3.6. Functions and Continuity#

The material in this section is primarily based on [2, 41].

Let (X,d) and (Y,ρ) be metric spaces.

Recall that for a function f:(X,d)(Y,ρ):

  1. If f is a partial function, then domfX.

  2. If f is a total function, then domf=X.

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#

Definition 3.35 (Open mapping)

A (total) function f:(X,d)(Y,ρ) is called an open mapping if f(A) is open whenever A is open.

In other words, f maps open sets to open sets.

Definition 3.36 (Closed mapping)

A (total) function f:(X,d)(Y,ρ) is called a closed mapping if f(A) is closed whenever A is closed.

In other words, f maps closed sets to closed sets.

Example 3.14 (A closed map need not be open)

Let f:XY be defined as f(x)=y0 for every xX where y0Y is some fixed point.

Then, for every AX, f(A)={y0}. Since {y0} is a singleton, hence

  1. f maps every closed set in X to a (fixed) closed set in Y.

  2. But f doesn’t map open sets in X to open sets in Y.

Example 3.15 (An open map need not be closed)

Consider the function f:R2R given by

f((x,y))=x.

Thus, f is a projection function which projects a point in R2 to its first coordinate.

We first show that f is an open mapping.

  1. Now, let A be an open set in R2.

  2. Let xf(A).

  3. Let yR such that (x,y)A. Thus, f((x,y))=x.

  4. Since A is open, hence (x,y) is an interior point of A.

  5. Thus, there exists r>0 such that B((x,y),r)A.

  6. But then, f(B((x,y),r))f(A).

  7. Note that f(B((x,y),r))=(xr,x+r).

  8. Thus, (xr,x+r)f(A).

  9. Thus, x is an interior point of f(A).

  10. Thus, every point in f(A) is its interior point.

  11. Thus, f(A) is open.

  12. Thus, f is an open mapping.

We now show that f is not a closed mapping.

  1. Consider the set $L = { (x, y) \in \RR^2 | x> 0 \text{ and } xy = 1}.

  2. This is a curve in R2 (visualize).

  3. It is thus a closed set.

  4. f(L)=(0,).

  5. f(L) is not a closed set in R. It is rather an open interval.

  6. Thus, f doesn’t map every closed set to a closed set. Although, it does map some closed sets to closed sets.

  7. Thus, f is not a closed mapping.

3.6.2. Bounded Functions#

Definition 3.37 (Bounded functions)

A (total) function f:(X,d)(Y,ρ) is called a bounded if its image f(X) is bounded (in the metric space sense).

In other words, there exists a number M>0 such that

ρ(f(x),f(y))Mx,yX.

Compare this definition with the definition for bounded real valued functions in Definition 2.50.

3.6.3. Continuous Functions#

Definition 3.38 (Continuous function)

A function f:XY between the two metric spaces is said to be continuous at a point adomf if for every ϵ>0, there exists δ>0 (depending on ϵ and a) such that for all xdomf

d(x,a)<δρ(f(x),f(a))<ϵ

holds true.

f is said to be continuous on Adomf if f is continuous at every point of A.

In other words, there is an open ball BX(a,δ) around a which maps within the open ball BY(f(a),ϵ) around f(a).

The “continuity at a point” definition above is valid for both partial and total functions.

  1. If a is an isolated point in domf, then f is continuous at a.

  2. If a is an accumulation point in domf then, limxaf(x)=f(a) must hold true for f to be continuous at x=a.

Remark 3.7

If f is continuous on A then Adomf. If f is continuous on X then domf=X.

Note

If f:(X,d)(Y,ρ) is not defined for all of X, then we can consider the restriction of f to A=domf and work with f~:(A,d)(Y,ρ) given by f~(x)=f(x)xA where (A,d) is a metric subspace of (X,d). Then f is a total function.

Example 3.16 (Distance from a set)

Recall from Definition 3.5 that the distance from a nonempty set AX of any point xX is defined as:

d(x,A)=inf{d(x,a)aA}.

If we fix A, then we can define a function dA:XR as:

dA(x)=d(x,A).

