## Cut points, continuous paths and increasing sequence of sets

currently I have a question about the following point-set topology problem. Everything takes place in $(mathbb C,|cdot|)$ and we are given the following setup.

- One has a sequence of closed connected sets of points in the complex plane ${L_t}_{t geq 0}$ (the precise definition for this problem should be irrelavant) such that $L_t subseteq L_{t+s}$ for each $s > 0$. So basically as $t$ increases the set grows aswell.
- There exist continuous curves $gamma$ such that for each $t geq 0$ one has $gamma : [-t,t] to mathbb C$ such that $gamma([-t,t]) subseteq L_t$. Moreover we know especially that $gamma(t+s)$ and $gamma(-(t+s))$ are not in $L_t$ and together with another given property one can actually proof that $gamma(t)$ and $gamma(-t)$ are cut points of $L_{t+s}$ for $s > 0$. Using the definition found here this tells us (for example) that $L_{t+s} setminus gamma(t)$ is disconnected.

**Question**: We already have for each $t geq 0 : gamma([-t,t]) subseteq L_t$ I want to proof that one actually has equality, i.e. $L_t = gamma([-t,t])$.

**Idea**: Take for example $[a,b]$ in $mathbb R$ then every point $c in (a,b)$ is a cutpoint, the important take away here is that the ”end-points” of the interval $[a,b]$ are not cutpoints. Here something similiar should be possible. