On one hand, in order to be a Euclidean a space should be equipped with a concept of "an angle". And the angle on a line has only two values (0 and 180 degrees), while in higher dimensional spaces it can have a continuum of values. So, is it Euclidean? A simple test would be: are there theorems proven for a Euclidean space of arbitrary many dimensions, that fail in case of a 1D line?
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No restriction; see definition of Euclidean space – Mauro ALLEGRANZA Nov 08 '23 at 14:26
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See e.g. Jerry Shurman, Calculus and Analysis in Euclidean Space (Springer, 2019), page 23: "Let $n$ be a positive integer. The set of all ordered n-tuples of real numbers, $\mathbb R^n = {( x_1, . . . , x_n) ∶ xi ∈ \mathbb R, i = 1, . . . ,n }$, constitutes n-dimensional Euclidean space. When $n = 1$, the parentheses and subscript in the notation $(x_1)$ are superfluous, so we simply view the elements of $\mathbb R^1$ as real numbers ..." – Mauro ALLEGRANZA Nov 08 '23 at 14:28
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A Euclidean space is a vector space $\mathbb R^n$ equipped with a scalar product that is positive definite. By putting $\langle x,y\rangle = xy$ in $\mathbb R$, you get your scalar product, so $(\mathbb R,\langle{}\cdot{},{}\cdot{}\rangle)$ is definitely Euclidean.
Gibbs
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$\mathbb{R}$ is indeed a Euclidean space.
Angles between nonzero vectors $x$ and $y$ in Euclidean space are defined by $$\theta = \arccos\dfrac{x\cdot y}{\| x\|\|y\|}\in [0,\pi].$$ In $\mathbb{R}$, $x\cdot y = xy$ and $\|x\|=|x|$, so necessarily $\dfrac{x\cdot y}{\| x\|\|y\|} = \dfrac{xy}{|xy|}=\pm 1$ and therefore these angles are necessarily either $\theta=0$ or $\theta=\pi$ as you mention. Nothing wrong with that.
MPW
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