# Proof

The proof consists of three steps:

### $$(1)$$ We show that $$\mathbb R^n$$ is a vector space over $$\mathbb R$$.

In order to show that the cartesian product of the $$n$$ sets of real numbers. $\mathbb R^n:=\underbrace{\mathbb R\times \mathbb R\times\cdots\times \mathbb R}_{n\text{ times}}.$

constitutes a vector space over $$\mathbb R$$, we have to verify the following rules:

1. $$(\mathbb R^n, + )$$ is an Abelian group.
2. For $$\alpha,\beta\in \mathbb R$$ and $$x,y\in \mathbb R^n$$ the following axioms of scalar multiplication hold:

# 1. $$1\cdot x=x$$.

With the $\text{vector addition "+"}:=\pmatrix{ x_{1}\cr x_{2}\cr \vdots \cr x_{n} \cr }+\pmatrix{ y_{1}\cr y_{2}\cr \vdots \cr y_{n} \cr }=\pmatrix{ x_{1}+y_1\cr x_{2}+y_2\cr \vdots \cr x_{n}+y_n \cr }$ $\text{and scalar multiplication "}\cdot\text{"}:=\alpha\cdot\pmatrix{ x_{1}\cr x_{2}\cr \vdots \cr x_{n} \cr }=\pmatrix{ \alpha\cdot x_{1}\cr \alpha\cdot x_{2}\cr \vdots \cr \alpha\cdot x_{n} \cr }$

in $$\mathbb R^n$$ these rules follow immediately from the fact that $$(\mathbb R,\cdot, + )$$ is a field.

### $$(2)$$ We show that the Euclidean norm $$\|x\|$$ of a vector $$x\in\mathbb R^n$$ is a norm. Accordingly, the pair $$(\mathbb R^n,\|~\|)$$ constitutes a normed vector space.

The Euclidean norm of a vector $$x\in\mathbb R^n$$ is defined as the square root of the dot product of this vector with itself, i.e.:

$||x||:=\sqrt{\langle x,x\rangle}=\sqrt{x_1^2+\ldots+x_n^2}.$

In order to show that this definition constitutes a norm, we have to prove the following properties:

$$(i)$$ $$\|x\|=0$$ if and only if $$x=0$$. Clearly, if $$x=0$$, i.e. if all coordinates of the vector $$x_i=0$$ for $$1\le i \le n$$, then $$||x||=0$$. Conversely, if $$||x||=0$$, all coordinates must equal $$0$$, because $$\sqrt{\alpha^2} > 0$$ for all $$\alpha > 0$$.

$$(ii)$$ $$\|\lambda x\|=|\lambda|\cdot\|x\|$$ for all $$\lambda\in\mathbb R$$. $\begin{array}{rclcl} \|\lambda x\|&=&\sqrt{(\lambda\cdot x_1)^2+\ldots+(\lambda\cdot x_n)^2}&=&\\ &=&\sqrt{\lambda^2\cdot (x_1^2+\ldots+x_n)^2}&=&\\ &=&|\lambda|\cdot\sqrt{\cdot (x_1^2+\ldots+x_n)^2}&=&|\lambda|\cdot\|x\|\\ \end{array}$

$$(iii)$$ $$\|x+y\|\le \|x\|+\|y\|$$ for all $$x,y\in \mathbb R^n$$. According to the triangle inequality, for all pairs of coordinates $$x_i,y_i$$ , $$1\le i\le n$$ we have $\begin{array}{rclcl} |x_i+y_i|&=&\sqrt{(x_i+y_i)^2}&\le&\\ &\le&\sqrt{x_i^2}+\sqrt{y_i^2}&=&\\ &=&|x_i|+|y_i|. \end{array}$Therefore, we have $\begin{array}{rclcl} \|x+y\|&=&\sqrt{(x_1+y_1)^2+\ldots+(x_n+y_n)^2}&\le&\\ &\le&\sqrt{x_1^2+\ldots+x_n^2}+\sqrt{y_1^2+\ldots+y_n^2}&=&\|x\|+\|y\|.\end{array}$

It follows, that the pair $$(\mathbb R^n,||~||)$$ constitutes a normed vector space.

### $$(3)$$ We show that the Euclidean distance of two vectors $$x,y\in\mathbb R^n$$ is a distance. Accordingly, the pair $$(\mathbb R^n,d)$$ constitutes a metric space.

According to $$(1)$$, $$(\mathbb R^n,\|~\|)$$ is a normed vector space. Thus, according to the corresponding proposition, the norm of the difference of two vectors $$x,y\in\mathbb R^n$$ defines a distance:

$d(x,y):=||x-y||=\sqrt{\langle x-y,x-y\rangle}=\sqrt{(x_1-y_1)^2+\ldots+(x_n-y_n)^2}.$

Thus, the pair $$(\mathbb R^n,d)$$ constitutes a metric space.

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### References

#### Bibliography

1. Forster Otto: "Analysis 2, Differentialrechnung im $$\mathbb R^n$$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984