Every convex subset of ${\mathbb{R}}^n$, in particular every ball and ${\mathbb{R}}^n$ itself, is path-connected and hence connected

Let $n\in \mathbb{N}$ with $n\ge 1$ and give ${\mathbb{R}}^n$ the product topology, which is the metric topology of $d_{\mathrm{\infty }}$ (For $n\ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_{\mathrm{\infty }}$ on ${\mathbb{R}}^n$, and hence also of $d_1$ and $d_2$, so ${\mathbb{R}}^n$ as a product and ${\mathbb{R}}^n$ as a metric space are one space, ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement). Recall that ${\mathbb{R}}^n$ is a real vector space under coordinatewise operations (Vector space over a field, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

A subset $C\subseteq {\mathbb{R}}^n$ is convex when

$x,y\in C\text{ and }t\in [0,1]⟹(1-t)x+ty\in C$

(Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Then:

1. Every convex $C\subseteq {\mathbb{R}}^n$ is path-connected (Paths, path-connected spaces and path components), hence connected (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, Every path-connected space is connected, and every path component lies inside a component).
2. Every ball is convex, in each of the norms $\parallel \cdot {\parallel }_1$, $\parallel \cdot {\parallel }_2$, $\parallel \cdot {\parallel }_{\mathrm{\infty }}$ (The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$, Open ball, closed ball and sphere in a metric space); so every ball of ${\mathbb{R}}^n$ is path-connected and connected.
3. ${\mathbb{R}}^n$ itself is convex, hence path-connected and connected, and so is every half-space $\{x:x_k\le c\}$, and every box ${\prod }_{k<n}{J}_k$ with each $J_k$ an order-convex subset of $\mathbb{R}$.

• The path is the straight segment and nothing more is needed. Convexity is exactly the hypothesis that the segment between two points of the set stays in the set, so the definition of the path writes itself; the only work is that the segment is a continuous map, which is [A1] plus the continuity of an affine map of one real variable.

• Convexity is far from necessary. A circle is path-connected and not convex, and so is any set obtained from a convex one by bending it. Nothing above asserts a converse.

• The hypothesis $n\ge 1$ comes from $d_{\mathrm{\infty }}$. $\parallel \cdot {\parallel }_{\mathrm{\infty }}$ is a maximum over $n$ terms and is undefined at $n=0$ (The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$, ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it). At $n=0$ the product is a one-point space, which is path-connected outright.