[proofplan] The argument has two ingredients. First, because $U$ is open in $\mathbb{R}^n$, every tangent space $T_pU$ is canonically $\mathbb{R}^n$ (no chart choice is required), so the set of pointwise orientations $\mathrm{Or}_p$ is canonically the two-element set $\{+, -\}$, independent of $p$. An orientation on $U$ therefore reduces to a locally constant function $\mathcal{O} : U \to \{+, -\}$. Second, a locally constant function on a connected space is constant — this is the standard topological fact that the only continuous maps from a connected space to a discrete two-point space are the two constants. Restricting $\mathcal{O}$ to each connected component yields a constant function, giving claim (1); when $U$ itself is connected, the only two such constants are $+$ and $-$, giving claim (2). [/proofplan]

[step:Identify the pointwise orientation set $\mathrm{Or}_p$ canonically with $\{+,-\}$ for all $p \in U$]For $p \in U$, the tangent space $T_pU$ is canonically identified with $\mathbb{R}^n$ via the translation isomorphism \begin{align*} \tau_p : \mathbb{R}^n &\to T_pU, \\ v &\mapsto \tfrac{d}{dt}\Big|_{t=0}(p + tv). \end{align*} An orientation of an $n$-dimensional real [vector space](/page/Vector%20Space) $V$ is an equivalence class of ordered bases under the relation: $(b_1, \dots, b_n) \sim (b_1', \dots, b_n')$ iff the change-of-basis matrix has positive determinant. Since $\mathrm{GL}(n, \mathbb{R})$ has exactly two connected components, distinguished by the sign of the determinant, the set of orientations of any $n$-dimensional real [vector space](/page/Vector%20Space) has exactly two elements. Applying this to $T_pU$ via $\tau_p$, we declare $+_p \in \mathrm{Or}_p$ to be the orientation represented by $(\tau_p(e_1), \dots, \tau_p(e_n))$, where $(e_1, \dots, e_n)$ is the standard basis of $\mathbb{R}^n$, and $-_p$ to be its opposite. The labelling $\mathrm{Or}_p = \{+_p, -_p\}$ is canonical because $\tau_p$ does not depend on any choice of chart. Suppressing the subscript, we write $\mathrm{Or}_p = \{+, -\}$ uniformly across $p \in U$, and the disjoint union $\bigsqcup_{p \in U} \mathrm{Or}_p$ is canonically identified with $U \times \{+, -\}$.[/step]

[guided]The point of this step is to fix once and for all what we mean by "an orientation at $p$" for $p \in U \subseteq \mathbb{R}^n$. For a general smooth manifold $M$, the tangent space $T_pM$ has no canonical identification with $\mathbb{R}^n$ — one must choose a chart, and different charts give different identifications differing by an element of $\mathrm{GL}(n, \mathbb{R})$. For an [open set](/page/Open%20Set) $U \subseteq \mathbb{R}^n$, however, there *is* a canonical identification: \begin{align*} \tau_p : \mathbb{R}^n &\to T_pU, \\ v &\mapsto \tfrac{d}{dt}\Big|_{t=0}(p + tv). \end{align*} This map is the differential of the inclusion $U \hookrightarrow \mathbb{R}^n$ at $p$ (after identifying $T_p\mathbb{R}^n$ with $\mathbb{R}^n$ via the same construction). It is a linear isomorphism that depends only on the affine structure of $\mathbb{R}^n$, not on any chart choice. We now recall what an orientation on a finite-dimensional real [vector space](/page/Vector%20Space) is. Let $V$ be an $n$-dimensional real [vector space](/page/Vector%20Space). Two ordered bases $(b_1, \dots, b_n)$ and $(b_1', \dots, b_n')$ of $V$ are *equivalently oriented* if the unique [linear map](/page/Linear%20Map) sending $b_i \mapsto b_i'$ has positive determinant. This is an [equivalence relation](/page/Equivalence%20Relation); an *orientation* of $V$ is an equivalence class. The set of orientations has exactly two elements because $\mathrm{GL}(n, \mathbb{R})$ has exactly two connected components — those with $\det > 0$ and those with $\det < 0$. Applying this to $V = T_pU$ via the canonical isomorphism $\tau_p$, we define: - $+_p$ = orientation represented by $(\tau_p(e_1), \dots, \tau_p(e_n))$ (the "standard" orientation at $p$); - $-_p$ = the opposite orientation. Because $\tau_p$ is canonical, the labels $+$ and $-$ make sense uniformly in $p$. This is the key feature distinguishing open subsets of $\mathbb{R}^n$ from general manifolds: there is a globally defined "standard orientation" $+$ on $U$ a priori, without any further work. Consequently, the disjoint union of pointwise orientation sets is canonically a product: \begin{align*} \bigsqcup_{p \in U} \mathrm{Or}_p \;\cong\; U \times \{+, -\}. \end{align*}[/guided]

[step:Recast an orientation on $U$ as a locally constant map $\mathcal{O}: U \to \{+,-\}$] By the identification in the previous step, specifying an orientation $\mathcal{O}(p) \in \mathrm{Or}_p$ for each $p \in U$ is the same as specifying a function $\mathcal{O} : U \to \{+, -\}$. The defining condition of an orientation on $U$ — local constancy — translates verbatim: $\mathcal{O}$ is an orientation iff for every $p \in U$ there exists an open neighbourhood $V \subseteq U$ of $p$ on which $\mathcal{O}$ is constant. Equipping $\{+, -\}$ with the discrete topology, local constancy of $\mathcal{O}$ is equivalent to continuity of \begin{align*} \mathcal{O} : U &\to \{+, -\}_{\mathrm{disc}}. \end{align*} Indeed, both conditions assert that every point of $U$ has a neighbourhood mapped into a single point of $\{+, -\}$; for the discrete topology, the only neighbourhoods of $\pm \in \{+, -\}_{\mathrm{disc}}$ that we need are the singletons $\{+\}$ and $\{-\}$. [/step]

