Discontinuity of the indicator function

Question

Consider the function $q(x,\theta)=1\{ x \in \{x \text{ s.t. } \theta+x_i>0 \text{ }\forall i \}\}$ where 1 is the indicator function taking value 1 if the condition inside $\{ \}$ is satisfied and zero otherwise, $x \in \mathbb{R}^k$, $x_i$ is the ith component of the vector $x$ and $\theta \in \Theta \subseteq \mathbb{R^k}$. Is this function not continuous both in $x$ and $\theta$? How can I show it?

Answer 1

Let $u$ denote the vector of ones, then, for every real number $t$ and every vector $x$, one has $q(x+tu,-x)=\mathbf 1_{t\gt0}$ and $q(x,-x+tu)=\mathbf 1_{t\lt0}$. What does this tell you?