Problem Sheet 7 due today
Final Problem Sheet 8, due Thursday 4th May.
I will circulate solutions on 4th May.
No problem class next Monday (bank holiday).
Revision next week. Let me know (by email or in person) any topics,
mock exam questions etc. you would like to be covered.
Last time
Last time
We found two properties of a metric space X which imply that everycontinuous function f : X → R is bounded:
Last time
We found two properties of a metric space X which imply that everycontinuous function f : X → R is bounded:
• X is sequentially compact if every sequence in X has a convergentsubsequence.
Last time
We found two properties of a metric space X which imply that everycontinuous function f : X → R is bounded:
• X is sequentially compact if every sequence in X has a convergentsubsequence.
• X is compact if, given any collection of open subset of X which cover X,there is a finite number of them which also cover X.
Last time
We found two properties of a metric space X which imply that everycontinuous function f : X → R is bounded:
• X is sequentially compact if every sequence in X has a convergentsubsequence.
• X is compact if, given any collection of open subset of X which cover X,there is a finite number of them which also cover X.
(In fact, these two properties are equivalent for metric spaces (though notfor topological spaces).)
Last time
We found two properties of a metric space X which imply that everycontinuous function f : X → R is bounded:
• X is sequentially compact if every sequence in X has a convergentsubsequence.
• X is compact if, given any collection of open subset of X which cover X,there is a finite number of them which also cover X.
(In fact, these two properties are equivalent for metric spaces (though notfor topological spaces).)
We showed that, for subsets A of Rn, being sequentially compact is thesame as being closed and bounded.
Last time
We found two properties of a metric space X which imply that everycontinuous function f : X → R is bounded:
• X is sequentially compact if every sequence in X has a convergentsubsequence.
• X is compact if, given any collection of open subset of X which cover X,there is a finite number of them which also cover X.
(In fact, these two properties are equivalent for metric spaces (though notfor topological spaces).)
We showed that, for subsets A of Rn, being sequentially compact is thesame as being closed and bounded.
page 92 of notes
Examples of sequentially compact spaces
page 92 of notes
Examples of sequentially compact spaces
a) Any closed interval [a, b] is sequentially compact.
page 92 of notes
Examples of sequentially compact spaces
a) Any closed interval [a, b] is sequentially compact. (Closed bounded subset
of R.)
page 92 of notes
Examples of sequentially compact spaces
a) Any closed interval [a, b] is sequentially compact. (Closed bounded subset
of R.)
b) Other intervals such as (a, b), (a, b], [a,∞), (a,∞) are not sequentially
compact.
page 92 of notes
Examples of sequentially compact spaces
a) Any closed interval [a, b] is sequentially compact. (Closed bounded subset
of R.)
b) Other intervals such as (a, b), (a, b], [a,∞), (a,∞) are not sequentially
compact. (Not closed, or not bounded, or neither.)
page 92 of notes
Examples of sequentially compact spaces
a) Any closed interval [a, b] is sequentially compact. (Closed bounded subset
of R.)
b) Other intervals such as (a, b), (a, b], [a,∞), (a,∞) are not sequentially
compact. (Not closed, or not bounded, or neither.)
c) A shape in Rk is precisely a non-empty sequentially compact subset of Rk.
page 92 of notes
Examples of sequentially compact spaces
a) Any closed interval [a, b] is sequentially compact. (Closed bounded subset
of R.)
b) Other intervals such as (a, b), (a, b], [a,∞), (a,∞) are not sequentially
compact. (Not closed, or not bounded, or neither.)
c) A shape in Rk is precisely a non-empty sequentially compact subset of Rk.
page 92 of notes
d) Any metric space (X, d) with only finitely many elements is sequentially
compact.
page 92 of notes
d) Any metric space (X, d) with only finitely many elements is sequentially
compact.
Bare hands proof:
page 92 of notes
d) Any metric space (X, d) with only finitely many elements is sequentially
compact.
Bare hands proof: Let (xn) be any sequence in X.
page 92 of notes
d) Any metric space (X, d) with only finitely many elements is sequentially
compact.
