Euclidean Space is Path-Connected

Theorem
Let $\R^n$ be the $n$-dimensional Euclidean space for $n \in \N$ a natural number.

Then $\R^n$ is path-connected.

Proof
Let $\mathbf x, \mathbf y \in \R^n$ be arbitrary points of $\R^n$.

Define $l: \left[{0 \,.\,.\, 1}\right] \to \R^n$ by:


 * $l \left({t}\right) = \left({1 - t}\right) \mathbf x + t \mathbf y$

Then $l \left({0}\right) = 1 \mathbf x + 0 \mathbf y = \mathbf x$, whereas $l \left({1}\right) = 0 \mathbf x + 1 \mathbf y = \mathbf y$.

Finally, it remains to show that $l$ is continuous.

Fix $\epsilon > 0$ and suppose that $t, t' \in \left[{0 \,.\,.\, 1}\right]$ are such that $\left\vert{t - t'}\right\vert < \dfrac {\epsilon} {1 + \left\Vert{\mathbf x}\right\Vert + \left\Vert{\mathbf y}\right\Vert}$.

Then:

We can now estimate the norm of this last expression:

Since $\epsilon$ was arbitrary, we conclude that $l$ is continuous.

Therefore, it forms a path from $\mathbf x$ to $\mathbf y$.

Since $\mathbf x$ and $\mathbf y$ were arbitrary, it follows that $\R^n$ is path-connected.