Chapter 4: Continuity Flashcards
Thm 4.4
-
**Let f : (X, dX) → (Y, dY ) be a map between metric
spaces. Take x ∈ X. Then the following are equivalent:
(i) f is continuous at x in the sense of Definition 4.1.
(ii) f is continuous at x in the sense of Definition 4.2
Proof. (1) ⇒ (2). Suppose that f is continuous at x ∈ X in the
sense of Definition 4.1, and let xn → x. We need to show that
f(xn) → f(x). Choose ε > 0. Then there exists δ > 0 such that
dY (f(xn), f(x)) < ε whenever dX(xn, x) < δ. But, since xn → x,
there is an N such that dX(xn, x) < δ for all n > N. Combining
these, we see that for all n > N, dY (f(xn), f(x)) < ε, and so (f(xn))
tends to f(x).
(2) ⇒ (1). We suppose that f is continuous at x in the sense of
Definition 4.2, and suppose for a contradiction that Definition 4.1
fails. Then there is an ε > 0 such that:
For all possible δ > 0, there is a y ∈ X such that dX(x, y) <
δ, but dY (f(x), f(y)) > ε
We’ll now use this property to find a sequence that makes Definition 4.2 fail, giving us a contradiction. Let δ =1/n
. Then by the above
there is some yn such that dX(x, yn) <1/n while dY (f(x), f(yn)) > ε.
Now consider the sequence (yn); we have dX(x, yn) <1/n
, and so
yn → x. However, dY (f(x), f(yn)) > ε for all n, so f(yn) not → to f(x).
This gives the required contradiction to (2).
A function f : R → R is continuous if..
A function f : R → R is continuous at x ∈ R if, given any
sequence xn → x, then the sequence f(xn) → f(x).
A function f : R → R is continuous at x ∈ R if for all
ε > 0, there is a δ > 0 such that if |x − y| < δ, then
|f(x) − f(y)| < ε.
4.1 The Definition of Continuity. ε
A function f : X → Y between two metric spaces
(X, dX) and (Y, dY ) is continuous at x ∈ X if:
Given an ε > 0 we can find a δ > 0 such that, whenever
dX(x, y) < δ, we have dY (f(x), f(y)) < ε.
4.1 The Definition of Continuity
in terms of open balls:
in terms of open balls:
For all ε > 0, there exists δ > 0 such that f(B(x, δ)) ⊆ B(f(x), ε).
4.2 The Definition of Continuity.
sequences
A function f : X → Y between two metric spaces
(X, d_X) and (Y, d_Y ) is continuous at x ∈ X if:
Whenever we have a sequence x1, x2, . . . of elements of X
converging to x, then the sequence f(x1), f(x2), . . . in Y
converges to the limit f(x) ∈ Y .
Definition 4.3. continuous functions
We say that a function f : X → Y between metric
spaces X and Y is continuous if f is continuous at every x ∈ X
Theorem 4.4
Let f : (X, dX) → (Y, dY ) be a map between metric
spaces. Take x ∈ X. Then the following are equivalent:
(i) f is continuous at x in the sense of Definition 4.1.
(ii) f is continuous at x in the sense of Definition 4.2
Example:
The function f : R → R, defined by
f(x) =
{1 if x < 2,
{2 if 2 ≤ x,
continuous?
is not continuous. We can use either definition;
(4.1): there does
not exist δ > 0 such that f(B(2, δ)) ⊆ B(2,1/2),
(4.2): the sequence
xn = 2 −1/n → 2 but f(xn) = 1 → 1 ≠ f(2).
EXAMPLE 4.6:
Define θ : C[0, 1] → R by θ(g) = g(1). Then θ(fn) = fn(1) = 1 → 1
but θ(f) = f(1) = 0.
Recall from Example 2.16 that if f_n(x) = x^n
then, in C[0, 1] under d_1, (f_n) converges to f where f(x) = 0 for all x.
Define θ : C[0, 1] → R by θ(g) = g(1). Then θ(fn) = fn(1) = 1 → 1
but θ(f) = f(1) = 0. Hence θ is not continuous (at f) when C[0, 1]
has the metric d1 and R has its usual metric.
consider the same map θ when C[0, 1] has the metric d∞ and
R has its usual metric. Let (fn) be any convergent sequence in
(C[0, 1], d∞) with limit f. Then by Proposition 2.15, (f_n) converges
pointwise to f. In particular, fn(1) → f(1), that is θ(fn) → θ(f).
So here θ is continuous
EXAMPLE
Let f : (R^2, d_2) → R (usual metric on R) be defined
by f(x, y) = x^2 − y^3.
