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It remains to show that $S_u\not\subseteq\R_+^d$ for any unit vector $u$. Take indeed any such vector $u=(u_1,\dots,u_d)$. Without loss of generality, $|u_1|$ is the smallest of all the $|u_j|$'s, so that $$|u_1|-\sqrt{1-u_1^2}<0.$$ Let now $$v=v^*:=(v_1,\dots,v_d):=\Big(-\sqrt{1-u_1^2},\frac{u_1u_2}{\sqrt{1-u_1^2}},\dots,\frac{u_1u_d}{\sqrt{1-u_1^2}}\Big). \tag{1}\label{1}$$ Then $u+v\in S_u$. However, $u_1+v_1<0$, so that $S_u\not\subseteq\R_+^d$. $\quad\Box$

Added: Since the OP wanted an explicit finite set of vectors (rather than the infinite set $S_u$), let us provide such a set. So, for $d=3$, let $S_{u,N}$ be the set of all $N$ vertices of a regular $N$-gon inscribed into the circle $S_u$. Then the shortest distance from the point $v^*$ defined in \eqref{1} (and actually from any point on the circle $S_u$) to the set $S_{u,N}$ will be no greater than the length $2\sin\frac{2\pi}{4N}$ of a side of a regular $2N$-gon inscribed into the unit circle $S_u$. So, following lines of above reasoning, we see that, to get $w_1<0$ for at least one vertex $w=(w_1,\dots,w_3)$ of $S_{u,N}$, it is enough that $\frac1{\sqrt3}-\sqrt{1-(\frac1{\sqrt3})^2}+2\sin\frac{2\pi}{4N}<0$. The latter inequality holds for all natural $N\ge14$. Thus, we get a counterexample with a set of $14$ vectors. This is worse than $N=5$ in Nathaniel Johnston's answer; however, here the reasoning is quite elementary. (The number $N=14$ can apparently be improved even within this elementary framework, using a bit more complicated consideration.)

Details on the update: As in the paragraph just above, let $d=3$ and consider the set $S_{u,4}$, that is, the set $S_{u,N}$ with $N=4$, so that $S_{u,4}$ is the set of the vertices of a square inscribed into the circle $S_u$, defined in \eqref{10}. So, \begin{equation*} S_{u,4}=\{u+V_j\colon j\in\{0,1,2,3\}\}, \end{equation*} where \begin{equation*} V_j:=a\cos\Big(t+\frac{2\pi j}4\Big)+b\sin\Big(t+\frac{2\pi j}4\Big)=(V_{j,1},V_{j,2},V_{j,3})\in\R^3, \end{equation*} $t\in[0,\pi/2]$ is the angle of the rotation of the vertices of the square about the axis through the vector $u$, and $u,a,b$ are orthonormal vectors in $\R^3$. To be specific, let \begin{equation*} a=\Big(\sqrt{1-u_1^2},-\frac{u_1u_2}{\sqrt{1-u_1^2}},-\frac{u_1 \sqrt{1-u_1^2-u_2^2}}{\sqrt{1-u_1{}^2}}\Big) \end{equation*} and \begin{equation*} b=\Big(0,\frac{\sqrt{1-u_1^2-u_2^2}}{\sqrt{1-u_1^2}},-\frac {u_2}{\sqrt{1-u_1^2}}\Big), \end{equation*} with \begin{equation*} u=(u_1,u_2,u_3). \end{equation*}

Without loss of generality (wlog), \begin{equation*} u_1\le u_2\le u_3. \end{equation*}

It remains to show that $S_u\not\subseteq\R_+^d$ for any unit vector $u$. Take indeed any such vector $u=(u_1,\dots,u_d)$. Without loss of generality, $|u_1|$ is the smallest of all the $|u_j|$'s, so that $$|u_1|-\sqrt{1-u_1^2}<0.$$ Let now \begin{equation} v=v^* :=(v_1,\dots,v_d):=\Big(-\sqrt{1-u_1^2},\frac{u_1u_2}{\sqrt{1-u_1^2}},\dots,\frac{u_1u_d}{\sqrt{1-u_1^2}}\Big). \tag{15}\label{15} \end{equation} Then $u+v\in S_u$. However, $u_1+v_1<0$, so that $S_u\not\subseteq\R_+^d$. $\quad\Box$

