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 $l: \closedint 0 1 \to \R^n$ by:

$\map l t = \paren {1 - t} \mathbf x + t \mathbf y$

Then $\map l 0 = 1 \mathbf x + 0 \mathbf y = \mathbf x$, whereas $\map l 1 = 0 \mathbf x + 1 \mathbf y = \mathbf y$.


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

Fix $\epsilon > 0$ and suppose that $t, t' \in \closedint 0 1$ are such that $\size {t - t'} < \dfrac {\epsilon} {1 + \norm {\mathbf x} + \norm {\mathbf y} }$.

Then:

\(\ds \map l t - \map l {t'}\) \(=\) \(\ds \paren {1 - t} \mathbf x + t \mathbf y - \paren {\paren {1 - t'} \mathbf x + t' \mathbf y}\)
\(\ds \) \(=\) \(\ds \paren {\paren {1 - t} - \paren {1 - t'} } \mathbf x + \paren {t - t'} \mathbf y\)
\(\ds \) \(=\) \(\ds \paren {t' - t} \mathbf x + \paren {t - t'} \mathbf y\)

We can now estimate the norm of this last expression:

\(\ds \norm {\paren {t' - t} \mathbf x + \paren {t - t'} \mathbf y}\) \(\le\) \(\ds \norm {\paren {t' - t} \mathbf x} + \norm {\paren {t - t'} \mathbf y}\) Triangle Inequality for Vectors in Euclidean Space
\(\ds \) \(=\) \(\ds \size {t' - t} \norm {\mathbf x} + \size {t - t'} \norm {\mathbf y}\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \size {t' - t} \paren {\norm {\mathbf x} + \norm {\mathbf y} }\)
\(\ds \) \(<\) \(\ds \frac \epsilon {1 + \norm {\mathbf x} + \norm {\mathbf y} } \paren {\norm {\mathbf x} + \norm {\mathbf y} }\)
\(\ds \) \(<\) \(\ds \epsilon\)

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

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