Bare hands proof: Let (xn) be any sequence in X.
Since there are only finitely many possible values of xn, one of them, say x∗,must be taken infinitely often.
page 92 of notes
d) Any metric space (X, d) with only finitely many elements is sequentially
compact.
Bare hands proof: Let (xn) be any sequence in X.
Since there are only finitely many possible values of xn, one of them, say x∗,must be taken infinitely often.
Throw away all the xn 6= x∗ to leave a constant (and hence convergent)
subsequence.
page 92 of notes
d) Any metric space (X, d) with only finitely many elements is sequentially
compact.
Bare hands proof: Let (xn) be any sequence in X.
Since there are only finitely many possible values of xn, one of them, say x∗,must be taken infinitely often.
Throw away all the xn 6= x∗ to leave a constant (and hence convergent)
subsequence.
page 92 of notes
e) {0,1}N and {0,1}Z are sequentially compact.
page 92 of notes
e) {0,1}N and {0,1}Z are sequentially compact.
We shall just do {0,1}N: the argument for {0,1}Z is similar.
page 92 of notes
e) {0,1}N and {0,1}Z are sequentially compact.
We shall just do {0,1}N: the argument for {0,1}Z is similar.
Recall {0,1}N is the set of all sequences
x = x0x1x2x3x4 . . .
of 0s and 1s.
page 92 of notes
e) {0,1}N and {0,1}Z are sequentially compact.
We shall just do {0,1}N: the argument for {0,1}Z is similar.
Recall {0,1}N is the set of all sequences
x = x0x1x2x3x4 . . .
of 0s and 1s.
The distance between two sequences x and y is 0 if x = y,
and otherwise is 1/2n, where n is smallest with xn 6= yn.
page 92 of notes
e) {0,1}N and {0,1}Z are sequentially compact.
We shall just do {0,1}N: the argument for {0,1}Z is similar.
Recall {0,1}N is the set of all sequences
x = x0x1x2x3x4 . . .
of 0s and 1s.
The distance between two sequences x and y is 0 if x = y,
and otherwise is 1/2n, where n is smallest with xn 6= yn.
pages 92 – 93 of notes
Next we have several results which enable us to take sequentially compact
spaces we already know about and build new ones from them.
pages 92 – 93 of notes
Next we have several results which enable us to take sequentially compact
spaces we already know about and build new ones from them.
The first one is very similar to the corresponding result for complete metric
spaces (Theorem 2.10, which says that a subset A of a complete metric
space X is complete if and only if it is closed in X).
pages 92 – 93 of notes
Next we have several results which enable us to take sequentially compact
spaces we already know about and build new ones from them.
The first one is very similar to the corresponding result for complete metric
spaces (Theorem 2.10, which says that a subset A of a complete metric
space X is complete if and only if it is closed in X).
page 93 of notes
Theorem Let (X, d) be a sequentially compact metric space, and A be a
subset of X. Then the following are equivalent:
page 93 of notes
Theorem Let (X, d) be a sequentially compact metric space, and A be a
subset of X. Then the following are equivalent:
a) A is sequentially compact.
page 93 of notes
Theorem Let (X, d) be a sequentially compact metric space, and A be a
subset of X. Then the following are equivalent:
a) A is sequentially compact.
b) A is closed in X.
page 93 of notes
Theorem Let (X, d) be a sequentially compact metric space, and A be a
subset of X. Then the following are equivalent:
a) A is sequentially compact.
b) A is closed in X.
We use:
Lemma 1.11: “A is closed in X” is equivalent to
“if an → ` is a convergent sequence in X, and all the an lie in A, then ` lies in A”.
page 93 of notes
Theorem Let (X, d) be a sequentially compact metric space, and A be a
subset of X. Then the following are equivalent:
a) A is sequentially compact.
b) A is closed in X.
We use:
Lemma 1.11: “A is closed in X” is equivalent to
“if an → ` is a convergent sequence in X, and all the an lie in A, then ` lies in A”.
page 93 of notes
Theorem Let (X, d) be a sequentially compact metric space, and A be a
subset of X. Then the following are equivalent:
a) A is sequentially compact.
b) A is closed in X.
