Euclidean Space is Path-Connected

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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.

$\blacksquare$


Sources