Let $F:\Omega\rightarrow\mathbb{R}^n$ a continuous vector field.
Given that for any compact set $K\subset\Omega$, $\epsilon>0$ exists $\delta>0$ such that if $\gamma(t),\eta(t), t\in[a,b]$ are piecewise smooth curves contained in $K$ and $max_{t\in[a,b]}|\gamma(t)-\eta(t)|<\delta$ ,then $|\int_{\gamma}F\cdot dl-\int_{\eta}F\cdot dl|<\epsilon$.
Prove F is a locally conservative vector field.
My thoughts:
I wanted to find for every $p\in\mathbb{R}^n$ a neighborhood where $F$ is Conservative, but I cannot conclude $\int_{\gamma}F\cdot dl=0$ or any closed curve.
if F is not locally conservative, exists $x_0\in K$ (for some compact set K) which has no neighborhood for which F is conservative. In other words, For any open ball $B(x_0,\epsilon_n)$ and $\epsilon_n>0$ exists a curve $\gamma_n$ such that $\int_{\gamma_n}{F\cdot dl}\neq0$. If we look at a sequence of descending open balls as mentioned $B(x_0,\epsilon_n)\to\{x_0\}$, the sequence of curves $(\gamma_n)_{n=1}^\infty\to\Gamma(t)=x_0$ (uniformly convergence). so $lim_{n\to\infty}{\int_{\gamma_n}F\cdot dl}=0$... I think.
So suppose you have a closed curve $\gamma$ inside $B(x_0,\delta)$ for which $\oint_\gamma F \cdot dl = \alpha \ne 0$. Then let $N \gamma$ be the curve $\gamma$ concatenated with itself $N$ times. That is, if $\gamma:[0,1] \to \mathbb R^n$, then $N\gamma:[0,1]\to\mathbb R^n$ with $$ (N\gamma)(t) = \gamma((Nt)\bmod 1) $$ Then $\oint_{N\gamma} F \cdot dl = N\alpha$ can be arbitrarily large, but $N\gamma$ is still only a distance $\delta$ from the path $\eta$ that stays constantly at $x_0$. And of course $\oint_\eta F \cdot dl = 0$.