Sequential Compactness

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Section 4.1 Sequential Compactness

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Recordings on Re:View.

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The entire section

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Definition 4.1.

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A subset

K

of a metric space

(X, d)

is called sequentially compact if every sequence in

K

has subsequence which converges to a limit in

K .

K = X

is sequentially compact, we say the metric space

(X, d)

is sequentially compact.

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Example 4.2.

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An interval of the form

[a, b]

with

a < b,

considered as a subset of

R

with the usual metric, is sequentially compact by the theorem of Bolzano–Weierstrass.

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The following are some basic properties of sequentially compact sets.

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Lemma 4.3. Sequential compactness and metric subspaces.

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A subset

K

of a metric space

(X, d)

is sequentially compact if and only if the associated metric subspace

(K, d^{'})

is sequentially compact.

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Proof.

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Let

(x_{n})_{n \in N}

be a sequence of points in

K,

and let

x_{0} \in K .

Then a subsequence

(x_{n_{k}})_{k \in N}

converges to

x_{0}

as a sequence in

(X, d)

if and only if it converges to

x_{0}

as a sequence in

(K, d^{'}) .

The result follows.

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Theorem 4.4.

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Any sequentially compact metric space is complete.

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Proof.

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In a sequentially compact space

(X, d),

let

(x_{n})_{n \in N}

be a Cauchy sequence. Then there exists a convergent subsequence

(x_{n_{k}})_{k \in N} .

Let

x_{0} \in X

be its limit. We claim that

x_{n} \to x_{0}

n \to \infty .

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Let

ε > 0 .

Then there exists

N \in N

such that

d (x_{m}, x_{n}) < \frac{ε}{2}

when

m, n \geq N .

Moreover, as

x_{n_{k}} \to x_{0},

there exists

K \in N

such that

n_{K} \geq N

and

d (x_{n_{K}}, x_{0}) < \frac{ε}{2} .

Now

d (x_{n}, x_{0}) \leq d (x_{n}, x_{n_{K}}) + d (x_{n_{K}}, x_{0}) < ε

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for

n \geq N .

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Theorem 4.5.

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K

is a sequentially compact subset of a metric space

(X, d),

then

K

is closed.

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Proof.

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Consider a sequence

(x_{n})_{n \in N}

K

that converges in

X

to a limit

x_{0} \in X .

By Theorem 1.35, it suffices to show that

x_{0} \in K .

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K

is sequentially compact, there exists a subsequence

(x_{n_{k}})_{k \in N}

that converges to a limit

y_{0} \in K .

The same subsequence also converges to

x_{0},

and by the uniqueness of limits, we infer that

x_{0} = y_{0} \in K .

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Theorem 4.6.

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Let

(X, d)

be a sequentially compact metric space. Let

S \subset X

be closed. Then

S

is sequentially compact as well.

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Proof.

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Let

(x_{n})_{n \in N}

be a sequence in

S .

As it is also a sequence in

X,

there exists a subsequence

(x_{n_{k}})_{k \in N}

with a limit

x_{0} \in X .

Since

S

is closed, Theorem 1.35 then implies that

x_{0} \in S,

and the proof is complete.

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Exercises Exercises

1. (PS8) Compactness and minimisation.

Let

(X, d)

be a non-empty metric space and

f : X \to R

a function. A minimising sequence for

f

is a sequence

(x_{n})_{n \in N}

X

with

inf_{x \in X} f (x) = lim_{n \to \infty} f (x_{n}),

where we allow

lim_{n \to \infty} f (x_{n}) = - \infty

if necessary.

(a)

Show that every function

f : X \to R

has a minimising sequence.

Hint 1.

Exercise B.2 is useful here.

Hint 2.

Note that question asks you to consider the possibility that

inf_{x \in X} f (x) = - \infty .

Solution.

inf_{x \in X} f (x) > - \infty,

then this follows from Exercise B.2. Indeed, the definition of the infimum implies that for every

n \in N,

there exists a point

x_{n} \in X

such that

f (x_{n}) - \frac{1}{n} \leq inf_{x \in X} f (x) \leq f (x_{n}) .

Thus we obtain a sequence

(x_{n})_{n \in N}

with the required property. If

inf_{x \in X} f (x) = - \infty,

then for any

n \in N

there exists

x_{n} \in X

with

f (x_{n}) \leq - n .

Then

lim_{n \to \infty} f (x_{n}) = - \infty

as well. (Note that

inf_{x \in X} f (x) = + \infty

is not possible as

X \neq \emptyset .

)

Comment.

Note that the question explicitly mentions the possibility that

inf_{x \in X} f (x) = - \infty .

Many students did not treat this case; on an exam this would have cost marks.

(b)

Show that the infimum of

f

is attained, i.e., there exists a point

x_{0} \in X

with

f (x_{0}) = inf_{x \in X} f (x),

(X, d)

is sequentially compact and

f

is continuous.

Hint.

Argue that the minimising sequence from the previous part has a convergent subsequence.

Solution.

Let

(x_{n})_{n \in N}

be a minimising sequence. If

(X, d)

is sequentially compact, then there exists a convergent subsequence

(x_{n_{k}})_{k \in N} .

Let

x_{0}

be its limit. If

f

is continuous, then it follows (from Theorem 1.34 and Theorem 1.46) that

f (x_{0}) = lim_{k \to \infty} f (x_{n_{k}}) = lim_{n \to \infty} f (x_{n}) = inf_{x \in X} f (x) .

Comment.

It’s easy to see to find counterexamples if we weaken the requirement that

X

is compact. For instance, the function

x \mapsto x

does not attains its minimum over either

X = (0, 1)

X = R .

2. Bounding functions on compact spaces.

Let

(X, d)

be a metric space and

K \subseteq X

be sequentially compact. Suppose

f : X \to R

is continuous. Show that

f

is bounded on

K .

That is, show that there exists a real number

M > 0

such that

| f (x) | \leq M

for every

x \in K .

Hint.

Try an argument by contradiction.

3. Weierstrass extreme value theorem.

Let

X

be a metric space and

K \subseteq X

be compact. Let

f \in C_{b} (K) .

Show that there exists

x_{sup} \in K

and

x_{inf} \in K

such that

sup_{x \in K} f (x) = f (x_{sup})

and

inf_{x \in K} f (x) = f (x_{inf}) .

Hint.

Use Exercise 4.1.1.

4. Nested compact sets.

In a metric space

(X, d),

consider a sequence of sets

C_{n} \subseteq X

for

n \in N .

Suppose that

C_{n}

is non-empty and sequentially compact, and

C_{n + 1} \subseteq C_{n}

for every

n \in N .

Show that the intersection

⋂_{n \in N} C_{n}

is not empty.

Hint 1.

For each

n \in N,

pick a point

x_{n} \in C_{n} .

Argue that

(x_{n})_{n \in N}

is then a sequence in the sequentially compact set

C_{1} .

Hint 2.

Theorem 4.5 is useful.

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