The function is well defined since the set {d(x,a)xA} is bounded from below and thus has a unique greatest lower bound.

We now show that dA is continuous.

We first establish some basic inequalities. For any x,yX and aA we have:

d(a,x)d(a,y)+d(x,y)

Thus,

d(a,y)d(a,x)d(x,y).

Since d(a,x)dA(x), we get:

d(a,y)dA(x)d(x,y).

Since this inequality is valid for every aA, taking the infimum on the L.H.S, we get:

dA(y)dA(x)d(x,y)dA(x)dA(y)d(x,y).

Interchanging x with y, we get:

dA(y)dA(x)d(y,x)=d(x,y).

Combining the two, we get:

|dA(x)dA(y)|d(x,y).

Now for any ϵ>0, choosing δ=ϵ, we get that

d(x,y)<δ|dA(x)dA(y)|<δ=ϵ.

Hence, dA is continuous on X.

3.6.3.1. Continuity Characterization#

Theorem 3.42 (Characterization of continuous functions)

Let f:XY be a (total) function between two metric spaces. The following statements are equivalent:

  1. f is continuous on X.

  2. f1(O) is open subset of X whenever O is an open set of Y.

  3. If limxn=x holds in X, then limf(xn)=f(x) holds in Y.

  4. f(clA)clf(A) holds for every subset A of X.

  5. f1(C) is a closed subset of X whenever C is a closed subset of Y.

Proof. (1) (2). We prove this by showing that every point in f1(O) is its interior point.

  1. Let O be open in Y. Let af1(O).

  2. Thus, f(a)O.

  3. Since O is open, there exists r>0 such that BY(f(a),r)O.

  4. Since f is continuous at a, there exists δ>0 (depending on r) such that

    d(x,a)<δρ(f(x),f(a))<r.
  5. i.e., for any xBX(a,δ), f(x)BY(f(a),r)O.

  6. Thus, xf1(O) for any xBX(a,δ).

  7. This means that BX(a,δ)f1(O).

  8. Thus, a is an interior point of f1(O).

  9. Since every point in f1(O) is its interior point, hence f1(O) is open.

(2) (3)

  1. Let limxn=x in X and r>0.

  2. Let O=BY(f(x),r).

  3. By (2), f1(O) is open in X.

  4. Since xf1(O), there exists some δ>0 such that BX(x,δ)f1(O).

  5. Since limxn=x, there exists kN such that xnBX(x,δ) for all n>k.

  6. Thus, xnf1(O) for all n>k.

  7. Thus, f(xn)O for all n>k.

  8. i.e., for every r>0, there exists kN such that f(xn)BY(f(x),r) for all n>k.

  9. Thus, limf(xn)=f(x).

(3) (4). We will show that every point in f(clA) belongs to clf(A).

  1. Let A be a subset of X.

  2. Let yf(clA).

  3. Then, there exists xclA such that y=f(x).

  4. Since x is a closure point of A, there exists a sequence {xn} of A that converges to x.

  5. Since xnA, hence f(xn)f(A).

  6. Thus, {f(xn)} is a sequence of f(A).

  7. Since limxn=x, hence, by (3), limf(xn)=f(x)=y.

  8. Thus, y is a closure point of f(A).

  9. Thus, yf(clA)yclf(A).

  10. Thus, f(clA)clf(A).

(4) (5)

  1. Let C be a closed subset of Y.

  2. Thus, C=clC.

  3. Let A=f1(C).

  4. Using (4), f(clA)clf(A)=clC=C.

  5. Thus, clAf1(C)=A.

  6. Since AclA always, hence A=clA.

  7. Thus, A=f1(C) is a closed subset of X.

(5) (1). We will show that f is continuous at any xX.

  1. Let aX. Let ϵ>0.

  2. Consider the open ball BY(f(a),ϵ).

  3. Consider the closed set C=YBY(f(a),ϵ).

  4. C can be written as:

    C={yY|ρ(f(a),y)ϵ}.
  5. Using (5), f1(C) is closed in X.

