Exercise 1.1.3

(a)

take: $X=ℝ$ and $d(x,y)=1$ for all $x,y\in ℝ$. Clearly, property (a) is false. Also (b), and (c) are clearly true, and the triangle inequality is also satisfied as $1\le 1+1=2$

(b)

take: $X=ℝ$ and $d(x,y)=0$ for all $x,y\in ℝ$. Here, property (b) is false. And it is clear that properties (a), (c), and (d) are true.

(c)

(d)

For (d) take: $X=\{1,2,3\}$ and define $d$ as follows.

$$\begin{array}{ccc}\hfill & d(1,1)=0,d(1,2)=3,d(1,3)=1\hfill & \hfill \\ \hfill & d(2,1)=3,d(2,2)=0,d(2,3)=1\hfill \\ \hfill & d(3,1)=1,d(3,2)=1,d(3,3)=0\hfill \end{array}$$

By definition properties (a), (b), and (c) are true. However, the triangle inequality does not hold: Consider the following: $d(1,2)\pm ̸d(1,3)+d(3,2)$ as $3\nleqq 1+1=2$.

(a)

Let $X=ℝ,d(x,y)=1+|x-y|$. Property (a) is obviously false. And again (b) and (c) are clearly true. And the triangle inequality is also satisfied as follows: $d(x,z)=1+|x-z|$ but $|x-z|\le |x-y|+|y-z|$ so $d(x,z)\le$ $1+|x-y|+|y-z|\le 1+|x-y|+1+|y-z|=d(x,y)+d(y,z)$. Thus the triangle inequality is satisfied.

(b)

Let $X={ℝ}^2$ and $$d\left((x,y),(x^{\prime },y^{\prime })\right)=\left|y-y^{\prime }\right|$$

Property (a) is definitely true. (b) is false as if we take $x\ne x^{\prime }$ and $y=y^{\prime }$ then we get

$$d\left((x,y),(x^{\prime },y^{\prime })\right)=0.$$

(c) is true since

$$\left|y-y^{\prime }\right|=\left|y^{\prime }-y\right|.$$

(d) is also true since

$$d\left((x,x^{\prime }),(z,z^{\prime })\right)=\left|x^{\prime }-z^{\prime }\right|\le \left|x^{\prime }-y^{\prime }\right|+\left|y^{\prime }-z^{\prime }\right|=d\left((x,x^{\prime }),(y,y^{\prime })\right)+d\left((y,y^{\prime }),(z,z^{\prime })\right).$$

(c)

Take $X=ℝ$ and $$d(x,y)=\{\begin{array}{cc}0\hfill & \text{ if }x=y\hfill \\ a\hfill & \text{ if }x<y\hfill \\ b\hfill & \text{ if }x>y\hfill \end{array}$$

Where $a,b>0$ and $a\ne b$. First, without loss of generality, assume that $a>b$. By definition (a) and (b) are true. Also, (c) is false, consider: $d(2,4)=a$, but $d(4,2)=b$. And for (d) let us do the following: Let $x,y,z\in X$.

Case 1: $(x\ne y\ne z)$ Without loss of generality assume that $x<y<z$. Thus we have: $d(x,y)=a,d(x,z)=a,d(y,x)=b,d(y,z)=a,d(z,x)=b,d(z,y)=b$. So we get:

$$\begin{array}{ccc}\hfill & d(x,y)=a\le a+b=d(x,z)+d(z,y)\hfill & \hfill \\ \hfill & d(x,z)=a\le a+a=d(x,y)+d(y,z)\hfill \\ \hfill & d(y,x)=b\le a+b=d(y,z)+d(z,x)\hfill \\ \hfill & d(y,z)=a\le b+a=d(y,x)+d(x,z)\hfill \\ \hfill & d(z,x)=b\le b+b=d(z,y)+d(y,x)\hfill \\ \hfill & d(z,y)=b\le b+a=d(z,x)+d(x,y)\hfill \end{array}$$

In each case the triangle inequality holds.

Case 2: $(x=y$, and $x,y\ne z)$ Suppose that $x,y<z.$ Each of these assumptions is without loss of generality since we will consider every case below (and of course the letters can be switched around). So we have: $d(x,y)=0,d(x,z)=a,d(y,x)=0,d(y,z)=a,d(z,x)=b,d(z,y)=b$. And thus we get the following:

$$\begin{array}{ccc}\hfill & d(x,y)=0\le a+b=d(x,z)+d(z,y)\hfill & \hfill \\ \hfill & d(x,z)=a\le 0+a=d(x,y)+d(y,z)\hfill \\ \hfill & d(y,x)=0\le a+b=d(y,z)+d(z,x)\hfill \\ \hfill & d(y,z)=a\le 0+a=d(y,x)+d(x,z)\hfill \\ \hfill & d(z,x)=b\le b+0=d(z,y)+d(y,x)\hfill \\ \hfill & d(z,y)=b\le b+0=d(z,x)+d(x,y)\hfill \end{array}$$

Again the triangle inequality holds for each case (a similar argument shows for $z<x=y$ ).

Case 3: $(x=y=z)$

Trivial since each distance will be $0$, and of course $0\le 0+0$.

In each case, we get that the triangle inequality, property $(d)$, is true.

(d)

Let $X=ℝ,d(x,y)=|x-y{\text{|}}^2$. Clearly we get that properties (a),(b), and (c) are true. For (d) consider $x=1,y=4,z=5$. Then $$d(x,z)=|1-5{\text{|}}^2=|-4{\text{|}}^2=16,$$

$$d(x,y)=|1-4{\text{|}}^2=|-3{\text{|}}^2=9,$$

and

$$d(y,z)=|4-5{\text{|}}^2=|-1{\text{|}}^2=1,$$

but then we get that $d(x,z)\nless d(x,y)+d(y,z)$ as $16\nless 9+1=10$.

(d) Set $X=ℝ$ and

By definition (a), (b) and (c) are true. For (d) consider: $x=1,y=4,z=5$ then $d(x,z)\le d(x,y)+d(y,z)$ as $e^4\nleqq e^3+e$.