Download Advanced Calculus

Advanced Calculus
Proposition 0.1. Let S be a subset of a metric space X.
1. S = S.
2. int(int(S)) = int(S).
3.
\
S=
C
S⊂C
C
is closed
4.
int(S) =
[
U
U ⊂S
U open
Proof.
1. If p is a limit of S, there is a sequence {pn } in S that converges to p. Since pn is a
limit of S, there is qn ∈ S such that d(pn , qn ) < n1 . Then d(p, qn ) ≤ d(p, pn ) + d(pn , qn ) → 0
as n → ∞. Hence p is a limit of S and we have S ⊂ S. The other direction follows from
definition.
2. For p ∈ int(S), there is r > 0 such that Br (p) ⊂ S. For each q ∈ Br (p), since Br (p) is
open, there is r0 > 0 such that Br0 (q) ⊂ Br (p) ⊂ S. Hence Br (p) ⊂ int(S). Therefore
p ∈ int(int(S)).
3. Since S is closed, the set in the right hand side is a subset of S. For p ∈ S, if p is not in the
set on right hand side, there is some C closed, S ⊂ C such that p ∈
/ C. Since X C is open,
there is r > 0 such that Br (p) ∩ C = ∅. But S ⊂ C, Br (p) ∩ S = ∅ implies that p is not a
limit of S which is a contradiction.
S
4. Since int(S) is open and contained in S, it is a subset of U ⊂S U . For p ∈ U for some open
U open
set U contained in S, by definition, p ∈ int(S).
Remark 0.2. The closure S is the smallest closed set containing S, and the interior int(S) is the
largest open set contained in S.
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Metric subspaces
Definition 1.1. Let (X, d) be a metric space and A a subset of X. By restricting d to A × A, we
get a metric dA on A. The metric space (A, dA ) is called a metric subspace of X.
Remark 1.2. If Y is a metric subspace of X, and A ⊂ Y . The notation A is confusing since
we may consider A as a subset of X too. To avoid the confusion, we write clX (A) to indicate the
closure of A in X.
Proposition 1.3. (Inheritance principle) Given a metric space X and a metric subspace Y ⊂ X, a
subset A ⊂ Y is closed in Y if and only if there is some closed subset B in X such that A = Y ∩ B.
Proof. If A is closed in Y , let B = clX (A) be the closure of A in X. Then B is a closed set in X
and A ⊂ B ∩ Y . If p ∈ B ∩ Y , p is a limit of A in Y and A is closed in Y , hence p ∈ A. So we have
A = B ∩ Y . Conversely, if B is some closed subset of X and p is a limit of B ∩ Y in Y , then p is
also a limit of B in X, hence p ∈ B ∩ Y , therefore B ∩ Y is closed in Y .
By taking complement, we have the following result.
Corollary 1.4. A metric subspace inherits its open sets from the ambient metric space.
Corollary 1.5. Suppose that Y is a metric subspace of X which is also open in X. If A ⊂ Y is
open, then A is open in X.
Proof. By above result, there is an open set B in X such that A = Y ∩ B. Since Y, B are open in
X, Y ∩ B = A is open in X.
Definition 1.6. Suppose that d1 , d2 are two metrics on X. We say that d1 , d2 are equivalent if
there exists positive constants α and β such that for every x, y ∈ X,
αd1 (x, y) ≤ d2 (x, y) ≤ βd1 (x, y)
Proposition 1.7. Suppose that d1 , d2 are two equivalent metrics on X. A sequence {pn } converges
in (X, d1 ) if and only if it is convergent in (X, d2 ).
Proof. This follows directly from the definition.
Definition 1.8. Suppose that (X1 , d1 ), (X2 , d2 ) are metric spaces. Let X = X1 × X2 be the Cartesian product of X1 , X2 . For p = (p1 , p2 ), q = (q1 , q2 ) in X, define
p
dE (p, q) := d1 (p1 , q1 )2 + d2 (p2 , q2 )2
dmax (p, q) := max{d1 (p1 , q1 ), d2 (p2 , q2 )}
dsum (p, q) := d1 (p1 , q1 ) + d2 (p2 , q2 )
The following result is clear.
Proposition 1.9. The above functions dE , dmax , dsum are metrics on X. Furthermore, they are
equivalent and satisfy
dmax ≤ dE ≤ dsum ≤ 2dmax
Example 1.10. The discrete metric d on R is not equivalent to the Euclidean metric ||.
Proof. If d and || are equivalent, there exists positive constants α, β such that αd(x, y) ≤ |x − y| ≤
βd(x, y) for any x, y ∈ R. Let y = 0, we have αd( n1 , 0) = α ≤ n1 for any n ∈ N. This implies α = 0
which is a contradiction.
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Continuous functions
Definition 2.1. Let f : X → Y be a function between topological spaces. We say that f is
continuous if for any open set V in Y , f −1 (V ) is open in X.
Proposition 2.2. A function f : X → Y between topological spaces is continuous if and only if
f −1 (C) is closed for every closed set C in Y .
Proof. If f is continuous and C is closed in Y , then C c is open in Y and hence f −1 (C c ) = (f −1 (C))c
is open which implies f −1 (C) is closed. Conversely, let U be an open set in Y . Then U c is closed
and by hypothesis, f −1 (U c ) = (f −1 (U ))c is closed which implies that f −1 (U ) is open.
