Given $\textbf{x}, \textbf{y}\in\mathbb{R}^d,$ find $\delta$ s.t. $\textbf{x}\cdot\textbf{y}>0 \iff \textbf{x}\cdot(\textbf{y}+\delta)>0$ Let $\textbf{x}, \textbf{y}\in\mathbb{R}^d$ be two $d$-dimensional vectors. For which vectors $\delta\in\mathbb{R}^d$ we have that the dot products $\textbf{x}\cdot\textbf{y}$ and $\textbf{x}\cdot(\textbf{y}+\delta)$ have the same sign?
More precisely, for which $\delta\in\mathbb{R}^d$ we have that $\textbf{x}\cdot\textbf{y}>0$ if and only if $\textbf{x}\cdot(\textbf{y}+\delta)>0$?
As the sign of a dot product of two vectors is determined by the (cosine of the) angle between them,  then the condition $\textbf{x}\cdot\textbf{y}>0$ should imply that $\textbf{x}\cdot(\textbf{y}+\delta)>0$ for all sufficiently "small" vectors $\delta$. So the question is about describing such vectors $\delta$ in terms of the vectors $\textbf{x}$ and $\textbf{y}$.
 A: I'll answer first in the case $x \cdot y = 0$. Then $y$ is in the hyperplane normal to $x$, and $\delta$ must also be in this hyperplane; there is no sufficiently small $\delta$. In particular, if $\varepsilon > 0$, then $x \cdot (y + \varepsilon x) = \varepsilon \lVert x \rVert^2 \ne 0 = x \cdot y$. (Unless $x = 0$, in which case $\delta$ can be any vector at all.)
On the other hand if $x \cdot y \ne 0$, then there is an open half space $\mathbb{H}$ containing the origin such that all $\delta \in \mathbb{H}$ preserve the sign of $x \cdot y$. Note that $\mathbb{H}$ then contains a ball centered at the origin.
Specifically, to answer your question about how small $\delta$ has to be, if $\lVert \delta \rVert$ is less than the distance from $y$ to the hyperplane normal to $x$, then it preserves the sign of $x \cdot y$. And that distance is exactly the magnitude of the projection of $y$ onto $x$, $\frac{x \cdot y}{\lVert x \rVert}$. So if $\lVert \delta \rVert < \frac{x \cdot y}{\lVert x \rVert}$, then the sign of $x \cdot y$ and $x \cdot (y + \delta)$ are the same.