  6. Since f(a)C, hence af1(C).

  7. Thus, aXf1(C) which is an open subset of X.

  8. Thus, there exists δ>0 such that BX(a,δ)Xf1(C).

  9. Note that xXf1(C) implies that ρ(f(a),f(x))<ϵ. For if ρ(f(a),f(x))ϵ then f(x)C then xf1(C).

  10. Thus, xBX(a,δ) implies that f(x)BY(f(a),ϵ).

  11. Thus, for every ϵ>0, there exists δ>0 such that d(x,a)<δ implies ρ(f(x),f(a))<ϵ.

  12. Thus, f is continuous at a.

  13. Since a is arbitrary, hence f is continuous on X.

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.

Theorem 3.43 (Characterization of continuity for partial functions)

Let f:(X,d)(Y,ρ) be a partial function between two metric spaces with S=domfX. Let (S,d) denote the metric subspace with the distance function d restricted to S×S.

The following statements are equivalent:

  1. f is continuous on S.

  2. f1(O) is open relative to (S,d) whenever O is an open set of Y.

  3. If limxn=x holds in S, then limf(xn)=f(x) holds in Y.

  4. f(clA)clf(A) holds for every subset A of S where clA means closure of A relative to the subspace (S,d).

  5. f1(C) is a closed in (S,d) whenever C is a closed subset of (Y,ρ).

Proof. (1) (2). We prove this by showing that every point in f1(O) is its interior point relative to (S,d).

  1. Let O be open in Y. Let af1(O).

  2. Thus, f(a)O.

  3. Since O is open, there exists r>0 such that BY(f(a),r)O.

  4. Since f is continuous at a, there exists δ>0 (depending on r) such that for every xS

    d(x,a)<δρ(f(x),f(a))<r.
  5. In other words, for every xBX(a,δ)S, f(x)BY(f(a),r)O.

  6. Thus, xf1(O) for every xBX(a,δ)S.

  7. This means that BX(a,δ)Sf1(O).

  8. Thus, a is an interior point of f1(O) relative to (S,d).

  9. Since every point in f1(O) is its interior point, hence f1(O) is open in (S,d).

(2) (3)

  1. Let limxn=x in S and r>0.

  2. Let O=BY(f(x),r).

  3. By (2), f1(O) is open in (S,d).

  4. Since xf1(O), there exists some δ>0 such that BX(x,δ)Sf1(O).

  5. Since limxn=x, there exists kN such that xnBX(x,δ)S for all n>k.

  6. Thus, xnf1(O) for all n>k.

  7. Thus, f(xn)O for all n>k.

  8. i.e., for every r>0, there exists kN such that f(xn)BY(f(x),r) for all n>k.

  9. Thus, limf(xn)=f(x).

(3) (4). We will show that every point in f(clA) belongs to clf(A).

  1. Let A be a subset of S.

  2. Let yf(clA).

  3. Then, there exists xclA such that y=f(x).

  4. Since x is a closure point of A, there exists a sequence {xn} of A that converges to x.

  5. Since xnA, hence f(xn)f(A).

  6. Thus, {f(xn)} is a sequence of f(A).

  7. Since limxn=x, hence, by (3), limf(xn)=f(x)=y.

  8. Thus, y is a closure point of f(A).

  9. Thus, yf(clA)yclf(A).

  10. Thus, f(clA)clf(A).

(4) (5)

  1. Let C be a closed subset of Y.

  2. Thus, C=clC.

  3. Let A=f1(C).

  4. Using (4), f(clA)clf(A)=clC=C.

  5. Thus, clAf1(C)=A.

  6. Since AclA always, hence A=clA.

  7. Thus, A=f1(C) is a closed subset of S.

(5) (1). We will show that f is continuous at every xS.

  1. Let aS. Let ϵ>0.

  2. Consider the open ball BY(f(a),ϵ).

  3. Consider the closed set C=YBY(f(a),ϵ).

  4. C can be written as:

    C={yY|ρ(f(a),y)ϵ}.
  5. Using (5), f1(C) is closed in (S,d).

