Proof. By definition, the absolute value on \(\mathbb {R}\) is non-negative. Therefore \(d(x,y)=|x-y|\geq 0\). Since \(|x-y|=0\) if and only if \(x=y\), it follows that \(d(x,y)=0\) if and only if \(x=y\). To verify the third axiom, we have \[d(x,y)=|x-y|=|-(x-y)|=|y-x|=d(y,x).\] To verify the triangle inequality, suppose that \(x,y,z\in \mathbb {R}\). Then \[d(x,z)=|x-z|=|x-y+y-z|\leq |x-y|+|y-z|=d(x,y)+d(y,z).\] Thus, \((\mathbb {R},d)\) is a metric space. □

Proof.

i

Since \(d_1\) is the sum of the absolute value function on \(\mathbb {R}^2\), it is non-negative and \(d_1=|x_1-y_1|+|x_2-y_2| \geq 0\) for all \(x,y\in \mathbb {R}^2\).
To verify that \(d_1(x,y)=0\) if and only if \(x=y\), we have that \(|x_1-y_1|+|x_2-y_2|=0\) if and only if \(x_1=y_1\) and \(x_2=y_2\). This implies that \(d_1(x,y)=0\) if and only if \(x=y\).
We show that \(d_1(x,y)=d_1(y,x)\) for all \(x,y\in \mathbb {R}^2\). We have that \begin {align*} d_1(x,y) & =|x_1-y_1|+|x_2-y_2|=|-(y_1-x_1)|+|-(y_2-x_2)|\\ & =|y_1-x_1|+|y_2-x_2|\\ & = d_1(y,x). \end {align*}

Finally, we prove that the triangle inequality holds. Let \(x=(x_1,x_2), y=(y_1,y_2), z=(z_1,z_2)\in \mathbb {R}^2\). Then \begin {align*} d_1(x,z) &=|x_1-z_1|+|x_2-z_2|\\ &=|x_1-y_1+y_1-z_1|+|x_2-y_2+y_2-z_2|\\ &\leq |x_1-y_1|+|y_1-z_1|+|x_2-y_2|+|y_2-z_2|\\ &=(|x_1-y_1|+|x_2-y_2|) +(|y_1-z_1|+|y_2-z_2|)\\ &=d_1(x,y)+d_1(y,z). \end {align*}

ii

For any \(x=(x_1,x_2), y=(y_1,y_2)\in \mathbb {R}^2\). We have that \(\sqrt {(x_1-y_1)^2+(x_2-y_2)^2}\geq 0\) so that \(d_2(x,y)\geq 0\).
Also, \((x_1-y_1)^2+(x_2-y_2)^2=0\) if and only if \(x_1-y_1=0\) and \(x_2-y_2=0\). In other words \((x_1-y_1)^2+(x_2-y_2)^2=0\) if and only if \(x_1=y_1\) and \(x_2=y_2\). This means that \(d_2(x,y)=0\) if and only if \(x=y\).
To show that \(d_2(x,y)=d_2(y,x)\), we just observe that \((x_1-y_1)^2=(-(-x_1+y_1))^2=(y_1-x_1)^2\) and \((x_2-y_2)^2=(-(-x_2+y_2))^2=(y_2-x_2)^2\).This implies that

\begin {align*} d_2(x,y) &=\sqrt {(x_1-y_1)^2+(x_2-y_2)^2}\\ &=\sqrt {(y_1-x_1)^2+(y_2-x_2)^2}\\ &=d_2(y,x). \end {align*}

To verify the triangle inequality, let \(x=(x_1,x_2), (y_1,y_2), (z_1,z_2)\in \mathbb {R}^2\). Then \begin {align*} d_2(x,z) &=\sqrt {(x_1-z_1)^2+(x_2-z_2)^2}\\ &=\sqrt {(x_1-y_1+y_1-z_1)^2+(x_2-y_2+y_2-z_2)^2}\\ &=\sqrt {(a_1+b_1)^2+(a_2+b_2)^2} \end {align*}

where \(a_1=x_1-y_1\), \(b_1=y_1-z_1\), \(a_2=x_2-y_2\), \(b_2=y_2-z_2\). By applying Minkowski’s inequality, we obtain that \[\sqrt {(a_1+b_1)^2+(a_2+b_2)^2}\leq \sqrt {a^2_1 +a_2^2}+\sqrt {b_1^2+b_2^2}\] Substituting for \(a_1,a_2,b_1,b_2\) we obtain that \[d_2(x,z)\leq d_2(x,y)+d_2(y,z).\]