Added: Since the OP wanted an explicit finite set of vectors (rather than the infinite set $S_u$), let us provide such a set. So, for $d=3$, let $S_{u,N}$ be the set of all $N$ vertices of a regular $N$-gon inscribed into the circle $S_u$. Then the shortest distance from the point $v^*$ defined in \eqref{15} (and actually from any point on the circle $S_u$) to the set $S_{u,N}$ will be no greater than the length $2\sin\frac{2\pi}{4N}$ of a side of a regular $2N$-gon inscribed into the unit circle $S_u$. So, following lines of above reasoning, we see that, to get $w_1<0$ for at least one vertex $w=(w_1,\dots,w_3)$ of $S_{u,N}$, it is enough that $\frac1{\sqrt3}-\sqrt{1-(\frac1{\sqrt3})^2}+2\sin\frac{2\pi}{4N}<0$. The latter inequality holds for all natural $N\ge14$. Thus, we get a counterexample with a set of $14$ vectors. This is worse than $N=5$ in Nathaniel Johnston's answer; however, here the reasoning is quite elementary. (The number $N=14$ can apparently be improved even within this elementary framework, using a bit more complicated consideration.)

Details on the update: As in the paragraph just above, let $d=3$ and consider the set $S_{u,4}$, that is, the set $S_{u,N}$ with $N=4$, so that $S_{u,4}$ is the set of the vertices of a square inscribed into the circle $S_u$, defined in \eqref{10}. So, \begin{equation*} S_{u,4}=\{u+V_j\colon j\in\{0,1,2,3\}\}, \end{equation*} where \begin{equation*} V_j:=a\cos\Big(t+\frac{2\pi j}4\Big)+b\sin\Big(t+\frac{2\pi j}4\Big)=(V_{j,1},V_{j,2},V_{j,3})\in\R^3, \end{equation*} $t\in[0,\pi/2]$ is the angle of the rotation of the vertices of the square about the axis through the vector $u$, and $u,a,b$ are orthonormal vectors in $\R^3$. To be specific, let \begin{equation*} a=\Big(\sqrt{1-u_1^2},-\frac{u_1u_2}{\sqrt{1-u_1^2}},-\frac{u_1 \sqrt{1-u_1^2-u_2^2}}{\sqrt{1-u_1{}^2}}\Big) \end{equation*} and \begin{equation*} b=\Big(0,\frac{\sqrt{1-u_1^2-u_2^2}}{\sqrt{1-u_1^2}},-\frac {u_2}{\sqrt{1-u_1^2}}\Big), \end{equation*} with \begin{equation*} u=(u_1,u_2,u_3). \end{equation*} Here it is assumed that $u_1^2\ne1$, which can be done without loss of generality (wlog) -- otherwise, rearrange the coordinate axes.

Wlog, \begin{equation*} u_1\le u_2\le u_3. \end{equation*}

$\newcommand\R{\mathbb R}$The answer is no.

Now, for any vector $u$ in $\R^d$ ($d\ge3$) with Euclidean norm $|u|=1$, consider the set
$$S_u:=\{u+v\colon\, v\in\R^d,|v|=1,v\cdot u=0\},$$ with $\cdot$ denoting the dot product. (One may note that (say) for the $d=3$ the set $S_u$ is the unit-radius circle in $\R^d$ centered at $u$ and lying in the affine plane through $u$ perpendicular to $u$.)

Added: Since the OP wanted an explicit finite set of vectors (rather than the infinite set $S_u$), let us provide such a set. So, for $d=3$, let $S_{u,N}$ be the set of all $N$ vertices of a regular $N$-gon inscribed into the circle $S_u$. Then the shortest distance from the point $v^*$ defined in \eqref{1} (and actually from any point on the circle $S_u$) to the set $S_{u,N}$ will be no greater than the length $2\sin\frac{2\pi}{4N}$ of a side of a regular $2N$-gon inscribed into the unit circle $S_u$. So, following lines of above reasoning, we see that, to get $w_1<0$ for at least one vertex $w=(w_1,\dots,w_3)$ of $S_{u,N}$, it is enough that $\frac1{\sqrt3}-\sqrt{1-(\frac1{\sqrt3})^2}+2\sin\frac{2\pi}{4N}<0$. The latter inequality holds for all natural $N\ge14$. Thus, we get a counterexample with a set of $14$ vectors. This is worse than $N=5$ in Nathaniel Johnston's answer; however, here the reasoning is quite elementary. (The number $N=14$ can apparently be improved even within this elementary framework, using a bit more complicated consideration.)