  6. Since f(a)C, hence af1(C).

  7. Thus, aSf1(C) which is an open subset of (S,d).

  8. Thus, there exists δ>0 such that BX(a,δ)SSf1(C).

  9. Note that xSf1(C) implies that ρ(f(a),f(x))<ϵ. For if ρ(f(a),f(x))ϵ then f(x)C then xf1(C).

  10. Thus, xBX(a,δ)S implies that f(x)BY(f(a),ϵ).

  11. Thus, for every ϵ>0, there exists δ>0 such that for every xS, d(x,a)<δ implies ρ(f(x),f(a))<ϵ.

  12. Thus, f is continuous at a.

  13. Since a is arbitrary, hence f is continuous on S.

3.6.3.3. Closures#

Theorem 3.44 (Continuity and closure)

Let f:XY be a continuous function between two metric spaces. Let AX. Then,

clf(clA)=clf(A).

Proof. Since f is continuous, hence by Theorem 3.42 (4)

f(clA)clf(A).

Now, clf(A) is a closed set. Hence, due to Proposition 3.6

clf(clA)clf(A).

Now, note that:

f(A)f(clA)

since AclA.

But then, taking closure on both sides, we get:

clf(A)clf(clA).

Combining the inclusions, we obtain:

clf(clA)=clf(A).

3.6.3.4. Interiors#

Theorem 3.45 (Continuity characterization with interiors)

Let f:XY be a function between two metric spaces. f is continuous if and only if for every AY

f1(intA)intf1(A).

Proof. Assume f is continuous. Let AY be arbitrary.

  1. Let xf1(intA).

  2. Then, f(x)intA.

  3. Thus, there is an open ball U=B(f(x),r)A.

  4. Now, xf1(U) since f(x)U.

  5. But UA.

  6. Thus, xf1(A).

  7. Since f is continuous and U is open, hence f1(U) is an open set due to Theorem 3.42.

  8. But since UA, hence f1(U)f1(A).

  9. Thus, we have xf1(U)f1(A).

  10. Since f1(U) is open, hence x is an interior point of f1(U).

  11. Thus, x is an interior point of f1(A) also.

  12. Thus, xintf1(A).

  13. Thus, we have

    f1(intA)intf1(A)

    for every AY.

Now assume that for every AY

f1(intA)int(f1(A))

holds true.

  1. Let UY be an arbitrary open set.

  2. We have that

    f1(intU)=f1(U)intf1(U)f1(U).
  3. Therefore f1(U)=intf1(U) must hold (PQP means P=Q).

  4. Therefore, f1(U) is open.

  5. We have shown that for every open set UY, its pre-image f1(U) is open.

  6. Thus, by Theorem 3.42, f is continuous.

3.6.3.5. Function Compositions#

Theorem 3.46 (Continuity and composition)

Composition of continuous (total) functions is continuous.

Proof. Let (X1,d1),(X2,d2),(X3,d3) be metric spaces. Let f:X1X2 and g:X2X3 be continuous. Define h:X1X3 as h=gf.

Let {xn} be a sequence of X1.

  1. Then, {yn=f(xn)} is a sequence of X2.

  2. And, {zn=g(yn))} is a sequence of X3.

Assume that limxn=x. Let y=f(x) and z=g(y).

  1. Since f is continuous, hence limxn=xlimyn=y due to Theorem 3.42.

  2. Since g is continuous, hence limyn=ylimzn=z.

  3. Thus, limxn=xlimzn=z.

  4. Since this is valid for any convergent sequence of X, gf is continuous; again due to Theorem 3.42.

3.6.3.6. Level Sets#

Theorem 3.47 (Level sets of continuous functions)

Let f:XY be a continuous function between two metric spaces. Then, the level sets of f given by

Ay={xX|f(x)=y}

are closed.

Proof. By definition:

Ay=f1({y}).

Now, {y} is a singleton set in Y, hence closed in Y due to Theorem 3.8.

Then, by Theorem 3.42 (5), f1({y}) is closed.

3.6.4. Discontinuity#

Definition 3.39 (Discontinuity)

A function f:XY between the two metric spaces is said to be discontinuous at a point adomf if there exists ϵ>0, such that for every δ>0 there exists xdomf with d(x,a)<δ and ρ(f(x),f(a))ϵ.

