Connected Open Subset of Euclidean Space is Path-Connected

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Theorem

Let $\R^n$ be a Euclidean $n$-space.

Let $U$ be a connected open subset of $\R^n$.


Then $U$ is path-connected.


Proof

Let $a \in U$.

Let $H \subseteq U$ be the subset of points in $U$ which can be joined to $a$ by a path in $U$.

Let $K = U \setminus H$.


Let $x \in H$.

Then:

$\exists \epsilon > 0: B_\epsilon \left({x}\right) \subseteq U$

where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$.

Given any $y \in B_\epsilon \left({x}\right)$, there is a (straight line) path $g$ in $B_\epsilon \left({x}\right) \subseteq U$ connecting $x$ and $y$.

But since $x \in H$, there is a path $f$ in $U$ joining $a$ to $x$.

From Joining Paths makes Another Path, traversing $f$, and then $g$ forms a path from $a$ to $y$.

It follows that $y \in H$, and therefore $B_\epsilon \left({x}\right) \subseteq H$.

Thus $H$ is open.


By a similar argument, $K$ is also shown to be open:

If $x \in K$, then $B_\epsilon \left({x}\right) \subseteq U$ for some $\epsilon > 0$.

If any point in $B_\epsilon \left({x}\right)$ can be joined to $a$ by a path in $U$, then so could $x$.


It is clear that $H \cap K = \varnothing$ and $H \cup K = U$ by definition of set difference.

As, trivially, $a \in H$, we have $H \ne \varnothing$.

Knowing that $U$ is connected, it follows that $K = \varnothing$, and $H = U$.

Hence the result.

$\blacksquare$


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