Theorem 2.3. Let f : X → Y be a function between metric spaces. Then f is continuous if
and only if for each > 0 and p ∈ X, there is δ > 0 such that for q ∈ X, if dX (p, q) < δ, then
dY (f (p), f (q)) < .
Proof. Suppose that f is continuous. For p ∈ X and ε > 0, the open ball Bε (f (p)) is open
in Y , hence f −1 (Bε (f (p))) is open in X. Since p ∈ f −1 (Bε (f (p))), there is δ > 0 such that
Bδ (p) ⊂ f −1 (Bε (f (p))), in other words, for any x ∈ X with dX (x, p) < δ, dY (f (x), f (p)) < ε.
Hence f is continuous.
For the converse, let V ⊂ Y be an open set. For p ∈ f −1 (V ), since V is open, there is ε > 0 such
that Bε (f (p)) ⊂ V . By the hypothesis, there is δ > 0 such that for any x ∈ Bδ (p), dY (f (x), f (p)) <
ε. Hence f (x) ∈ Bε (f (p)). Therefore, f (Bδ (p)) ⊂ Bε (f (p)), and Bδ (p) ⊂ f −1 (Bε (f (p))) ⊂ f −1 (V )
which implies that f −1 (V ) is open.
Theorem 2.4. Given a function f : X → Y between metric spaces X, Y . The function f is
continuous if and only if it sends each convergent sequence in X to a convergent sequence Y , limits
being set to limits.
Proof. Suppose that f is continuous and {pn }∞
n=1 is a sequence in X convergent to p ∈ X. For
> 0, by the continuity of f , there is δ > 0 such that if dX (x, p) < δ, dY (f (x), f (p)) < . Since
pn → p, there is N ∈ N such that if n > N , dX (pn , p) < δ. Therefore dY (f (pn ), f (p)) < . By
definition, f (pn ) → f (p).
If f is not continuous, there exists p ∈ X and such that for each δn = n1 , there is xn ∈ X such
that dX (xn , p) < n1 but dY (f (xn ), f (p)) ≥ . Then xn → p but {f (xn )}∞
n=1 does not converges to
f (p).
Theorem 2.5. Suppose that X, Y, Z are topological spaces. If f : X → Y, g : Y → Z are continuous
functions, the composite g ◦ f : X → Z is continuous.
Proof. Let U be an open set in Z. From basic set theory, (g ◦ f )−1 (U ) = f −1 (g −1 (U )). Since g is
continuous, g −1 (U ) is open in Y , and by the continuity of f , f −1 (g −1 (U ) is open in X. So g ◦ f is
continuous.
Proposition 2.6. The four arithmetic operations +, −, ×, ÷ of R are continuous in their domains.
Proof. Let (x0 , y0 ) ∈ R × R and a = |x0 | + |y0 |. For each > 0,
• take δ = , then if dsum ((x, y), (x0 , y0 )) < δ, we have |(x+y)−(x0 +y0 )| = |(x−x0 )+(y−y0 )| ≤
|x − x0 | + |y − y0 | = dsum ((x, y), (x0 , y0 )) < .
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• take δ = , then if dsum ((x, y), (x0 , y0 )) < δ, we have |(x−y)−(x0 −y0 )| = |(x−x0 )−(y−y0 )| ≤
|x − x0 | + |y − y0 | = dsum ((x, y), (x0 , y0 )) < .
• take δ = min{1, 1+a
}, we have |xy − x0 y0 | = |(xy − x0 y) + (x0 y − x0 y0 )| ≤ |x − x0 ||y| + |x0 ||y −
y0 | ≤ |x − x0 |(|y − y0 | + |y0 |) + |x0 ||y − y0 | ≤ δ(1 + a) + aδ < 2.
y 2
0
• For the division, we need y0 6= 0. Take δ = min{ |y20 | , 1, 1+a
}. Then if dsum ((x, y), (x0 , y0 )) <
δ, we have |y − y0 | ≤ |y20 | , hence |y0 | ≤ |y0 − y| + |y| ≤ 21 |y0 | + |y| which implies |y| ≥
1
1
2
2 |y0 | > 0. Therefore (x, y) lies in the domain of the division. Since |yy0 | ≥ 2 |y0 | , and
|x| ≤ |x − x0 | + |x0 | ≤ δ + |x0 | + |y0 | ≤ 1 + a, we have
|
x x0
xy0 − x0 y
|x||y0 − y| + |x − x0 ||y|
2
|≤
((1 + a)δ + δ(1 + a)) ≤ 4
− |=|
≤
y
y0
yy0
|yy0 |
|y0 |2
Definition 2.7. Given topological spaces X, Y . We say that a function f : X → Y is a homeomorphism if f is bijective, continuous and f −1 is continuous.
Example 2.8.
1. The interval (0, 1) and R are homeomorphic. A homeomorphism is given by
tan( π2 (x − 12 )) where x ∈ (0, 1).
2. Let f : [0, 2π) → S 1 be defined by f (x) = (cos x, sin x) where S 1 is the unit circle in the plane.
Then f is continuous and bijective, but f −1 is not continuous.
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