3.6.5. Uniform Continuity#

Definition 3.40 (Uniform continuity)

A function f:(X,d)(Y,ρ) between two metric spaces is called uniformly continuous on Adomf if for every ϵ>0, there exists some δ>0 (depending on ϵ) such that

ρ(f(x),f(y))<ϵ whenever d(x,y)<δ and x,yA.

Remark 3.8

If f is uniformly continuous on Adomf, then f is continuous on A.

For real valued functions, the standard metric on R is d(x,y)=|xy|. The uniform continuity definition simplifies accordingly as:

Remark 3.9 (Uniform continuity for real valued functions)

A real valued function f:(X,d)R is called uniformly continuous on Adomf if for every ϵ>0, there exists some δ>0 (depending on ϵ) such that

|f(x)f(y)|<ϵ whenever d(x,y)<δ and x,yA.

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.

Definition 3.41 (Homeomorphism)

Let (X,d) and (Y,ρ) be two metric spaces. A function f:XY is called a homeomorphism if

  1. f is bijective (thus the inverse f1 exists).

  2. f is continuous.

  3. f1 is continuous.

If a homeomorphism exists between two metric spaces (X,d) and (Y,ρ), then the metric spaces are called homeomorphic.

Procedure to show that two metric spaces are homeomorphic:

  1. Pick a suitable bijective function f:XY.

  2. Show that f is continuous.

  3. Show that f1 is continuous.

Example 3.17 (1/x)

Let X=(0,1] and Y=[1,). Let f:XY be given by

f(x)=1x.
  1. f is continuous.

  2. f is bijective.

  3. f1 exists.

  4. f1=f. It is self inverse (involution).

  5. Thus, f is a homeomorphism between X and Y.

Further observations:

  1. (0,1] is bounded but [1,) is not.

  2. [1,) is complete but (0,1] is not.

Thus, homeomorphisms do not preserve boundedness or completeness.

3.6.6.1. An Equivalence Relation#

Theorem 3.48 (Homeomorphism is an equivalence relation)

Consider the family of all metric spaces denoted by M. Consider the relation where we say that AB for any A,BM if A and B are homeomorphic (i.e., there exists a homeomorphism between them).

Then, is an equivalence relation.

Proof. [Reflexivity]

  1. Let AM.

  2. Consider the identity mapping I:AA given by I(x)=x for every xA.

  3. Then, I is a homeomorphism.

  4. Thus, AA.

[Symmetry]

  1. Let A,BM such that AB.

  2. Thus, there exists a homeomorphism f:AB where f is bijective, f is continuous and f1 is continuous.

  3. Let g=f1.

  4. Then, g is a bijective mapping from B to A, g is continuous and g1=f is also continuous.

  5. Thus, g is a homeomorphism from B to A.

  6. Thus, AB implies BA.

[Transitivity]

  1. Let A,B,CM so that AB and BC.

  2. Thus, there exists a homeomorphism f:AB and another homeomorphism g:BC.

  3. Consider the function h=gf which is a mapping from A to C.

  4. Since f and g are bijective, hence h is also bijective.

  5. In fact, h1=(gf)1=f1g1.

  6. Since f and g are both continuous, hence h is also continuous.

  7. Since g1 and f1 are both continuous, hence h1 is also continuous.

  8. Thus, h:AC is bijective and both h and h1 are continuous.

  9. Thus, h is a homeomorphism from A to C.

  10. Thus, A and C are homeomorphic.

  11. Thus, AC.

3.6.6.2. Open and Closed Mappings#

Theorem 3.49 (Homeomorphisms are both open and closed)

Let f:(X,d)(Y,ρ) be a homeomorphism. Then f is both an open mapping as well as a closed mapping.

f1 is also both an open mapping and a closed mapping.

Proof. Let f be homeomorphism and g=f1. Then, g1=f.

We first show that f is an open mapping.

  1. Let AX be open.

  2. Since g is continuous, hence g1(A) is open due to by Theorem 3.42 (2).

  3. But g1(A)=f(A).

  4. Thus, f(A) is open whenever A is open.

  5. Thus, f maps open sets to open sets.

  6. Thus, f is an open mapping.

We next show that f is an open mapping.

