Let $f: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}$ be locally bounded, continuous in the first argument.
Consider a converging sequence $\{ x_k \}_{k=1}^{\infty}$, $x_k \rightarrow x$.
I am wondering if the following holds. $$ \limsup_{k \rightarrow \infty} \left\{ y \in \mathbb{R}^m \mid f(x_k, y) > 0 \right\} = \left\{y \in \mathbb{R}^m \mid \limsup_{k \rightarrow \infty} f(x_k,y) > 0 \right\}$$
Comments. The $\limsup$ of sets $S_k \subseteq \mathbb{R}^n$ is defined as follows. $$ \limsup_{k \rightarrow \infty} S_k := \bigcap_{k=1}^{\infty} \bigcup_{ j \geq k } S_j $$
The $\limsup$ of a function $g: \mathbb{R}^n \rightarrow \mathbb{R}$ is defined as follows. $$ \limsup_{k \rightarrow \infty} g(x_k) := \lim_{n \rightarrow \infty} \sup_{j \geq n} g(x_j) $$
I am not clear if the continuity assumption is actually needed. For sure it is to further claim $\limsup_{k \rightarrow \infty} f(x_k,y) = f(x,y)$.
$y$ is an element of the LHS iff the condition is met for infinitely many k. $f$ is continuous, so we have lim sup $f$ = lim $f$. Hence the RHS is contained in the LHS. The reverse inclusion does not hold due to the strict inequality on the RHS.