Does continuity on a union of compact sets imply overall local boundedness?

Question

Suppose $(F_{n})$ is a sequence of compact sets in $\mathbb{R}^{N}$ with $N\geq1$, and $f:\mathbb{R}^{N}\to(-\infty,\infty)$ is a function whose restriction to each $F_{n}$ is continuous. Can we say that for all $x\in A=\cup_{n\geq1}F_{n}$ there exists an open set $V$ in $\mathbb{R}^{N}$ that contains $x$ and such that $f$ is bounded on $V\cap A$?

Answer 1

No. For example, with $N=1$ you might have $F_0 = \{0\}$ and $F_n = \{1/n\}$ for $n \ge 1$, with $f = n$ on $F_n$. There is no open set $V$ containing $0$ such that $f$ is bounded on $V \cap A$.