y ((x1, y1), (x2, y2)) = √(d_X (x1, x2)² + d_Y (y1, y2)²)
DEFN: Open and Closed Balls - ANSWER Open ball: B(a,r) = {x∈X : d(x, a)
< r}
Closed ball: B⁻(a,r) same thing but ≤.
Unit ball is B(0,1).
In the discrete metric, the open ball B(a, 1) is just a, the closed ball B⁻(a, 1) is
the entire space, since everything not a is exactly distance 1 from a.
B_Y (a, r) = Y ∩ B_X (a, r)
DEFN: Bounded, and its eqv defns. - ANSWER Let X be a m.s. and Y⊆X. Y
is bounded if Y is contained in some open ball.
EQVs:
Y is contained in some closed ball.
The set {d(y1, y2) : y1, y2 ∈ Y} is a bounded subset of Reals. (∆ ineq from
above)
From 3rd to 1st, sps d(y1, y2) ≤ K for all y1, y2 ∈ Y. If empty, bounded. O/W,
pick a in Y, then d(a, y) < K + 1 ∀ y ∈ Y, and so Y ⊆ B(a, K+1)
, DEFN: Limit, proof limits are unique - ANSWER Sps that (xₙ) is a sequence
of elements in a metric space (X, d). x∈X. Then xₙ → x or lim n→∞ xₙ = x if
for every ε>0, ∃N s.t. ∀ n≥N, d(xₙ, x) < ε.
Limits are unique. Sps a,b are both limits of xₙ. Let δ := d(a,b), and let ε = δ/2,
then d(xₙ, a), d(xₙ,b) < δ/2 for suff. large n, but δ = d(a,b) ≤ d(a, xₙ) + d(b,xₙ) < δ
by ∆ ineq. Contradiction.
DEFN: Continuity - ANSWER Let (X, d_X) and (Y, d_Y) be metric spaces. f :
X → Y is cts at a ∈ X if ∀ ε > 0, ∃ δ > 0 s.t. ∀ x ∈ X where d_X (a, x) < δ, d_Y
(f(x), f(a)) < ε.
f is continuous if it continuous at every a in X.
DEFN: Uniform Continuity - ANSWER f : X → Y is unif cts if for any ε > 0,
∃ δ > 0 s.t. ∀ x₁, x₂ ∈ X where d_X (x₁, x₂) < δ, d_Y (f(x₁), f(x₂)) < ε
LEMMA: Continuity written with Limits and proof - ANSWER f : X → Y. f is
cts at a iff for any sequence (xₙ) with xₙ → a, we have f(xₙ) → f(a)
pf: 1st, Sps f is cts at a. ∴∀ε>0, ∃δ>0 s.t ∀x∈X where d(x,a)<δ, d(f(x),f(a))<ε.
Now if (xₙ) is a seq with limit a, then ∃N s.t. d(a,xₙ)<δ ∀n≥N.
∴ d(f(a), f(xₙ))<ε, ∴ f(xₙ)→f(a)
2nd, Sps f not cts at a. ∴∃ ε>0 s.t. ∀ δ > 0 ∃ x ∈ X where d(x,a) < δ but d(f(x),
f(a)) ≥ ε.
Taking δ = 1/n, for each n ∃ xₙ ∈ X where d(xₙ,a) < 1/n but d(f(xₙ), f(a)) ≥ ε.
∴ xₙ → a but f(xₙ) does not limit to f(a).
LEMMA: Let f : V → W be a linear map between normed v.s. f is cts iff {‖f(x)‖
: ‖x‖ ≤ 1} is bounded. Proof - ANSWER (→) Sps f is cts, and so is cts at 0.
Taking ε = 1, ∃ δ > 0 s.t. d(f(x), 0) < 1 when ‖x‖ < δ. (f(0) = 0).
∴ ‖f(x)‖ ≤ 1 for ‖x‖ < δ.
Take ‖v‖ = 1, so ‖δv/2‖ < δ, and so ‖f(δv/2)‖ ≤ 1.
∴ ‖f(v)‖ ≤ 2/δ, and hence is bounded.
(←) Sps ‖f(v)‖ < M ∀ v s.t. ‖v‖ ≤ 1.
Let ε > 0 and set δ = ε/M.