iii

If \(x=(x_1,x_2), y=(y_1,y_2) \in \mathbb {R}^2\), we see that \(|x_1-y_1|\geq 0\) and \(|x_2-y_2|\geq 0\) so that \(d_3(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}\geq 0\).
Also, \(\max \{|x_1-y_1|,|x_2-y_2|\}=0\) if and only if \(|x_1-y_1|=0\) and \(|x_2-y_2|=0\), which means that \(\max \{|x_1-y_1|,|x_2-y_2|\}=0\) if and only if \(x_1=y_1\) and \(x_2=y_2\). Thus \(d_3=0\) if and only if \(x=y\).
We show that \(d_3(x,y)=d_3(y,x)\). Since \[\max \{|x_1-y_1|,|x_2-y_2|\}=\max \{|y_1-x_1|,|y_2-x_2|\}\] we have that \(d_3(x,y)=d_3(y,x)\).
To show the triangle inequality holds, let \(x=(x_1,x_2), y=(y_1,y_2), z=(z_1,z_2)\in \mathbb {R}^2\). Then \begin {align*} d_3(x,z) &=\max \{|x_1-z_1|,|x_2-z_2|\}\\ &=\max \{|x_1-y_1+y_1-z_1|,|x_2-y_2+y_2-z_2|\}\\ &\leq \max \{|x_1-y_1|+|y_1-z_1|,|x_2-y_2|+|y_2-z_2|\}\\ &\leq \max \{|x_1-y_1|,|x_2-y_2|\} +\max \{|y_1-z_1|,|y_2-z_2|\}\\ &=d_3(x,y) +d_3(y,z). \end {align*}

Proof. Let \(x=(x_n),y=(y_n)\in \ell \). Since \(|x_n-y_n|\geq 0\) for all \(n\in \mathbb {N}\), we obtain that \(d(x,y)=\sum \limits ^{\infty }_{n=1} |x_n-y_n|\geq 0\). Also since \(|x_n-y_n|=0\) if and only if \(x_n=y_n\), it follows that \(d(x,y)=0\) if and only if \(x=y\).
From \(|x_n-y_n|=|-(-x_n+y_n)|=|y_n-x_n|\), we get that \(d(x,y)=d(y,x).\)
To verify the triangle inequality, let \(x=(x_n), y=(y_n), z=(z_n)\in \ell \). From properties of the absolute value on the set of real numbers, we have that \begin {align*} \sum ^n_{k=1} |x_k-z_k| &=\sum _{k=1}^n |x_k-y_k+y_k-z_k|\\ &\leq \sum ^n_{k=1} |x_k-y_k| +\sum ^n_{k=1} |y_k-z_k|. \end {align*}

Taking limits on both sides as \(n\rightarrow \infty \), we obtain that \[\sum \limits ^{\infty }_{k=1} |x_k-z_k|\leq \sum \limits _{k=1}^{\infty } |x_k-y_k|+\sum \limits ^{\infty }_{k=1} |y_k-z_k|.\] Hence \(d(x,z)\leq d(x,y)+d(y,z)\). □

Proof. Let \(x=(x_n), y=(y_n)\in \mathbb {R}^{\mathbb {N}}\). Using the properties of the absolute value on the set of real numbers, the properties \(d(x,y)\geq 0\), \(d(x,y)=0\) if and only if \(x=y\) and \(d(x,y)=d(y,x)\) are easily established.
Let \(x=(x_n), y=(y_n), z=(z_n)\in \mathbb {R}^{\mathbb {N}}\). To verify that the triangle inequality holds we prove the inequality \[\dfrac {|a+b|}{1+|a+b|}\leq \dfrac {|a|}{1+|a|}+\dfrac {|b|}{1+|b|}.\] From the fact that \(\frac {1}{|a|+|b|}<\frac {1}{|a+b|}\), it follows that \begin {align*} \frac {|a+b|}{1+|a+b|} &=\frac {1}{1+\dfrac {1}{|a+b|}}\\ &\leq \frac {1}{1+\dfrac {1}{|a|+|b|}}=\frac {|a|+|b|}{1+|a|+|b|}\\\\ &=\frac {|a|}{1+|a|+|b|}+\frac {|b|}{1+|a|+|b|}\\\\ &\leq \frac {|a|}{1+|a|}+\frac {|b|}{1+|b|} \end {align*}

Therefore, \(\frac {|a+b|}{1+|a+b|}\leq \frac {|a|}{1+|a|}+\frac {b}{1+b}\), if we take \(a=x_n-y_n\) and \(b=y_n-z_n\) in this inequality, we obtain that \[\frac {|x_n-z_n|}{1+|x_n-z_n|}\leq \frac {|x_n-y_n|}{1+|x_n-y_n|}+\frac {|y_n-z_n|}{1+|y_n-z_n|}\] from which it follows that \(d(x,z)\leq d(x,y)+d(y,z)\). □

Proof. By definition \(d(x,y)\geq 0\), \(d(x,y)=0\), if and only if \(x=y\) and \(d(x,y)=d(y,x)\) for all \(x,y\in X\).
To show that the triangle inequality holds, let \(x, y,z\in X\). if \(x=z\), then \(d(x,z)=0\) and \(d(x,z)\leq d(x,y)+d(y,z)\) holds. Suppose that \(x\neq z\). If \(y=z\), then \(x\neq y\), so that \(d(x,z)=1\), \(d(x,y)=1\) and \(d(y,z)=0\). Thus \(d(x,z)\leq d(x,y)+d(y,z)\). If \(y\neq z\) then \(d(x,z)=1\), \(d(y,z)=1\), which imply that \(d(x,z)\leq d(x,y)+d(y,z)\). □