# 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:

 $\displaystyle l \left({t}\right) - l \left({t'}\right)$ $=$ $\displaystyle \left({1 - t}\right) \mathbf x + t \mathbf y - \left({\left({1 - t'}\right) \mathbf x + t' \mathbf y}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\left({1 - t}\right) - \left({1 - t'}\right)}\right) \mathbf x + \left({t - t'}\right) \mathbf y$ $\displaystyle$ $=$ $\displaystyle \left({t' - t}\right) \mathbf x + \left({t - t'}\right) \mathbf y$

We can now estimate the norm of this last expression:

 $\displaystyle \left\Vert{\left({t' - t}\right) \mathbf x + \left({t - t'}\right) \mathbf y}\right\Vert$ $\le$ $\displaystyle \left\Vert{\left({t' - t}\right) \mathbf x}\right\Vert + \left\Vert{\left({t - t'}\right) \mathbf y}\right\Vert$ Triangle Inequality $\displaystyle$ $=$ $\displaystyle \left\vert{t' - t}\right\vert \left\Vert{\mathbf x}\right\Vert + \left\vert{t - t'}\right\vert \left\Vert{\mathbf y}\right\Vert$ Axiom $(N2)$ for norms $\displaystyle$ $=$ $\displaystyle \left\vert{t' - t}\right\vert \left({\left\Vert{\mathbf x}\right\Vert + \left\Vert{\mathbf y}\right\Vert}\right)$ $\displaystyle$ $<$ $\displaystyle \frac {\epsilon} {1 + \left\Vert{\mathbf x}\right\Vert + \left\Vert{\mathbf y}\right\Vert} \left({\left\Vert{\mathbf x}\right\Vert + \left\Vert{\mathbf y}\right\Vert}\right)$ $\displaystyle$ $<$ $\displaystyle \epsilon$

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.

$\blacksquare$