[step:Show that a continuous map from a connected space to $\{+,-\}_{\mathrm{disc}}$ is constant]Let $X$ be a nonempty connected topological space and let $f : X \to \{+, -\}_{\mathrm{disc}}$ be continuous. The preimages $A := f^{-1}(\{+\})$ and $B := f^{-1}(\{-\})$ are open in $X$ (since $\{+\}$ and $\{-\}$ are open in the discrete topology) and satisfy $A \cap B = \varnothing$, $A \cup B = X$. By the definition of [connectedness](/page/Connected%20Space), $X$ cannot be written as a disjoint union of two nonempty open subsets; hence either $A = \varnothing$ (so $f \equiv -$) or $B = \varnothing$ (so $f \equiv +$).[/step]

[guided]We isolate the pure topological fact that drives the theorem: continuous maps from connected spaces into the discrete two-point set $\{+, -\}_{\mathrm{disc}}$ are constant. Let $X$ be a nonempty connected topological space and let $f : X \to \{+, -\}_{\mathrm{disc}}$ be continuous. Set \begin{align*} A := f^{-1}(\{+\}), \qquad B := f^{-1}(\{-\}). \end{align*} Why are $A$ and $B$ open in $X$? Because in the discrete topology on $\{+, -\}$, *every* subset is open — in particular, the singletons $\{+\}$ and $\{-\}$ are open. Continuity of $f$ means preimages of open sets are open, so $A$ and $B$ are open in $X$. These two sets are disjoint (an element cannot map to both $+$ and $-$) and cover $X$ (every element maps to one of the two). Thus $X = A \sqcup B$ with $A, B$ open. By the definition of [connectedness](/page/Connected%20Space), a connected space cannot be partitioned into two *nonempty* disjoint open subsets. Therefore one of $A$, $B$ is empty. If $A = \varnothing$, then $f \equiv -$; if $B = \varnothing$, then $f \equiv +$. In either case $f$ is constant. What would fail if we dropped connectedness? Nothing topological in the argument changes — we simply could not conclude that one of $A$, $B$ is empty. The map could legitimately take different values on different connected components, which is precisely what the next step exploits.[/guided]

[step:Restrict to each connected component to conclude part (1)] Let $C \subseteq U$ be a connected component of $U$. By definition $C$ is connected and nonempty. The restriction $\mathcal{O}|_C : C \to \{+, -\}_{\mathrm{disc}}$ is continuous (the restriction of a continuous map to a subspace is continuous). By the previous step, $\mathcal{O}|_C$ is constant: either $\mathcal{O}|_C \equiv +$ or $\mathcal{O}|_C \equiv -$. This is exactly claim (1). [/step]

[step:Specialize to connected $U$ and exhibit exactly two orientations]Suppose now that $U$ is nonempty and connected. Then $U$ is its own unique connected component, so by part (1) every orientation $\mathcal{O}$ on $U$ satisfies $\mathcal{O} \equiv +$ or $\mathcal{O} \equiv -$. This gives *at most* two orientations. Conversely, both constant functions $\mathcal{O}_+ \equiv +$ and $\mathcal{O}_- \equiv -$ are locally constant: for every $p \in U$, the set $U$ is an open neighbourhood of $p$ in the [subspace topology](/page/Subspace%20Topology) on $U$, and each constant function is constant on this neighbourhood. Therefore each is a valid orientation. They are distinct because $U$ is nonempty: for any $p \in U$, $\mathcal{O}_+(p) = + \neq - = \mathcal{O}_-(p)$. Therefore $U$ admits *exactly two* orientations, namely $\mathcal{O}_+$ (the standard orientation) and $\mathcal{O}_-$ (the opposite orientation). This is claim (2), completing the proof.[/step]

[guided]Combining the previous steps, we obtain part (2) by specialization. If $U$ is nonempty and connected, then $U$ itself is the unique connected component of $U$. By part (1) applied to $C = U$, every orientation $\mathcal{O}$ on $U$ is one of the two constant functions $\mathcal{O}_+ \equiv +$ or $\mathcal{O}_- \equiv -$. This gives the upper bound: at most two orientations. For the lower bound, we must verify that *both* constants actually are orientations. The constant function $\mathcal{O}_+ : U \to \{+, -\}$, $p \mapsto +$, is locally constant: for any $p \in U$, take $V = U$; then $\mathcal{O}_+$ is identically $+$ on $V$. The same applies to $\mathcal{O}_-$. So both are orientations. Finally, $\mathcal{O}_+ \neq \mathcal{O}_-$ as functions, because $U$ is nonempty: pick any $p \in U$, and $\mathcal{O}_+(p) = + \neq - = \mathcal{O}_-(p)$. Where is nonemptiness consumed? If $U = \varnothing$, the *only* function $U \to \{+, -\}$ is the empty function — there is just one orientation, not two. The nonemptiness hypothesis rules out this degenerate case. Thus $U$ has exactly two orientations: the standard one $\mathcal{O}_+$ and its opposite $\mathcal{O}_-$. This completes the proof of (2) and of the theorem.[/guided]