We use:
Lemma 1.11: “A is closed in X” is equivalent to
“if an → ` is a convergent sequence in X, and all the an lie in A, then ` lies in A”.
Note that when we showed that a) =⇒ b), we didn’t make any use of the
assumption that X is sequentially compact. So this result is true whether
or not X is sequentially compact.
page 93 of notes
Theorem Let (X, d) be a sequentially compact metric space, and A be a
subset of X. Then the following are equivalent:
a) A is sequentially compact.
b) A is closed in X.
We use:
Lemma 1.11: “A is closed in X” is equivalent to
“if an → ` is a convergent sequence in X, and all the an lie in A, then ` lies in A”.
Note that when we showed that a) =⇒ b), we didn’t make any use of the
assumption that X is sequentially compact. So this result is true whether
or not X is sequentially compact.
Lemma Let (X, d) be a metric space, and A be a subset of X. If A is
sequentially compact, then A is closed in X.
page 93 of notes
Theorem Let (X, d) be a sequentially compact metric space, and A be a
subset of X. Then the following are equivalent:
a) A is sequentially compact.
b) A is closed in X.
We use:
Lemma 1.11: “A is closed in X” is equivalent to
“if an → ` is a convergent sequence in X, and all the an lie in A, then ` lies in A”.
Note that when we showed that a) =⇒ b), we didn’t make any use of the
assumption that X is sequentially compact. So this result is true whether
or not X is sequentially compact.
Lemma Let (X, d) be a metric space, and A be a subset of X. If A is
sequentially compact, then A is closed in X.
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X.
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∪ B. We need
to show that (xn) has a convergent subsequence.A
B
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∪ B. We need
to show that (xn) has a convergent subsequence.
Suppose that infinitely many of the xn are in A
(if not, then infinitely many are in B and a similar
argument will work).
A
B
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∪ B. We need
to show that (xn) has a convergent subsequence.
Suppose that infinitely many of the xn are in A
(if not, then infinitely many are in B and a similar
argument will work).
Throw away the xn which aren’t in A to obtain a
subsequence (xni) contained in A.
A
B
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∪ B. We need
to show that (xn) has a convergent subsequence.
Suppose that infinitely many of the xn are in A
(if not, then infinitely many are in B and a similar
argument will work).
Throw away the xn which aren’t in A to obtain a
subsequence (xni) contained in A.
Since A is sequentially compact, there is a sub-
sequence (xnij) which converges to ` ∈ A.
A
B
l
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∪ B. We need
to show that (xn) has a convergent subsequence.
Suppose that infinitely many of the xn are in A
(if not, then infinitely many are in B and a similar
argument will work).
Throw away the xn which aren’t in A to obtain a
subsequence (xni) contained in A.
Since A is sequentially compact, there is a sub-
sequence (xnij) which converges to ` ∈ A.
This is the required subsequence of (xn) which
converges to ` ∈ A ∪B.
A
B
l
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∩ B. We need
to show that (xn) has a convergent subsequence
xni → ` ∈ A ∩B.
A
B
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∩ B. We need
to show that (xn) has a convergent subsequence
xni → ` ∈ A ∩B.
Since (xn) is a sequence in A, which is sequen-
tially compact, it has a subsequence (xni) which
converges to ` ∈ A.
A
B
l
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∩ B. We need
to show that (xn) has a convergent subsequence
xni → ` ∈ A ∩B.
Since (xn) is a sequence in A, which is sequen-
tially compact, it has a subsequence (xni) which
converges to ` ∈ A.
We need to show that ` ∈ B also.
A
B
l
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∩ B. We need
to show that (xn) has a convergent subsequence
xni → ` ∈ A ∩B.
Since (xn) is a sequence in A, which is sequen-
tially compact, it has a subsequence (xni) which
converges to ` ∈ A.
We need to show that ` ∈ B also.
But B is closed in X (Lemma 3.3).
A
B
l
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∩ B. We need
to show that (xn) has a convergent subsequence
xni → ` ∈ A ∩B.
