Functions and Continuity
Contents
3.6. Functions and Continuity#
The material in this section is primarily based on [2, 41].
Let
Recall that for a function
We will primarily focus on total functions while some definition and results are valid for partial functions too.
3.6.1. Open and Closed Mappings#
(Open mapping)
A (total) function
In other words,
(Closed mapping)
A (total) function
In other words,
(A closed map need not be open)
Let
Then, for every
maps every closed set in to a (fixed) closed set in .But
doesn’t map open sets in to open sets in .
(An open map need not be closed)
Consider the function
Thus,
We first show that
Now, let
be an open set in .Let
.Let
such that . Thus, .Since
is open, hence is an interior point of .Thus, there exists
such that .But then,
.Note that
.Thus,
.Thus,
is an interior point of .Thus, every point in
is its interior point.Thus,
is open.Thus,
is an open mapping.
We now show that
Consider the set $L = { (x, y) \in \RR^2 | x> 0 \text{ and } xy = 1}.
This is a curve in
(visualize).It is thus a closed set.
. is not a closed set in . It is rather an open interval.Thus,
doesn’t map every closed set to a closed set. Although, it does map some closed sets to closed sets.Thus,
is not a closed mapping.
3.6.2. Bounded Functions#
(Bounded functions)
A (total) function
In other words, there exists a number
Compare this definition with the definition for bounded real valued functions in Definition 2.50.
3.6.3. Continuous Functions#
(Continuous function)
A function
holds true.
In other words, there is an open ball
The “continuity at a point” definition above is valid for both partial and total functions.
If
is an isolated point in , then is continuous at .If
is an accumulation point in then, must hold true for to be continuous at .
If
Note
If
(Distance from a set)
Recall from Definition 3.5
that the distance from a nonempty set
If we fix
The function is well defined since the set
We now show that
We first establish some basic inequalities.
For any
Thus,
Since
Since this inequality is valid for every
Interchanging
Combining the two, we get:
Now for any
Hence,
3.6.3.1. Continuity Characterization#
(Characterization of continuous functions)
Let
is continuous on . is open subset of whenever is an open set of .If
holds in , then holds in . holds for every subset of . is a closed subset of whenever is a closed subset of .
Proof. (1)
Let
be open in . Let .Thus,
.Since
is open, there exists such that .Since
is continuous at , there exists (depending on ) such thati.e., for any
, .Thus,
for any .This means that
.Thus,
is an interior point of .Since every point in
is its interior point, hence is open.
(2)
Let
in and .Let
.By (2),
is open in .Since
, there exists some such that .Since
, there exists such that for all .Thus,
for all .Thus,
for all .i.e., for every
, there exists such that for all .Thus,
.
(3)
Let
be a subset of .Let
.Then, there exists
such that .Since
is a closure point of , there exists a sequence of that converges to .Since
, hence .Thus,
is a sequence of .Since
, hence, by (3), .Thus,
is a closure point of .Thus,
.Thus,
.
(4)
Let
be a closed subset of .Thus,
.Let
.Using (4),
.Thus,
.Since
always, hence .Thus,
is a closed subset of .
(5)
Let
. Let .Consider the open ball
.Consider the closed set
. can be written as:Using (5),
is closed in .Since
, hence .Thus,
which is an open subset of .Thus, there exists
such that .Note that
implies that . For if then then .Thus,
implies that .Thus, for every
, there exists such that implies .Thus,
is continuous at .Since
is arbitrary, hence is continuous on .
3.6.3.2. Continuity Characterization for Partial Functions#
We can extend the characterization in Theorem 3.42 for partial functions with slight changes. The primary change is that on the function domain side we have to work with the subspace topology. Readers may skip this subsection on first reading.
(Characterization of continuity for partial functions)
Let
The following statements are equivalent:
is continuous on . is open relative to whenever is an open set of .If
holds in , then holds in . holds for every subset of where means closure of relative to the subspace . is a closed in whenever is a closed subset of .
Proof. (1)
Let
be open in . Let .Thus,
.Since
is open, there exists such that .Since
is continuous at , there exists (depending on ) such that for everyIn other words, for every
, .Thus,
for every .This means that
.Thus,
is an interior point of relative to .Since every point in
is its interior point, hence is open in .
(2)
Let
in and .Let
.By (2),
is open in .Since
, there exists some such that .Since
, there exists such that for all .Thus,
for all .Thus,
for all .i.e., for every
, there exists such that for all .Thus,
.
(3)
Let
be a subset of .Let
.Then, there exists
such that .Since
is a closure point of , there exists a sequence of that converges to .Since
, hence .Thus,
is a sequence of .Since
, hence, by (3), .Thus,
is a closure point of .Thus,
.Thus,
.
