The set is defined as:
$\mathbb{S} = \{\frac{\sum_{i=1}^Nx_iy_i}{\sum_{i=1}^Nx_i^2} |x_i \in [\underline{x}_i,\overline{x}_i] \hspace{0.1cm} \text{for all $i$, and }\overline{x}_i, \underline{x}_i, y_i \in \mathbb{R}^1, \sum_{i=1}^N x_i^2 \neq 0 \}$.
I am interested in characterizing the properties of this set. I am thinking of this like a regression. The function defining the set is the formula of the regression coefficient. I want to know what this set looks like for a fixed $y_i$ that I get from the data but where my $x_i$ are interval censored but my interval bounds i.e. $\underline{x}_i$ and $\overline{x}_i$ are fixed. In other words I am thinking of this set as conditional on the values of $y_i, \underline{x}_i, \overline{x}_i$ and so these numbers cannot be varied. Only the $x_i$ can vary. In particular, I want to know if this set is convex, does it have holes (is it simply connected or like a torus), is it compact? In light of John Hughes answer, I think I am looking for ranges of $\underline{x}_i$ $\overline{x}_i$ and $y_i$, where these properties will hold true or not (unless they always hold true).
I think it should be okay to consider the case where N=2 because the results should generalize. Couple of observations that I made:
1) If $0\in [\underline{x}_i,\overline{x}_i] \hspace{0.1cm} \text{for all $i$}$, then the set is not defined. Hence the added condition, suggested by one of the responses, that $\sum_{i=1}^N x_i^2 \neq 0$.
2) For the N=2 case, the gradient is given by the following:
$$[\frac{(-y_1)x_1^2 - (2x_2y_2)x_1 - x_2^2y_1}{(x_1^2+x_2^2)^2}, \frac{(-y_2)x_2^2 - (2x_1y_1)x_2 - x_1^2y_2}{(x_1^2+x_2^2)^2}]$$
This shows that the numerator is essentially a quadratic. If I want to solve for a local optima, fixing $x_{-i}$, I can solve a quadratic in $x_i$ to get the maximum or minimum (a constrained problem because of the bounds).
Questions:
1) Is this set convex? My hunch is that it is not. I tried the following:
We know that $\frac{\underline{x}_1y_1+\underline{x}_2y_2}{\underline{x}_1^2+\underline{x}_2^2} \in \mathbb{S}$ and $\frac{\overline{x}_1y_1+\overline{x}_2y_2}{\overline{x}_1^2+\overline{x}_2^2} \in \mathbb{S}$. Now consider the weighted sum of these two, where each of them get a weight of 0.5. We get the following expression: $$0.5 \times \frac{(\overline{x}_1^2 \underline{x}_1 + \overline{x}_2^2 \underline{x}_1)y_1 + (\overline{x}_1^2 \underline{x}_2 + \overline{x}_2^2 \underline{x}_2)y_2}{(\overline{x}_1^2 \underline{x}_1 + \overline{x}_2^2 \underline{x}_1)\underline{x}_1 + (\overline{x}_1^2 \underline{x}_2 + \overline{x}_2^2 \underline{x}_2)\underline{x}_2}$$ I am not sure how to show that this does not lie inside $\mathbb{S}$.
2) Is it correct that otherwise the set is compact? My intuition is that its bounded and closed, if $0$ is not in both intervals, so it must be compact in this case.
3)I am not sure how to show that this set does not contain holes (like torus) of the nature shown in this picture on wikipedia: https://en.wikipedia.org/wiki/Simply_connected_space#/media/File:Runge_theorem.svg
My intuition is that these are not present. The only problematic case is when $0$ is in both intervals but if this is not true then these holes shouldn't be there.
How do I show that this set is simply connected? Is this the same as connected? Except for the case where the $0$ is in both intervals, plots of this function look somewhat spiky but connected in the sense that the plots don't really show holes. Though I don't know if this rules out the case shown in wikipedia.
If $0$ is in all the intervals, then the set is not well-defined, so I'll ignore that case.
If $0$ is not in all the intervals, then there is some interval --- we might as well make it the first --- containing some number $s$ other than $0$. Now we can pick $x_1 = s$ and pick $x_i = \overline{x_i}$ for $i = 2, 3, \ldots$. Pick $y_i = 0 $ for $x_i = 2, 3, \ldots$ as well. Then the sum becomes $$ \frac{s_1y_1}{s_1^2} = \frac{1}{s_1} y_1 $$ Because $k = 1/s_1 \ne 0$, these products $ky_1$ range over all real numbers, so the set $S$ contains all of $\Bbb R$ -- it's connected, noncompact, convex, has no holes. I think that's the whole story.