Since (xn) is a sequence in A, which is sequen-
tially compact, it has a subsequence (xni) which
converges to ` ∈ A.
We need to show that ` ∈ B also.
But B is closed in X (Lemma 3.3).
And all of the xni are in B.
A
B
l
page 93 of notes
Theorem Let (X, d) be a metric space, and let A and B be sequentiallycompact subsets of X. Then both A∪B and A∩B are sequentially compact.
Let (xn) be any sequence in A ∩ B. We need
to show that (xn) has a convergent subsequence
xni → ` ∈ A ∩B.
Since (xn) is a sequence in A, which is sequen-
tially compact, it has a subsequence (xni) which
converges to ` ∈ A.
We need to show that ` ∈ B also.
But B is closed in X (Lemma 3.3).
And all of the xni are in B.
Hence ` ∈ B by Lemma 1.11.
A
B
l
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact.
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact. The same is true for products: the proof of the next result
is essentially identical to that of Theorem 2.8 (for complete spaces).
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact. The same is true for products: the proof of the next result
is essentially identical to that of Theorem 2.8 (for complete spaces).
Theorem Let (X, d) and (Y, e) be sequentially compact.
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact. The same is true for products: the proof of the next result
is essentially identical to that of Theorem 2.8 (for complete spaces).
Theorem Let (X, d) and (Y, e) be sequentially compact.
Then the product space (X × Y, D) (where D is the product metric) is also
sequentially compact.
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact. The same is true for products: the proof of the next result
is essentially identical to that of Theorem 2.8 (for complete spaces).
Theorem Let (X, d) and (Y, e) be sequentially compact.
Then the product space (X × Y, D) (where D is the product metric) is also
sequentially compact.
Let ((xn, yn)) be any sequence in X × Y . We need to show that it has a
convergent subsequence.
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact. The same is true for products: the proof of the next result
is essentially identical to that of Theorem 2.8 (for complete spaces).
Theorem Let (X, d) and (Y, e) be sequentially compact.
Then the product space (X × Y, D) (where D is the product metric) is also
sequentially compact.
Let ((xn, yn)) be any sequence in X × Y . We need to show that it has a
convergent subsequence.
Since X is sequentially compact, (xn) has a subsequence xni → ` ∈ X.
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact. The same is true for products: the proof of the next result
is essentially identical to that of Theorem 2.8 (for complete spaces).
Theorem Let (X, d) and (Y, e) be sequentially compact.
Then the product space (X × Y, D) (where D is the product metric) is also
sequentially compact.
Let ((xn, yn)) be any sequence in X × Y . We need to show that it has a
convergent subsequence.
Since X is sequentially compact, (xn) has a subsequence xni → ` ∈ X.
Since Y is sequentially compact, (yni) has a subsequence ynij→ m ∈ Y .
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact. The same is true for products: the proof of the next result
is essentially identical to that of Theorem 2.8 (for complete spaces).
Theorem Let (X, d) and (Y, e) be sequentially compact.
Then the product space (X × Y, D) (where D is the product metric) is also
sequentially compact.
Let ((xn, yn)) be any sequence in X × Y . We need to show that it has a
convergent subsequence.
Since X is sequentially compact, (xn) has a subsequence xni → ` ∈ X.
Since Y is sequentially compact, (yni) has a subsequence ynij→ m ∈ Y .
Then ((xnij, ynij
)) → (`, m) ∈ X × Y as required (Lemma 1.3).
pages 93 – 94 of notes
Thus unions and intersections of sequentially compact spaces are sequen-
tially compact. The same is true for products: the proof of the next result
is essentially identical to that of Theorem 2.8 (for complete spaces).
Theorem Let (X, d) and (Y, e) be sequentially compact.
Then the product space (X × Y, D) (where D is the product metric) is also
sequentially compact.
Let ((xn, yn)) be any sequence in X × Y . We need to show that it has a
convergent subsequence.
Since X is sequentially compact, (xn) has a subsequence xni → ` ∈ X.
Since Y is sequentially compact, (yni) has a subsequence ynij→ m ∈ Y .