$\newcommand\R{\mathbb R}$Update: Within the previously established framework, we now show that it is enough to have just $N=4$ points. This improves what seems to be the previous record, of $N=5$. It should be clear that this new record cannot be further improved. Details on this are given at the end of the answer.

The answer is no.

Now, for any vector $u$ in $\R^d$ ($d\ge3$) with Euclidean norm $|u|=1$, consider the set
\begin{equation*} S_u:=\{u+v\colon\, v\in\R^d,|v|=1,v\cdot u=0\}, \tag{10}\label{10} \end{equation*} with $\cdot$ denoting the dot product. (One may note that (say) for the $d=3$ the set $S_u$ is the unit-radius circle in $\R^d$ centered at $u$ and lying in the affine plane through $u$ perpendicular to $u$.)

Added: Since the OP wanted an explicit finite set of vectors (rather than the infinite set $S_u$), let us provide such a set. So, for $d=3$, let $S_{u,N}$ be the set of all $N$ vertices of a regular $N$-gon inscribed into the circle $S_u$. Then the shortest distance from the point $v^*$ defined in \eqref{1} (and actually from any point on the circle $S_u$) to the set $S_{u,N}$ will be no greater than the length $2\sin\frac{2\pi}{4N}$ of a side of a regular $2N$-gon inscribed into the unit circle $S_u$. So, following lines of above reasoning, we see that, to get $w_1<0$ for at least one vertex $w=(w_1,\dots,w_3)$ of $S_{u,N}$, it is enough that $\frac1{\sqrt3}-\sqrt{1-(\frac1{\sqrt3})^2}+2\sin\frac{2\pi}{4N}<0$. The latter inequality holds for all natural $N\ge14$. Thus, we get a counterexample with a set of $14$ vectors. This is worse than $N=5$ in Nathaniel Johnston's answer; however, here the reasoning is quite elementary. (The number $N=14$ can apparently be improved even within this elementary framework, using a bit more complicated consideration.)

Details on the update: As in the paragraph just above, let $d=3$ and consider the set $S_{u,4}$, that is, the set $S_{u,N}$ with $N=4$, so that $S_{u,4}$ is the set of the vertices of a square inscribed into the circle $S_u$, defined in \eqref{10}. So, \begin{equation*} S_{u,4}=\{u+V_j\colon j\in\{0,1,2,3\}\}, \end{equation*} where \begin{equation*} V_j:=a\cos\Big(t+\frac{2\pi j}4\Big)+b\sin\Big(t+\frac{2\pi j}4\Big)=(V_{j,1},V_{j,2},V_{j,3})\in\R^3, \end{equation*} $t\in[0,\pi/2]$ is the angle of the rotation of the vertices of the square about the axis through the vector $u$, and $u,a,b$ are orthonormal vectors in $\R^3$. To be specific, let \begin{equation*} a=\Big(\sqrt{1-u_1^2},-\frac{u_1u_2}{\sqrt{1-u_1^2}},-\frac{u_1 \sqrt{1-u_1^2-u_2^2}}{\sqrt{1-u_1{}^2}}\Big) \end{equation*} and \begin{equation*} b=\Big(0,\frac{\sqrt{1-u_1^2-u_2^2}}{\sqrt{1-u_1^2}},-\frac {u_2}{\sqrt{1-u_1^2}}\Big), \end{equation*} with \begin{equation*} u=(u_1,u_2,u_3). \end{equation*}

We want to show that \begin{equation*} S_{u,4}\overset{\text{(?)}}{\not\subseteq}\R_+^3 \tag{20}\label{20} \end{equation*} for any described choices of $u$ and $t$.

Without loss of generality (wlog), \begin{equation*} u_1\le u_2\le u_3. \end{equation*}

Suppose for a moment that $u_1\le0$. Clearly, $V_{j,1}<0$ for at least one $j\in\{0,1,2,3\}$, and then $u+V_j\in S_{u,4}\setminus\R_+^3$, so that \eqref{20} holds.

