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MATH 163, SHEET 13: THE EUCLIDEAN SPACE Rn
For the next three sheets we will be studying multivariable calculus, which is essentially
“calculus on Rn ”. First we need to understand the space Rn .
Definition 13.1. The Euclidean n-space Rn is the n-fold cartesian product of R. In other
words,
Rn = {(x1 , x2 , . . . , xn ) | x1 , x2 , . . . , xn ∈ R}
is the set of n-tuples of real numbers. We often write
x = (x1 , x2 , . . . , xn )
to denote an element, which is also referred to as a vector, in Rn and
0 = (0, 0, . . . , 0).
Although Rn is not necessarily a field, it inherits some of the structure of R.
Definition 13.2. Let x = (x1 , x2 , · · · , xn ), y = (y1 , y2 , · · · , yn ) ∈ Rn and λ ∈ R. We define
the following operations.
1. (Addition) x + y = (x1 + y1 , x2 + y2 , · · · , xn + yn ),
2. (Scalar Multiplication) λx = (λx1 , λx2 , · · · , λxn ).
Exercise 13.3 (Homework). Prove that the addition on Rn satisfies FA1-FA4. Moreover,
prove that:
1. (Associativity of Scalar Multiplication) If λ, µ ∈ R, and x ∈ Rn , then
(λµ)x = λ(µx).
2. (Distributivity of Scalars) If λ, µ ∈ R, and x ∈ Rn , then
(λ + µ)x = λx + µx.
3. (Distributivity of Vectors) If λ ∈ R, and x, y ∈ Rn , then
λ(x + y) = λx + λy.
4. (Scalar Multiplicative Identity)
1x = x.
These eight properties together are called the vector space axioms.
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Remark 13.4. Since Rn with the two operations defined as above satisfies these eight
axioms, we call Rn a vector space.
Definition 13.5. Let x ∈ Rn . The norm of x is defined as
q
kxk = x21 + x22 + · · · + x2n .
Remark 13.6. If n = 1, the norm coincides with the definition of the absolute value in R.
Definition 13.7. We call ky − xk the distance between x and y.
Theorem 13.8. If x, y ∈ Rn and λ ∈ R, then
1. kxk ≥ 0. Moreover, kxk = 0 if and only if x = 0.
2. (Cauchy-Schwarz Inequality) |x1 y1 + x2 y2 + · · · + xn yn | ≤ kxk · kyk.
(Hint: Consider the quadratic polynomial p(λ) = (x1 λ − y1 )2 + (x2 λ − y2 )2 + · · · +
(xn λ − yn )2 . Does it have a root?)
3. (Triangle Inequality) kx + yk ≤ kxk + kyk.
4. (Reverse Triangle Inequality) |kxk − kyk| ≤ kx − yk.
5. kλxk = |λ| · kxk.
The next goal is to “topologize” Rn . To discuss topology on Rn , we first need to introduce
notions of open and closed intervals in Rn , analogous to those for R.
Remark 13.9. For A ⊂ Rn and B ⊂ Rm , we define A × B ⊂ Rn × Rm to be the following
subset of Rn+m
(x1 , . . . , xn , xn+1 , . . . , xn+m ) ∈ Rn+m | (x1 , . . . , xn ) ∈ A, (xn+1 , . . . , xn+m ) ∈ B .
For simplicity we often write A × B ⊂ Rn+m .
Notice that for C ⊂ Rk , (A × B) × C and A × (B × C) correspond to the same subset of
n+m+k
R
under this identification; we write A × B × C for this common set.
Definition 13.10. An open rectangle in Rn is a set of the form
(a1 , b1 ) × (a2 , b2 ) × · · · × (an , bn ).
Similarly, a closed rectangle in Rn is a set of the form
[a1 , b1 ] × [a2 , b2 ] × · · · × [an , bn ].
Definition 13.11. A subset U ⊂ Rn is open if for all a ∈ U , there exists an open rectangle
R such that a ∈ R ⊂ U . A subset C ⊂ Rn is closed if its complement is open.
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Exercise 13.12 (Homework). Show that if R1 , R2 , . . . , Rm are open rectangles containing
x in Rn , then R1 ∩ R2 ∩ · · · ∩ Rm is an open rectangle containing x in Rn .
Exercise 13.13 (Homework). Let T = {U ⊂ Rn | U is open in Rn }, where “open” is
defined in Definition 13.11. Prove that T is a topology (as defined in Sheet 3) on Rn .
Definition 13.14. The open ball in Rn with center p and radius r > 0 is defined as
B(p, r) = {x ∈ Rn | kx − pk < r} .
Remark 13.15. Note that in R1 an open rectangle is also an open ball, and vice versa. The
following results illustrate how open rectangles and open balls in Rn are “compatible” with
each other.
Lemma 13.16. Fix a ∈ Rn .