(4)
Let
be a closed subset of .Thus,
.Let
.Using (4),
.Thus,
.Since
always, hence .Thus,
is a closed subset of .
(5)
Let
. Let .Consider the open ball
.Consider the closed set
. can be written as:Using (5),
is closed in .Since
, hence .Thus,
which is an open subset of .Thus, there exists
such that .Note that
implies that . For if then then .Thus,
implies that .Thus, for every
, there exists such that for every , implies .Thus,
is continuous at .Since
is arbitrary, hence is continuous on .
3.6.3.3. Closures#
(Continuity and closure)
Let
Proof. Since
Now,
Now, note that:
since
But then, taking closure on both sides, we get:
Combining the inclusions, we obtain:
3.6.3.4. Interiors#
(Continuity characterization with interiors)
Let
Proof. Assume
Let
.Then,
.Thus, there is an open ball
.Now,
since .But
.Thus,
.Since
is continuous and is open, hence is an open set due to Theorem 3.42.But since
, hence .Thus, we have
.Since
is open, hence is an interior point of .Thus,
is an interior point of also.Thus,
.Thus, we have
for every
.
Now assume that for every
holds true.
Let
be an arbitrary open set.We have that
Therefore
must hold ( means ).Therefore,
is open.We have shown that for every open set
, its pre-image is open.Thus, by Theorem 3.42,
is continuous.
3.6.3.5. Function Compositions#
(Continuity and composition)
Composition of continuous (total) functions is continuous.
Proof. Let
Let
Then,
is a sequence of .And,
is a sequence of .
Assume that
Since
is continuous, hence due to Theorem 3.42.Since
is continuous, hence .Thus,
.Since this is valid for any convergent sequence of
, is continuous; again due to Theorem 3.42.
3.6.3.6. Level Sets#
(Level sets of continuous functions)
Let
are closed.
Proof. By definition:
Now,
Then, by Theorem 3.42 (5),
3.6.4. Discontinuity#
(Discontinuity)
A function
3.6.5. Uniform Continuity#
(Uniform continuity)
A function
If
For real valued functions, the standard metric on
(Uniform continuity for real valued functions)
A real valued function
3.6.6. Homeomorphism#
Homeomorphism is the fusion of the ideas of continuity and bijection. We are interested in bijective mappings where the function and its inverse are both continuous.
Homeomorphisms characterize what are known as topological properties. Properties of sets in metric spaces which are preserved by homeomorphisms are known as topological properties. For example, homeomorphisms preserve openness, closedness, compactness. But they don’t preserve boundedness or completeness.
Homeomorphisms can be thought of as continuous deformations which are reversible.
If two spaces are connected through a homeomorphism, they are called homeomorphic. Once we prove that two spaces are homeomorphic, we need to study only one of them for their topological properties.
(Homeomorphism)
Let
is bijective (thus the inverse exists). is continuous. is continuous.
If a homeomorphism exists between two metric spaces
Procedure to show that two metric spaces are homeomorphic:
Pick a suitable bijective function
.Show that
is continuous.Show that
is continuous.
(1/x)
Let
is continuous. is bijective. exists. . It is self inverse (involution).Thus,
is a homeomorphism between and .
Further observations:
is bounded but is not. is complete but is not.
Thus, homeomorphisms do not preserve boundedness or completeness.
3.6.6.1. An Equivalence Relation#
(Homeomorphism is an equivalence relation)
Consider the family of all metric spaces denoted by
Then,
Proof. [Reflexivity]
Let
.Consider the identity mapping
given by for every .Then,
is a homeomorphism.Thus,
.
[Symmetry]
Let
such that .Thus, there exists a homeomorphism
where is bijective, is continuous and is continuous.Let
.Then,
is a bijective mapping from to , is continuous and is also continuous.Thus,
is a homeomorphism from to .Thus,
implies .
[Transitivity]
Let
so that and .Thus, there exists a homeomorphism
and another homeomorphism .Consider the function
which is a mapping from to .Since
and are bijective, hence is also bijective.In fact,
.Since
and are both continuous, hence is also continuous.Since
and are both continuous, hence is also continuous.Thus,
is bijective and both and are continuous.Thus,
is a homeomorphism from to .Thus,
and are homeomorphic.Thus,
.
3.6.6.2. Open and Closed Mappings#
(Homeomorphisms are both open and closed)
Let
Proof. Let
We first show that
Let
be open.Since
is continuous, hence is open due to by Theorem 3.42 (2).But
.Thus,
is open whenever is open.Thus,
maps open sets to open sets.Thus,
is an open mapping.