  1. Let AX be closed.

  2. Since g is continuous, hence g1(A) is closed due to by Theorem 3.42 (5).

  3. But g1(A)=f(A).

  4. Thus, f(A) is closed whenever A is closed.

  5. Thus, f maps closed sets to closed sets.

  6. Thus, f is a closed mapping.

A similar reasoning establishes that g=f1 is also both a closed and an open mapping.

3.6.6.3. Metric Equivalence as homeomorphism#

Theorem 3.50 (Metric equivalence and identity homeomorphism)

Two metrics d1 and d2 on X are equivalent if and only if the identity mapping I:(X,d1)(X,d2) given by

I(x)=xxX

is a homeomorphism.

Proof. Identity function is a bijection and is the inverse of itself. Thus, I1=I.

Recall from Definition 3.25 that two metrics are equivalent if they generate the same topology.

Assume d1,d2 to be equivalent.

  1. Let A(X,d2) be open.

  2. Then, I1(A)=A.

  3. But A is open in (X,d1) also since the metrics are equivalent.

  4. Thus, for every open set A in (X,d2) I1(A) is open in (X,d1).

  5. Therefore, I is continuous due to Theorem 3.42.

  6. Similarly, by starting with an open set in (X,d1), we can show that I1 is also continuous.

  7. Thus, I is bijective, and both I and I1 are continuous.

  8. Thus, I is a homeomorphism.

Now, assume I to be a homeomorphism.

  1. We first show that if A is an open set in (X,d1) then A is an open set in (X,d2) also.

  2. By Theorem 3.49, I is both an open mapping and a closed mapping.

  3. Thus, if A is an open set in (X,d1), then I(A)=A is an open set in (X,d2) also.

  4. We now show that if A is an open set in (X,d2) then A is an open set in (X,d1) also.

  5. Since I1=I, and I is an open mapping, hence, if A is an open set in (X,d2), then I1(A)=I(A)=A is an open set in (X,d1) also.

  6. Thus, every set in (X,d1) is open if and only if it is open in (X,d2).

  7. Thus, both metric spaces have same topology.

  8. Thus, the metrics d1 and d2 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 {xn} be a sequence of X.

Assume that the two metrics are equivalent.

  1. Then, limd1(xn,x)=0limd2(xn,x)=0.

  2. Thus, if limd1(xn,x)=0 then limd2(xn,x)=limd2(I(xn),I(x))=0 means that I is continuous.

  3. Similarly, if limd2(xn,x)=0 then limd1(xn,x)=limd1(I1(xn),I1(x))=0 means that I1 is continuous.

  4. Thus, I is a homeomorphism.

Assume that I is a homeomorphism.

  1. I is continuous. Hence limd1(xn,x)=0 implies limd2(I(xn),I(x))=d2(xn,x)=0.

  2. I1 is continuous. Hence limd2(xn,x)=0 implies limd1(I1(xn),I1(x))=d1(xn,x)=0.

  3. Hence, the metrics d1 and d2 are equivalent.

3.6.6.4. Closures#

Theorem 3.51 (Homeomorphisms preserve closures)

Let f:(X,d)(Y,ρ) be a homeomorphism. Let AX. Then,

f(clA)=clf(A).

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 f is a homeomorphism, hence f is bijective, f1 exists and both f and f1 are continuous.

Since f is continuous, hence by Theorem 3.42 (4)

f(clA)clf(A).

We showed in Theorem 3.44 that

clf(clA)=clf(A).

Thus, if we can show that f(clA) is closed, then we are done.

By Theorem 3.49, f is a closed mapping.

Hence, f(clA) is a closed set. Thus,

clf(clA)=f(clA)=clf(A).

3.6.6.5. Interiors#

Theorem 3.52 (Homeomorphisms preserve interiors)

Let f:(X,d)(Y,ρ) be a homeomorphism. Let AX. Then,

f(intA)=intf(A).

In other words, a homeomorphism preserves interiors.

Proof. Since f is a homeomorphism, it is bijective and both f and g=f1 are continuous.

We first show that f(intA)intf(A).