Then ((xnij, ynij
)) → (`, m) ∈ X × Y as required (Lemma 1.3).
page 94 of notes
The next result has no analogue for complete metric spaces.
page 94 of notes
The next result has no analogue for complete metric spaces.
Theorem Let (X, d) be sequentially compact, (Y, e) be any metric space,
and f : X → Y be continuous.
page 94 of notes
The next result has no analogue for complete metric spaces.
Theorem Let (X, d) be sequentially compact, (Y, e) be any metric space,
and f : X → Y be continuous. Then f(X) is sequentially compact.
page 94 of notes
The next result has no analogue for complete metric spaces.
Theorem Let (X, d) be sequentially compact, (Y, e) be any metric space,
and f : X → Y be continuous. Then f(X) is sequentially compact.
We use Lemma 1.5:
If f : X → Y is continuous and xn → ` ∈ X, then f(xn) → f(`) ∈ Y .
page 94 of notes
The next result has no analogue for complete metric spaces.
Theorem Let (X, d) be sequentially compact, (Y, e) be any metric space,
and f : X → Y be continuous. Then f(X) is sequentially compact.
We use Lemma 1.5:
If f : X → Y is continuous and xn → ` ∈ X, then f(xn) → f(`) ∈ Y .
page 94 of notes
The next result has no analogue for complete metric spaces.
Theorem Let (X, d) be sequentially compact, (Y, e) be any metric space,
and f : X → Y be continuous. Then f(X) is sequentially compact.
We use Lemma 1.5:
If f : X → Y is continuous and xn → ` ∈ X, then f(xn) → f(`) ∈ Y .
We finish with three properties of sequentially compact metric spaces.
page 94 of notes
The next result has no analogue for complete metric spaces.
Theorem Let (X, d) be sequentially compact, (Y, e) be any metric space,
and f : X → Y be continuous. Then f(X) is sequentially compact.
We use Lemma 1.5:
If f : X → Y is continuous and xn → ` ∈ X, then f(xn) → f(`) ∈ Y .
We finish with three properties of sequentially compact metric spaces.
page 94 of notes
Lemma Let (X, d) be sequentially compact. Then it is bounded.
page 94 of notes
Lemma Let (X, d) be sequentially compact. Then it is bounded.
Recall: (X, d) is bounded if there is some K with d(x, y) ≤ K for all x, y ∈ X.
page 94 of notes
Lemma Let (X, d) be sequentially compact. Then it is bounded.
Recall: (X, d) is bounded if there is some K with d(x, y) ≤ K for all x, y ∈ X.
Example: C[0,1] isn’t sequentially compact.
page 94 of notes
Lemma Let (X, d) be sequentially compact. Then it is bounded.
Recall: (X, d) is bounded if there is some K with d(x, y) ≤ K for all x, y ∈ X.
Example: C[0,1] isn’t sequentially compact.
For let fn ∈ C[0,1] be the constant function
fn(x) = n.
0 1
n
n
fn
f0
page 94 of notes
Lemma Let (X, d) be sequentially compact. Then it is bounded.
Recall: (X, d) is bounded if there is some K with d(x, y) ≤ K for all x, y ∈ X.
Example: C[0,1] isn’t sequentially compact.
For let fn ∈ C[0,1] be the constant function
fn(x) = n.
Then d(f0, fn) = n for all n, and so C[0,1] isn’t
bounded.
0 1
n
n
fn
f0
page 95 of notes
Theorem Let (X, d) be sequentially compact, and f : X → R be continuous.
page 95 of notes
Theorem Let (X, d) be sequentially compact, and f : X → R be continuous.
Then f is bounded and attains its bounds.
page 95 of notes
Theorem Let (X, d) be sequentially compact, and f : X → R be continuous.
Then f is bounded and attains its bounds.
We use:
Theorem 3.6: If X is sequentially compact and f : X → Y is continuous, then f(X) is
sequentially compact.
page 95 of notes
Theorem Let (X, d) be sequentially compact, and f : X → R be continuous.
Then f is bounded and attains its bounds.
We use:
Theorem 3.6: If X is sequentially compact and f : X → Y is continuous, then f(X) is
sequentially compact.
