Research Article
On Directed Metric Spaces
Mathematics Science Unit, Institute of Education, Obafemi Awolowo University, Ile-lfe, Nigeria
Philip Olu
Mathematics Science Unit, Institute of Education, Obafemi Awolowo University, Ile-lfe, Nigeria
We proceed by defining first a metric space
Definition 1: Let X be a set, a function d defined on XHX into the positive real numbers is a metric on X. If the following are satisfied.
(a) | d(x,y) ≥ 0 x, y∈X (non-negativity) |
(b) | d(x,y)=0 iff x=y (identity) |
(c) | d(x,y)=d(y,x) (symmetry) |
(d) | d(x,y)≤d(x,z)+d(z,y), x,y,z∈X (triangular inequality) |
(X,d) is said to be a metric space. Semi-metric and Pseudo-metric spaces have emerged from this and extensive work have been done on them[1-5]. We consider here an entirely different space, an abstraction of the displacement function which we define thus.
Definition 2: Let X be a set, suppose we define dd on X x X such that dd:X x X→ satisfies the followings:
(a) | dd(x,y)=-dd(y,x) |
(b) | |dd(x,y)|≤|dd(x,z)|+|dd(z,y)|x,y,z∈X |
If we call dd a directed-metric we say (X,dd) is a directed-metric space.
The fundamental notion here is the non-symmetry property i.e dd(x,y)=-dd(y,x). We observe that the identity property is deducible from the non-symmetry for if x=y we have that dd(x,y)=-dd(y,x) reduces to dd(x,x)=0 which is the identity property. The function dd as a directed distance function indicates that the sign depends on the direction.
Example 1: Consider R the set of all real number with dd(x,y)=xn-ynx,y ∈Rn ∈N then dd is a directed metric on R and (R, dd) is a directed metric space.
Proof:
dd(x,y)=xn-ynx,y ∈R, n ∈N
(a) | We have that dd(x,y)=xn-yn=-(yn-xn)=-dd(y,x) thus dd(x,y)=-dd(y,x) |
(b) | |dd(x,y)|=|xn-yn|=|xn-zn+zn-yn| |
Example 2: Let X be any set, if we define
x,y,∈X, (X,dd) make a discrete directed metric space. This example shows that an ordered set can be made into a directed-metric space.
Proof:
(a) | Suppose x=y, dd(x,y)=0 and there is nothing to prove. Now if x>y, dd(x,y)=-1 thus we have that dd(y,x)=1 (since y<x and dd(x,y)=-dd(y,x)). Also if x<y, dd(x,y)=1, thus dd(y,x)=-1 (i.e. y>x) which implies dd(x,y)=-dd(y,x) |
(b) | (b) Suppose x=y, dd(x,y)=0 |
thus dd so define is a directed-metric.
Example 3: Consider the set of points of a complex plane C. We define the directed-metric
dd(z1,z2)=(|z1|n-|z2|n) z1,z2∈C, n∈Z
(C,dd) is a directed metric space.
Proof:
(a) | dd(z1,z2)=(|z1|n-|z2|n)=-(|z2|n-|z1|n)=-dd(z2,z1) |
(b) | dd(z1,z2)=(|z1|n-|z2|n) |
dd so defined is a directed metric.
Example 4: dd(x,y)=ab (x(t)-y(t))dt defines a directed-metric space on the space C[a,b].
Proof:
(a) | Now dd(x,y)=ab (x(t)-y(t))dt=-ab (y(t)-x(t))dt=-dd(y,x) |
(b) | If suffices to prove that |dd(x,z)|≤|dd(x,y)|+|dd(y,z)| |
Thus dd is a directed metric.
ORDERING OF POINTS IN THE PLANE
Definition 3: Let x,y be points in the n-dimensional Euclidean plane i.e. x=(x1,x2, ... , xn);
Example 5: The set of all ordered n-tuples of real numbers (i.e. n-dimensional Euclidean space Rn) and the metric
defines a directed metric on (Rn,dd)
Proof
(a) | To prove that dd(x,y)=-dd(y,x) |
Suppose x<y
also
since
If x=y, dd(x,y)=0 and there is nothing to prove
Now if x>y,
since
(b) | The problem reduces to proving that |dd(x,z)|≤|dd(x,y)|+|dd(y,z)| |
if x<z,
Now if we suppose that x=z, 0≤0+0 and the result follows.
Suppose x>z,
TOPOLOGICAL CONCEPT IN DIRECTED METRIC SPACE
Definition 4: Open, closed balls and boundary points.
Let (X,dd) be a directed-metric space and a a fixed point, r a positive real number (r>a) we define an open ball.
(I) | B(a,r)={x∈X|dd(a,x)<r} and |
(ii) | A closed ball B(a,r)={x∈X|dd(a,x)≤r} and the boundary points as |
(iii) | Bp={x∈X|dd(a,x)=r}. |
We see that relationship Bp=B(a,r)-B(a,r) is preserved.
Definition 5: A sunset Y of a directed-metric space (X,dd) is said to be dense in X if the closure of Y is the same as X.
Definition 6: A directed-metric space (X,dd) is said to be seperable if it contains a countable dense subsets.
Observe that, in R1
Remark 1: An open ball is an open set
Proof:
From the definition, we need to show that every point of the ball has a neighborhood which is contained in it i.e. Let B(a;r) be an open ball, we now have to show that B(x,∈)⊂B(a;r)
Let y∈B(x,∈), then dd(x,y)<∈-r-dd(a,x) and from triangle inequality. dd(a,y)∈dd(a,x)+dd(x,y)
<dd(a,x)+r-dd(a,x)<r. This show that y∈B(a;r) and hence B(x,∈)⊂B(a,r) and the result follows.
Alternatively: We know that an open ball is defined as B(a;r)=(-∞,a+r) and hence for each point q∈B(a;r) we may choose Sq=B(a;r) which is contained in the open ball.
Remark 2: A closed ball is neither close nor open
Proof:
B(a;r) is not open since a+r is not an interior point of (-∞,a+r). Also if we set a real number q=a+r+0, qóB(a;r) we know that B(a;r)∩(G\{q})≠φ for any open ball G, thus q is an accumulation point of B(a;r). Hence B(a;r) is not close and the result follows.
Remark 3: Z is dense in R
Proof:
If suffices to show that Z∩(B(a;r)-{q})≠φq∈R
Suppose we assume on the contrary that Z1(B(a;r))-{q})=φ
We have that z∉(B(a;r)-{q}),q∈R
i.e. (B(a;r)-{q}) has no integers which is a contradiction.
Remark 4: Q is dense in R
Proof:
Since Z⊂Q and Z∩(B(a;r)-{q})≠φq∈R
definitely Q1(B(a; r)-{q})≠φq∈R and Q is dense in R
Remark 5: R is separable in Z and Q
Proof:
The set R is separable since it contains countable dense subsets Z and Q.