So, wlog \begin{equation*} 0<u_1\le u_2\le u_3=\sqrt{1-u_1^2-u_2^2}. \tag{30}\label{30} \end{equation*} Note that \begin{equation*} \begin{aligned} &\{u_1+V_{j,1}\colon j\in\{1,2\}\} =\left\{u_1-\sqrt{1-u_1^2}\,\sin t,u_1-\sqrt{1-u_1^2}\,\cos t\right\}. \end{aligned} \end{equation*} So, if \eqref{20} fails to hold, then \begin{equation*} u_1\ge \sqrt{1-u_1^2}\,\max(\sin t,\cos t)\ge\frac1{\sqrt2}\,\sqrt{1-u_1^2} \tag{40}\label{40} \end{equation*} and hence $u_1\ge\frac1{\sqrt3}$, which implies \begin{equation*} u_1=u_2=u_3=\frac1{\sqrt3}, \end{equation*} in view of \eqref{30}. Moreover, if we had $t\in[0,\pi/2]\setminus\{\pi/4\}$, then \eqref{40} would imply $u_1>\frac1{\sqrt3}$, which would contradict \eqref{30}.

So, if \eqref{20} fails to hold, then \begin{equation*} u_1=u_2=u_3=\frac1{\sqrt3}\quad\text{and}\quad t=\pi/4. \end{equation*} But then \begin{equation*} V_{0,3}=\frac{\sqrt{1-u_1^2-u_2^2} \left(\sqrt{1-u_1^2}-u_1 \cos t\right)-u_2 \sin t}{\sqrt{1-u_1^2}}=-0.211\ldots<0. \end{equation*} Thus, in all cases, if \eqref{20} fails to hold, then it necessarily holds. $\quad\Box$

$\newcommand\R{\mathbb R}$The answer is no.

Let us use the terms "acute angle" and "positive" in the nonstrict sense, meaning "angle $\le\pi/2$" and $\ge0$, respectively. The "strict" modification should be straightforward. By approximation, we may also assume that the set of vectors is allowed to be infinite.

Now, for any vector $u$ in $\R^d$ ($d\ge3$) with Euclidean norm $|u|=1$, consider the set
$$S_u:=\{u+v\colon\, v\in\R^d,|v|=1,v\cdot u=0\},$$ with $\cdot$ denoting the dot product. (One may note that (say) for the $d=3$ the set $S_u$ is the unit-radius circle in $\R^d$ centered at $u$ and lying in the affine plane through $u$ perpendicular to $u$.)

Then for any vector $u+v\in S_u$ the angle between $u+v$ and $u$ is $\pi/4$ and hence, by the subadditivity of the angle measure, the angle between any two vectors in $S_u$ is acute. One can also show this algebraically: For any $u+v$ and $u+v'$ in $S_u$, we have $(u+v)\cdot(u+v')=1+v\cdot v'\ge0$, by the Cauchy--Schwarz inequality.

Moreover, any rotation of $S_u$ is of the same form, $S_u$, but maybe for another unit vector $u$.

It remains to show that $S_u\not\subseteq\R_+^d$ for any unit vector $u$. Take indeed any such vector $u=(u_1,\dots,u_d)$. Without loss of generality, $|u_1|$ is the smallest of all the $|u_j|$'s, so that $$|u_1|-\sqrt{1-u_1^2}<0.$$ Let now $$v=v^*:=(v_1,\dots,v_d):=\Big(-\sqrt{1-u_1^2},\frac{u_1u_2}{\sqrt{1-u_1^2}},\dots,\frac{u_1u_d}{\sqrt{1-u_1^2}}\Big). \tag{1}\label{1}$$ Then $u+v\in S_u$. However, $u_1+v_1<0$, so that $S_u\not\subseteq\R_+^d$. $\quad\Box$

The picture below shows the set $[0,1]S_u=\{tw\,\colon t\in[0,1],w\in S_u\}$ for $d=3$ and (the apparently "best") unit vector $u=\frac1{\sqrt3}\,(1,1,1)$. It is kind of seen that $[0,1]S_u\not\subseteq\R_+^d$ and hence $S_u\not\subseteq\R_+^d$.

