Sequences and Continuity in Metric Spaces

Section Sequences and Continuity in Metric Spaces

We have seen that there are different ways to characterize. For example, there is the

ϵ - δ

definition and a characterization in terms of neighborhoods. In this section we investigate sequences and limits of sequences in metric spaces, and then provide a characterization of continuous functions in terms of sequences.

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Activity 9.2.

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A reasonable question to ask is if a limit of a sequence is unique. We will answer that question in this activity. Let

(X, d)

be a metric space and

(a_{n})

a sequence in

X .

Assume the sequence

(a_{n})

has a limit in

X .

To show that a limit of the sequence

(a_{n})

is unique, we need to show that if

lim a_{n} = a

and

lim a_{n} = a^{'}

for some

a, a^{'} \in X,

then

a = a^{'} .

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Suppose

lim a_{n} = a

and

lim a_{n} = a^{'}

for some

a, a^{'} \in X .

Without much to go on it might appear that proving

a = a^{'}

is a difficult task. However, if

d (a, a^{'}) < ϵ

for any

ϵ > 0,

then it will have to be the case that

a = a^{'} .

So let

ϵ > 0 .

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(a)

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Why must there exist a positive integer

N

so that

d (a_{n}, a) < \frac{ϵ}{2}

for all

n \geq N ?

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(b)

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Why must there exist a positive integer

N^{'}

so that

d (a_{n}, a^{'}) < \frac{ϵ}{2}

for all

n \geq N^{'} ?

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(c)

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Now let

m = max {N, N^{'}} .

What can we say about

d (a_{m}, a)

and

d (a_{m}, a^{'}) ?

Why?

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(d)

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Use the triangle inequality to conclude that

d (a, a^{'}) < ϵ .

What else can we conclude?

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Now we will examine how continuity can be described in terms of sequences. The basic idea is this. Suppose that

f : R \to R

is continuous at a point

a .

This means that

f

has a limit (as a continuous function) at

a .

So if we were to take any sequence

(a_{n})

that converges to

a,

then the continuity of

f

implies that

f (a) = f (lim a_{n}) = lim f (a_{n}) .

That this is both a necessary condition and a sufficient condition for continuity is given in the next theorem.

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Theorem 9.7.

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Let

(X, d_{X})

and

(Y, d_{Y})

be metric spaces, and let

a \in X .

A function

f : X \to Y

is continuous at

a

if and only if

lim f (a_{n}) = f (a)

for any sequence

(a_{n})

X

that converges to

a .

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Proof.

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Let

(X, d_{X})

and

(Y, d_{Y})

be metric spaces, let

a \in X,

and let

f : X \to Y

be a function. Assume that

f

is continuous at

a .

We will show that

lim f (a_{n}) = f (a)

for any sequence

(a_{n})

X

that converges to

a .

Let

(a_{n})

be a sequence in

X

that converges to

a

(we know such a sequence exists, namely the sequence

(a)

). To verify that

lim f (a_{n}) = a,

let

ϵ > 0 .

The fact that

f

is continuous at

a

means that there is a

δ > 0

so that

d_{Y} (f (x), f (a)) < ϵ

whenever

d_{X} (x, a) < δ .

Since

(a_{n})

converges to

a,

we know that there exists a positive integer

N

such that

d_{X} (a_{n}, a) < δ

whenever

n \geq N .

This implies that

d_{Y} (f (a_{n}), f (a)) < ϵ whenever n \geq N .

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We conclude that if

f

is continuous at

a,

then

lim f (a_{n}) = f (a)

for any sequence

(a_{n})

X

that converges to

a .

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The proof of the reverse implication is contained in the next activity.

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Activity 9.3.

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Let

(X, d_{X})

and

(Y, d_{Y})

be metric spaces, let

a \in X,

and let

f : X \to Y

be a function. We prove the remaining implication of Theorem 9.7, that

f

is continuous at

a

lim f (a_{n}) = f (a)

for any sequence

(a_{n})

X

that converges to

a,

in this activity.

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(a)

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To have an additional assumption with which to work, let us proceed by contradiction and assume that

f

is not continuous at

a .

Why can we then say that there is an

ϵ > 0

so that there is no

δ > 0

with the property that

d_{X} (x, a) < δ

implies

d_{Y} (f (x), f (a)) < ϵ ?

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(b)

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To create a contradiction, we will construct a sequence

(a_{n})

that converges to

a

while

(f (a_{n}))

does not converge to

f (a) .

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(i)

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Explain why we can find a positive integer

K

such that

\frac{1}{K} < ϵ .

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(ii)

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k > K,

explain why there is an element

a_{k} \in B (a, \frac{1}{k})

so that

d_{Y} (f (a_{k}), f (a)) \geq ϵ .

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(iii)

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For

k \leq K,

let

a_{k}

be any element in

B (a, \frac{1}{k}) .

Explain why

a

is a limit of

(a_{n}) .

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(iv)

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Explain why

f (a)

is not a limit of the sequence

(f (a_{n})) .

What conclusion can we draw, and why?

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One way that Theorem 9.7 is often used is illustrated in the next activity.

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Activity 9.4.

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Let

f

be the function from

R

R,

both with the Euclidean metric, defined by

f (x) = {\begin{cases} \sin (\frac{1}{x}) & if x \neq 0 \\ 0 & if x = 0. \end{cases}

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We consider the

f

continuity of

f

0

in this activity.

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(a)

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Draw a graph of

f

on some small interval centered at

0 .

Based on the graph, do you think

f

has a limit at

0 ?

Explain. (There is no right answer here, just your intuition based on the graph.)

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(b)

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At which inputs is

f (x) = 1 ?

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(c)

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Use the result of (b) to find a sequence

(a_{n})

that converges to

0

for which

f (a_{n}) = 1

for every

n .

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(d)

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What does the result of (c) tell us about the continuity of

f

0 ?

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While it can sometimes be difficult to prove a fact about all sequences that converge to a point, Activity 9.4 shows that we can use Theorem 9.7 to prove that a function

f

is not continuous at an input

a

be finding just one sequence

(a_{n})

that converges to

a

for which

lim f (a_{n}) \neq f (a) .

We conclude this section with one final note.

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IMPORTANT NOTE: Theorem 9.7 tells us that if

f : X \to Y

is a continuous function, then

f

commutes with limits. That is, if

(a_{n})

is a sequence in

X

that converges to

a \in X,

then

f (a) = f (lim a_{n}) = lim f (a_{n}) .

Topology: An Inquiry-Based Approach

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Section Sequences and Continuity in Metric Spaces

Activity 9.2.

(a)

(b)

(c)

(d)

Theorem 9.7.

Proof.

Activity 9.3.

(a)

(b)

(i)

(ii)

(iii)

(iv)

Activity 9.4.

(a)

(b)

(c)

(d)