Continuous Functions Between Metric Spaces

Section Continuous Functions Between Metric Spaces

In our preview activity we saw how to formally define what it means for a function

f : R \to R

to be continuous at a point.

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Note that Definition 6.2 depends only on being able to measure how close points are to each other. Since that is precisely what a metric does, we can extend this notion of continuity to define continuity for functions between metric spaces. Continuity is an important idea in topology, and we will work with this idea extensively throughout the semester.

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If we let

d_{E} : R \times R \to R

be defined by

d_{R} (x, y) = | x - y |,

then we have seen that

d_{E}

is a metric on

R

(note that

d_{E}

is the Euclidean metric on

R

). Using this metric we can reformulate what it means for a function

f : R \to R

to be continuous at a point.

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Definition 6.3. Alternate Definition.

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A function

f : R \to R

is continuous at a point

a

if, given any

ϵ > 0,

there exists a

δ > 0

so that

d_{E} (x, a) < δ

implies

d_{E} (f (x), f (a)) < ϵ .

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This alternate definition depends on the metric

d_{E} .

We could easily replace the metric

d

with any other metric we choose. This allows us to define continuity at a point for functions between metric spaces.

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Definition 6.4.

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Let

(X, d_{X})

and

(Y, d_{Y})

be metric spaces. A function

f : X \to Y

is continuous at

a \in X

if, given any

ϵ > 0,

there exists a

δ > 0

so that

d_{X} (x, a) < δ

implies

d_{Y} (f (x), f (a)) < ϵ .

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Once we have defined continuity at a point, we can define continuous functions.

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Definition 6.5.

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Let

(X, d_{X})

and

(Y, d_{Y})

be metric spaces. A function

f : X \to Y

is continuous if

f

is continuous at every point in

X .

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Example 6.6.

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In general, to prove that a function

f : X \to Y

is continuous, where

(X, d_{X})

and

(Y, d_{Y})

are metric spaces, we begin by choosing an arbitrary element

a

X .

Then we let

ϵ

be a number greater than

0

and show that there is a

δ > 0

so that

d_{Y} (f (x), f (a)) < ϵ

whenever

d_{X} (x, a) < δ .

The

δ

we need cannot depend on

x

(since

x

isn’t known), but can depend on the value of

a

that we choose, and will likely depend on

ϵ

as well. That is, there is a function

C

of only the independent variables

a

and

ϵ

that produces the

δ,

δ = C (a, ϵ) .

As an example, let

X = R

and let

d_{X}

be defined as

d_{X} (x, y) = min {| x - y |, 1} .

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The proof that

d_{X}

is a metric is left for Exercise 9. Consider

f : X \to Y

defined by

f (x) = x^{2},

where

(Y, d_{Y}) = (R, d_{E}) .

To show that

f

is continuous, we let

a \in R

and let

ϵ > 0 .

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Scratch work. What happens next is not part of the proof, but shows how we go about finding a $δ$ we need. We are looking for $δ > 0$ such that $d_{X} (x, a) < δ$ implies that $d_{E} (f (x), f (a)) < ϵ .$ That is, we want to make

$d_{E} (f (x), f (a)) = \sqrt{(f (x) - f (a))^{2}} = | f (x) - f (a) | = | x^{2} - a^{2} | < ϵ$

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whenever

$d_{X} (x, a) = min {| x - a |, 1} < δ .$

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Now $| x^{2} - a^{2} | = | (x - a) (x + a) | = | x - a | | x + a | .$ If $d_{X} (x, a) < δ,$ then $min {| x - a |, 1} < δ .$ If we choose $δ < 1,$ then $d_{X} (x, a) < δ < 1$ implies that $| x - a | < 1$ and so $d_{X} (x, a) = | x - a | .$ Now

$| x + a | = | (x - a) + 2 a | \leq | x - a | + 2 | a | < 1 + 2 | a | .$

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It follows that

$| x - a | | x + a | < δ (1 + 2 | a |) .$

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To make this product less that $ϵ,$ we can choose $δ$ such that $δ (1 + 2 | a |) < ϵ$ or $δ < \frac{ϵ}{1 + 2 | a |} .$ That is, there is a function $C$ of $ϵ$ that gives us the $δ$ we want, namely $δ = C (a, ϵ) = {1, \frac{ϵ}{1 + 2 | a |}} .$ Now we ignore this paragraph and present the proof, which is essentially reversing the steps we just made. If the steps can’t be reversed, then we have to rethink our argument. The next step in the proof might seem like magic to the uninitiated reader, but we have seen behind the curtain so it isn’t a mystery to us.