Show that f is continuous
Let ((x_n, y_n)) → (x, y) be a convergent sequence in R^2.
Then (x_n) → x and y_n → y by the d_2 version of Proposition 2.9.
By the algebra of limits, x^2_n − y^3_n → x^2 − y^3
, that is, f((x_n, y_n)) → f((x, y)). Thus f is continuous.
EXAMPLE
Let A be any 2 × 2 matrix of real numbers, and
regard it as a function A: R
2 → R
2
, where R
2 has the d∞ metric.
This is a continuous function, as we’ll now prove
Write A
[a b]
[c d]
let x=(x_1,x_2), y=(y_1,y_2) be elements of R^2.
Then d∞(Ax, Ay) ≤ 2M d∞(x, y), where M = max{|a|, |b|, |c|, |d|}.
proving continuity:
Let x∈ R^2 and let ɛ > 0. Then set δ = ɛ/2M, so that when
d∞(x, y) < δ, we have d∞(Ax, Ay) ≤ 2M d∞(x, y) <2Mδ = 2M ɛ/2M = ɛ. This proves that A is continuous at x, and it works for all choices of x, so that A is continuous.
proof fails when A is 0 matrix, can’t divide by =0 but A is obvs continuous
EXAMPLE 4.9:
show that the integration map is a continuous
map from C[a, b] to R.
Consider C[a, b] with the supremum metric d∞(f, g) = sup{|f(t) −
g(t)| : t ∈ [a, b]}, and define I : C[a, b] → R to be the function that
sends f ∈ C[a, b] to
I(f) = integral over[a,b] of f(t) dt ∈ R.
Thus, given a function f ∈ C[a, b], the value I(f) is the ‘area under
the graph’ of f. We will prove that I is continuous.
Now fix a function f ∈ C[a, b]. We show that I is continuous at f.
Let ε > 0, and take any δ such that 0 < δ < ε/(b−a)
. If g ∈ C[a, b] is
such that d∞(f, g) < δ, then |f(t) − g(t)| < δ for all t ∈ [a, b], and
so
d(I(f), I(g)) = |I(f) − I(g)|
=
| integral over [a,b] of f(t) dt −
integral over [a,b] of g(t) dt |
= | integral over [a,b] of (f(t) − g(t)) dt |
≤ integral over [a,b] of |(f(t) − g(t))| dt
≤
integral over [a,b] of δ dt = δ(b − a)
Proposition 4.10
constant funct
composition inclusion etc
. Let (X, dX), (Y, dY ), (Z, dZ) be metric spaces
(i) (Constant function) If f : X → Y maps all of X to a single
point y of Y then f is continuous.
(ii) (Composition) If f : X → Y , g : Y → Z are continuous then
the composite g ◦ f : X → Z is continuous.
(iii) (Inclusion) Let A ⊂ X. Then the inclusion j : A → X (defined
by j(a) = a, for all a ∈ A) is a continuous function (with A
considered as a subspace, as in Definition 1.24).
THM 4.11 metric space and mapping a → d(a, x)
Let (X, d) be a metric space and fix x ∈ X. Then
the mapping a → d(a, x) is continuous from (X, d) to R with its usual metric.
definition 4.12
preimage or inverse image
D. Let f : X → Y be a function between sets X and
Y , and let U be a subset of Y . Then the preimage or inverse image
of U under f, written f^−1(U), is defined to be
f^−1(U) = {x ∈ X : f(x) ∈ U}
Theorem 4.13.
equivalent
Let f : X → Y be a function between metric spaces.
The following are equivalent:
(i) f is continuous.
(ii) For every closed subset A ⊆ Y , f^−1(A) is closed.
(iii) For every open subset A ⊆ Y , f^−1(A) is open.
EXAMPLE
In (R^2, d2), let A = {(x, y) ∈ R^2: 3 < x^2 − y^3 <21}. Show that A is open in R^2 under d2.
A = f^−1((3, 21)), where f : R^2 → R is the continuous function from Example 4.7. As (3, 21) is open in R, A is open in R^2 by Theorem 4.13.
EXAMPLE
consider the maps
f : R → R, x → e^−x
g : R → R, x → x^2
These are continuous, R is closed and open but f(R) = (0,∞),
which is not closed in R and g(R) = [0,∞) which is not open.
if A is closed (resp. open)
in X, and if f : X → Y is continuous, then f(A) need not be closed
(resp. open) in Y .
preimage or inverse image properties
Thus f^−1(U) consists of all the points in X that f sends into U.
Note the following useful properties:
• f^−1(Y ) = X.
• f^−1(Y \ A) = X \ f^−1(A).