  1. Let AX be arbitrary.

  2. Due to Theorem 3.45,

    f(intA)intf(A)

    since g is continuous and f=g1.

We next show that intf(A)f(intA).

  1. Let yintf(A).

  2. Then, there exists an open ball in Y around y such that

    yU=B(y,r)f(A).
  3. Therefore,

    f1(y)f1(U)f1(f(A))=A

    since f is bijective.

  4. Now, since f is a homeomorphism, hence f1 is an open mapping (Theorem 3.49).

  5. Thus, since U is open, hence f1(U) is open.

  6. Thus, f1(U) is an open neighborhood of f1(y) contained in A.

  7. Thus, f1(y) is an interior point of A.

  8. Thus, f1(y)intA.

  9. Thus, yf(intA).

  10. We have shown that yintf(A)yf(intA).

  11. Thus, intf(A)f(intA).

Combining the two inclusions:

intf(A)=f(intA).

3.6.7. Isometry#

Definition 3.42 (Isometry)

A function f:(X,d)(Y,ρ) is an isometry if for all x,ydomf, we have:

ρ(f(x),f(y))=d(x,y).

Theorem 3.53 (Isometries are injective)

Let f:(X,d)(Y,ρ) be an isometry. Then, f is injective.

Proof. We proceed as follows:

  1. Let x1,x2domf with x1x2.

  2. Then, d(x1,x2)>0 since d is a metric.

  3. Since f is an isometry, hence ρ(f(x1),f(x2))=d(x1,x2)>0.

  4. Thus, f(x1)f(x2) since ρ is a metric.

  5. Thus, x1x2f(x1)f(x2).

  6. Thus, f is injective.

Theorem 3.54 (Isometries are uniformly continuous)

Let f:(X,d)(Y,ρ) be an isometry. Then, f is uniformly continuous.

Proof. Let ϵ>0. Choose δ=ϵ. Then, for any x,ydomf,

d(x,y)<δρ(f(x),f(y))<δ=ϵ

since f is an isometry. Thus, f is uniformly continuous on domf.

Definition 3.43 (Isometric spaces)

Two metric spaces (X,d) and (Y,ρ) are called isometric if there exists an isometry f:XY such that domf=X and rangef=Y.

Such an isometry is bijective.

Theorem 3.55

Two metric spaces which are isometric are necessarily homeomorphic.

Proof. Let (X,d) and (Y,ρ) be isometric. Then, there exists an isometry f from X to Y which is bijective. Since f is an isometry, it is continuous. It is easy to see that f1 is also an isometry and is continuous. Thus, the metric spaces are homeomorphic.

3.6.8. Bounded Metric#

This section is dedicated to the development of a bounded metric on any metric space.

Theorem 3.56 (Bounded metric)

If d is a metric on a set X, then the function ρ given by

ρ(x,y)=d(x,y)1+d(x,y)

is also a metric on X. Besides, (X,ρ) is bounded and ρ is equivalent to d.

We structure the proof into three parts:

  1. Show that ρ is a metric.

  2. Show that (X,ρ) is bounded.

  3. Show that ρ and d are equivalent.

Proof. ρ is a metric.

(1) Non-negativity: Since d(x,y)0, hence ρ(x,y)0 as it is a ratio of a non-negative number with a positive number.

(2) Identity of indiscernibles:

  1. Assume ρ(x,y)=0.

  2. Then d(x,y)=0.

  3. Thus x=y since d is a metric.

  4. Now, assume x=y.

  5. Then d(x,y)=0.

  6. Thus, ρ(x,y)=0.

(3) Symmetry:

ρ(y,x)=d(y,x)1+d(y,x)=d(x,y)1+d(x,y)=ρ(x,y).

(4) Triangle inequality. This will require some work.

Consider the function f:RR:

f(t)=t1+t

with domf=R+.

Its derivative is

f(t)=1(1+t)2.

f(t)0 for tR+. Thus, f is an increasing function on t0. In particular:

d(x,y)d(x,z)+d(z,y)f(d(x,y))f(d(x,z)+d(z,y)).