1. If R is an open rectangle containing a, then there exists r > 0 such that B(a, r) ⊂ R.
2. If r > 0, then there exists an open rectangle R containing a such that R ⊂ B (a, r).
Corollary 13.17. (Homework) Open balls are open.
Corollary 13.18. A set U ⊂ Rn is open if and only if for every a ∈ U , there exists r > 0
such that B(a, r) ⊂ U .
Remark 13.19. If X ⊂ Rn , then X is also a topological space with the subspace topology.
That is, A ⊂ X is open in X if there exists an open set U ⊂ Rn such that X ∩ U = A. (See
Sheet 5).
We now discuss functions between Euclidean spaces. In Sheet 5, we gave a definition of
continuity that generalizes to this case:
Definition 13.20. Let X and Y be topological spaces. A function f : X −→ Y is continuous
if for every open set U ⊂ Y , the preimage f −1 (U ) = {x ∈ X | f (x) ∈ U } is open in X.
The function f : X → Y is continuous at x ∈ X if, for every open set U containing f (x),
there exists an open set S containing x such that S ∩ X ⊂ f −1 (U ).
Remark 13.21. As before, one can prove that a function f : X → Y is continuous if and
only if it is continuous at every x ∈ X.
Lemma 13.22. Let A ⊂ Rn and a ∈ A. A function f : A → Rm is continuous at a if and
only if, for every > 0, there exists δ > 0 such that
if x ∈ A and kx − ak < δ, then kf (x) − f (a)k < .
Definition 13.23. Let A ⊂ Rn and a be a limit point of A. Recall that a is a limit point of
A if for every open set U containing a, A ∩ (U \{a}) 6= Ø. Let f : A → Rm . We say L ∈ Rm
is the limit of f at a if for every > 0, there exists δ > 0 such that
if x ∈ A and 0 < kx − ak < δ, then kf (x) − Lk < .
In this case, we write lim f (x) = L.
x→a
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x21 − x22
does not have a limit at (0, 0).
x21 + x22
(Hint: Consider g(x) = f (x, 0) and h(y) = f (0, y) and show that lim g(x) 6= lim h(y).)
Exercise 13.24. Prove that the function f (x1 , x2 ) =
x→0
y→0
Theorem 13.25. Let A ⊂ Rn and a ∈ A be a limit point of A. Then f : A → Rm is
continuous at a if and only if lim f (x) = f (a).
x→a
Proposition 13.26. Fix p ∈ Rn . Define f : Rn → R as f (x) = kx − pk. Prove that f is
continuous. (Hint: Use the reverse triangle inequality.)
Definition 13.27. Let m ∈ N. Suppose I = {i1 , i2 , . . . , ik } ⊂ [m] with i1 < i2 < · · · < ik .
We define the projection function πI : Rm → Rk as
πI (x) = (xi1 , xi2 , . . . , xik ),
If I = {i} has only one element, we write πi instead of π{i} .
Exercise 13.28. Prove that πI is continuous.
Definition 13.29. Let f : A → Rm . Its i-th component function fi : A → R is defined as
fi = πi ◦ f.
In other words,
f (x) = (f1 (x), f2 (x), . . . , fm (x)).
Exercise 13.30. Prove that f is continuous if and only if f1 , f2 , . . . , fm are all continuous.
Now we revisit compactness in Rn .
Definition 13.31. Let A ⊂ Rn . Then A is compact if every open cover G of A has a finite
subcover.
Theorem 13.32. Let A ⊂ Rn be compact and f : A → R be continuous. Prove that f is
uniformly continuous. That is, show that for every > 0, there exists δ > 0 such that,
if x, y ∈ A and kx − yk < δ, then kf (x) − f (y)k < .
Lemma 13.33. Let x ∈ Rn . If B is a compact subset of Rm , then {x} × B is a compact
subset of Rn+m . (Hint: Prove that if U is an open subset of Rn+m , then
πI (U ) = {πI (y) | y ∈ U }
is an open subset of Rm for I = {n + 1, n + 2, . . . , n + m}.)
Lemma 13.34. Let x ∈ Rn and B ⊂ Rm . If B is compact and G is an open cover of
{x} × B ⊂ Rn+m , then there exists an open set U ⊂ Rn containing x such that U × B is
covered by a finite number of sets in G .
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Theorem 13.35. If A ⊂ Rn and B ⊂ Rm are compact, then A × B ⊂ Rn+m is also compact.
Corollary 13.36. If A1 , A2 , . . . , An are all compact, then so is A1 × A2 × · · · × An . In
particular, a closed rectangle is compact.
Lemma 13.37. If A ⊂ X with X compact and A closed, then A is compact.
Theorem 13.38. A closed ball with center p and radius r > 0 is defined as
{x ∈ Rn | kx − pk ≤ r} .
Closed balls are compact. (Hint: First show that closed balls are closed.)
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