Added: Since the OP wanted an explicit finite set of vectors (rather than the infinite set $S_u$, let us provide such a set. So, for $d=3$, let $S_{u,N}$ be the set of all $N$ vertices of a regular $N$-gon inscribed into the circle $S_u$. Then the shortest distance from the point $v^*$ defined in \eqref{1} (and actually from any point ion the circle $S_u$) will be no greater than the length $2\sin\frac{2\pi}{4N}$ of a side of a regular $2N$-gon inscribed into the unit circle $S_u$. So, following lines of above reasoning, we see that, to get $w_1<0$ for at least one vertex $w=(w_1,\dots,w_3)$ of $S_{u,N}$, it is enough that $\frac1{\sqrt3}-\sqrt{1-(\frac1{\sqrt3})^2}+2\sin\frac{2\pi}{4N}<0$. The latter inequality holds for all natural $N\ge14$. Thus, we get a counterexample with a set of $14$ vectors. This is worse than $N=5$ in Nathaniel Johnston's answer; however, here the reasoning is quite elementary. (The number $N=14$ can apparently be improved even within this elementary framework, using a bit more complicated consideration.)

$\newcommand\R{\mathbb R}$The answer is no.

Let us use the terms "acute angle" and "positive" in the nonstrict sense, meaning "angle $\le\pi/2$" and $\ge0$, respectively. The "strict" modification should be straightforward. By approximation, we may also assume that the set of vectors is allowed to be infinite.

Now, for any vector $u$ in $\R^d$ ($d\ge3$) with Euclidean norm $|u|=1$, consider the set
$$S_u:=\{u+v\colon\, v\in\R^d,|v|=1,v\cdot u=0\},$$ with $\cdot$ denoting the dot product. (One may note that (say) for the $d=3$ the set $S_u$ is the unit-radius circle in $\R^d$ centered at $u$ and lying in the affine plane through $u$ perpendicular to $u$.)

Then for any vector $u+v\in S_u$ the angle between $u+v$ and $u$ is $\pi/4$ and hence, by the subadditivity of the angle measure, the angle between any two vectors in $S_u$ is acute. One can also show this algebraically: For any $u+v$ and $u+v'$ in $S_u$, we have $(u+v)\cdot(u+v')=1+v\cdot v'\ge0$, by the Cauchy--Schwarz inequality.

Moreover, any rotation of $S_u$ is of the same form, $S_u$, but maybe for another unit vector $u$.

It remains to show that $S_u\not\subseteq\R_+^d$ for any unit vector $u$. Take indeed any such vector $u=(u_1,\dots,u_d)$. Without loss of generality, $|u_1|$ is the smallest of all the $|u_j|$'s, so that $$|u_1|-\sqrt{1-u_1^2}<0.$$ Let now $$v=v^*:=(v_1,\dots,v_d):=\Big(-\sqrt{1-u_1^2},\frac{u_1u_2}{\sqrt{1-u_1^2}},\dots,\frac{u_1u_d}{\sqrt{1-u_1^2}}\Big). \tag{1}\label{1}$$ Then $u+v\in S_u$. However, $u_1+v_1<0$, so that $S_u\not\subseteq\R_+^d$. $\quad\Box$

The picture below shows the set $[0,1]S_u=\{tw\,\colon t\in[0,1],w\in S_u\}$ for $d=3$ and (the apparently "best") unit vector $u=\frac1{\sqrt3}\,(1,1,1)$. It is kind of seen that $[0,1]S_u\not\subseteq\R_+^d$ and hence $S_u\not\subseteq\R_+^d$.

Added: Since the OP wanted an explicit finite set of vectors (rather than the infinite set $S_u$), let us provide such a set. So, for $d=3$, let $S_{u,N}$ be the set of all $N$ vertices of a regular $N$-gon inscribed into the circle $S_u$. Then the shortest distance from the point $v^*$ defined in \eqref{1} (and actually from any point on the circle $S_u$) to the set $S_{u,N}$ will be no greater than the length $2\sin\frac{2\pi}{4N}$ of a side of a regular $2N$-gon inscribed into the unit circle $S_u$. So, following lines of above reasoning, we see that, to get $w_1<0$ for at least one vertex $w=(w_1,\dots,w_3)$ of $S_{u,N}$, it is enough that $\frac1{\sqrt3}-\sqrt{1-(\frac1{\sqrt3})^2}+2\sin\frac{2\pi}{4N}<0$. The latter inequality holds for all natural $N\ge14$. Thus, we get a counterexample with a set of $14$ vectors. This is worse than $N=5$ in Nathaniel Johnston's answer; however, here the reasoning is quite elementary. (The number $N=14$ can apparently be improved even within this elementary framework, using a bit more complicated consideration.)