We next show that
Let
be closed.Since
is continuous, hence is closed due to by Theorem 3.42 (5).But
.Thus,
is closed whenever is closed.Thus,
maps closed sets to closed sets.Thus,
is a closed mapping.
A similar reasoning establishes that
3.6.6.3. Metric Equivalence as homeomorphism#
(Metric equivalence and identity homeomorphism)
Two metrics
is a homeomorphism.
Proof. Identity function is a bijection and is the inverse of itself.
Thus,
Recall from Definition 3.25 that two metrics are equivalent if they generate the same topology.
Assume
Let
be open.Then,
.But
is open in also since the metrics are equivalent.Thus, for every open set
in is open in .Therefore,
is continuous due to Theorem 3.42.Similarly, by starting with an open set in
, we can show that is also continuous.Thus,
is bijective, and both and are continuous.Thus,
is a homeomorphism.
Now, assume
We first show that if
is an open set in then is an open set in also.By Theorem 3.49,
is both an open mapping and a closed mapping.Thus, if
is an open set in , then is an open set in also.We now show that if
is an open set in then is an open set in also.Since
, and is an open mapping, hence, if is an open set in , then is an open set in also.Thus, every set in
is open if and only if it is open in .Thus, both metric spaces have same topology.
Thus, the metrics
and are equivalent.
We can construct another proof by using an equivalent definition of equivalent metrics. In Theorem 3.37, we showed that two metrics are equivalent if and only if their convergent sequences are identical. This proof is from [2].
Proof. Let
Assume that the two metrics are equivalent.
Then,
.Thus, if
then means that is continuous.Similarly, if
then means that is continuous.Thus,
is a homeomorphism.
Assume that
is continuous. Hence implies . is continuous. Hence implies .Hence, the metrics
and are equivalent.
3.6.6.4. Closures#
(Homeomorphisms preserve closures)
Let
In other words, a homeomorphism preserves closures.
Compare this with the result in Theorem 3.44. We no longer have to take another closure on the L.H.S..
Proof. Since
Since
We showed in Theorem 3.44 that
Thus, if we can show that
By Theorem 3.49,
Hence,
3.6.6.5. Interiors#
(Homeomorphisms preserve interiors)
Let
In other words, a homeomorphism preserves interiors.
Proof. Since
We first show that
Let
be arbitrary.Due to Theorem 3.45,
since
is continuous and .
We next show that
Let
.Then, there exists an open ball in
around such thatTherefore,
since
is bijective.Now, since
is a homeomorphism, hence is an open mapping (Theorem 3.49).Thus, since
is open, hence is open.Thus,
is an open neighborhood of contained in .Thus,
is an interior point of .Thus,
.Thus,
.We have shown that
.Thus,
.
Combining the two inclusions:
3.6.7. Isometry#
(Isometry)
A function
(Isometries are injective)
Let
Proof. We proceed as follows:
Let
with .Then,
since is a metric.Since
is an isometry, hence .Thus,
since is a metric.Thus,
.Thus,
is injective.
(Isometries are uniformly continuous)
Let
Proof. Let
since
(Isometric spaces)
Two metric spaces
Such an isometry is bijective.
Two metric spaces which are isometric are necessarily homeomorphic.
Proof. Let
3.6.8. Bounded Metric#
This section is dedicated to the development of a bounded metric on any metric space.
(Bounded metric)
If
is also a metric on
We structure the proof into three parts:
Show that
is a metric.Show that
is bounded.Show that
and are equivalent.
Proof.
(1) Non-negativity: Since
(2) Identity of indiscernibles:
Assume
.Then
.Thus
since is a metric.Now, assume
.Then
.Thus,
.
(3) Symmetry:
(4) Triangle inequality. This will require some work.
Consider the function
with
Its derivative is
Now, we proceed as follows:
Proof.
It is easy to see that
Thus,
Proof.
We first show that the identity mapping
Let
Thus, if
Now, consider the case
Now, choosing
Thus, for every
Hence,
Thus,
Hence, due to Theorem 3.50, the two metrics are equivalent.
3.6.9. Lipschitz Continuity#
(Local Lipschitz continuity at a point)
Let
We say that
for every
In other words, for every
(Lipschitz function)
Let
We say that
for every
Sometimes, we also say that
Let
Choose some
There is no single value of
However, if we consider
.Thus,
.Thus,
for every .Thus,
is locally Lipschitz continuous at with .We can see that both
and depend on .
Now, restrict the domain of
Let
.Then,
.Thus,
.Thus,
.Choosing
, we see that is a Lipschitz function.
Every Lipschitz function is uniformly continuous.
Proof. Let
for every
Let
.Let
.Let
such that .Then,
Thus, for every
, there exists given by such that for everyThus,
is uniformly continuous.