Theorem 3.1: A sequentially compact subset of R is closed and bounded.
page 95 of notes
Theorem Let (X, d) be sequentially compact, and f : X → R be continuous.
Then f is bounded and attains its bounds.
We use:
Theorem 3.6: If X is sequentially compact and f : X → Y is continuous, then f(X) is
sequentially compact.
Theorem 3.1: A sequentially compact subset of R is closed and bounded.
page 95 of notes
Recall that a homeomorphism f : X → Y is an invertible function (bijection)such that both f and f−1 are continuous.
page 95 of notes
Recall that a homeomorphism f : X → Y is an invertible function (bijection)such that both f and f−1 are continuous.
Is it possible to have a continuous bijection f : X → Y for which f−1 is notcontinuous?
page 95 of notes
Recall that a homeomorphism f : X → Y is an invertible function (bijection)such that both f and f−1 are continuous.
Is it possible to have a continuous bijection f : X → Y for which f−1 is notcontinuous? Our intuition for functions f : R → R (that they’re continuousif we can draw their graphs without taking our pens off the paper) suggeststhat if f is continous, f−1 must be too.
f
page 95 of notes
Recall that a homeomorphism f : X → Y is an invertible function (bijection)such that both f and f−1 are continuous.
Is it possible to have a continuous bijection f : X → Y for which f−1 is notcontinuous? Our intuition for functions f : R → R (that they’re continuousif we can draw their graphs without taking our pens off the paper) suggeststhat if f is continous, f−1 must be too. (The graph of f−1 is just the graphof f reflected in the diagonal y = x.)
f
f -1
pages 95 – 96 of notes
This intuition is wrong.
pages 95 – 96 of notes
This intuition is wrong. For (X, d), take C[0,1] with the (usual) L∞ metric.
pages 95 – 96 of notes
This intuition is wrong. For (X, d), take C[0,1] with the (usual) L∞ metric.
For (Y, e), take C[0,1] with the L1 metric e(f, g) =∫ 10 |f(x)− g(x)|dx.
d
e
pages 95 – 96 of notes
This intuition is wrong. For (X, d), take C[0,1] with the (usual) L∞ metric.
For (Y, e), take C[0,1] with the L1 metric e(f, g) =∫ 10 |f(x)− g(x)|dx.
d
e
Let F : X → Y be the identity function F(f) = f .
pages 95 – 96 of notes
This intuition is wrong. For (X, d), take C[0,1] with the (usual) L∞ metric.
For (Y, e), take C[0,1] with the L1 metric e(f, g) =∫ 10 |f(x)− g(x)|dx.
d
e
Let F : X → Y be the identity function F(f) = f . Then
– F is a bijection (it is its own inverse).
pages 95 – 96 of notes
This intuition is wrong. For (X, d), take C[0,1] with the (usual) L∞ metric.
For (Y, e), take C[0,1] with the L1 metric e(f, g) =∫ 10 |f(x)− g(x)|dx.
d
e
Let F : X → Y be the identity function F(f) = f . Then
– F is a bijection (it is its own inverse).
– F is continuous.
pages 95 – 96 of notes
This intuition is wrong. For (X, d), take C[0,1] with the (usual) L∞ metric.
For (Y, e), take C[0,1] with the L1 metric e(f, g) =∫ 10 |f(x)− g(x)|dx.
d
e
Let F : X → Y be the identity function F(f) = f . Then
– F is a bijection (it is its own inverse).
– F is continuous. For if d(f, g) is very small, then the graphs of f and g
are very close to each other, and hence e(f, g) is very small.
pages 95 – 96 of notes
This intuition is wrong. For (X, d), take C[0,1] with the (usual) L∞ metric.
For (Y, e), take C[0,1] with the L1 metric e(f, g) =∫ 10 |f(x)− g(x)|dx.
d
e
Let F : X → Y be the identity function F(f) = f . Then
– F is a bijection (it is its own inverse).
– F is continuous. For if d(f, g) is very small, then the graphs of f and g
are very close to each other, and hence e(f, g) is very small.