Now, we proceed as follows:

ρ(x,y)=f(d(x,y))f(d(x,z)+d(z,y))=d(x,z)+d(z,y)1+d(x,z)+d(z,y)=d(x,z)1+d(x,z)+d(z,y)+d(z,y)1+d(x,z)+d(z,y)d(x,z)1+d(x,z)+d(z,y)1+d(z,y)=ρ(x,z)+ρ(z,y).

Proof. ρ is a bounded

It is easy to see that

supρ(x,y)=1.

Thus, (X,ρ) is bounded.

Proof. ρ is equivalent to d

We first show that the identity mapping I:(X,d)(X,ρ) is continuous.

Let aX and choose ϵ>0. Recall that ρ is bounded with ρ(x,y)<1.

Thus, if ϵ1, we can choose any δ>0 leading to ρ(x,y)<ϵ whenever d(x,y)<δ.

Now, consider the case ϵ<1.

ρ(x,a)<ϵd(x,a)1+d(x,a)<ϵ1d(x,a)1+d(x,a)>1ϵ11+d(x,a)>1ϵ1+d(x,a)<11ϵd(x,a)<ϵ1ϵ.

Now, choosing δ=ϵ1ϵ, we note that δ>0 for 0<ϵ<1.

Thus, for every ϵ>0, there exists δ>0 such that ρ(x,a)<ϵ whenever d(x,a)<δ.

Hence, I is continuous. A similar argument also shows that I1 is continuous.

Thus, I is homeomorphism and the metric spaces (X,d) and (X,ρ) are homeomorphic.

Hence, due to Theorem 3.50, the two metrics are equivalent.

3.6.9. Lipschitz Continuity#

Definition 3.44 (Local Lipschitz continuity at a point)

Let f:(X,d)(Y,ρ) be a (partial) function with S=domf.

We say that f is locally Lipschitz continuous at some aS if there is a constant L (depending on a) and δ>0 (depending on a) such that

ρ(f(x),f(a))Ld(x,a)

for every xB(a,δ)S.

In other words, for every xS

d(x,x)<δρ(f(x),f(a))Ld(x,a).

Definition 3.45 (Lipschitz function)

Let f:(X,d)(Y,ρ) be a (partial) function with S=domf.

We say that f is a Lipschitz function if there is a constant L>0 such that

ρ(f(x),f(y))Ld(x,y)

for every x,yS.

Sometimes, we also say that f is Lipschitz continuous on its domain.

Example 3.18

Let f(x)=1x with domf=(0,).

Choose some a>0. Then, for any x>0,

|f(x)f(a)|=|1x1a|=|ax|xa=1xa|ax|.

There is no single value of L such that L1xa for every x,a>0. Thus, f is not a Lipschitz function.

However, if we consider δ=a2, and consider the neighborhood (aδ,a+δ) then

  1. x>aδ=a2.

  2. Thus, 1xa<2a2.

  3. Thus, |f(x)f(a)|<2a2|ax| for every x(aδ,a+δ).

  4. Thus, f is locally Lipschitz continuous at a with L=2a2.

  5. We can see that both L and δ depend on a.

Now, restrict the domain of f to (c,) for some c>0.

  1. Let x,y(c,).

  2. Then, xy>c2.

  3. Thus, 1xy<1c2.

  4. Thus, |f(x)f(y)|<1c2|xy|.

  5. Choosing L=1c2, we see that f is a Lipschitz function.

Theorem 3.57 (Lipschitz uniform continuity)

Every Lipschitz function is uniformly continuous.

Proof. Let f:(X,d)(Y,ρ) be a function with S=domf. Assume that f is Lipschitz. Thus, there exists L>0 such that

ρ(f(x),f(y))Ld(x,y)

for every x,yS.

  1. Let ϵ>0.

  2. Let δ=ϵL.

  3. Let x,yS such that d(x,y)<δ.

  4. Then,

    ρ(f(x),f(y))Ld(x,y)<LϵL=ϵ.
  5. Thus, for every ϵ>0, there exists δ>0 given by δ=ϵL such that for every x,yS

    d(x,y)<δρ(f(x),f(y))<ϵ.
  6. Thus, f is uniformly continuous.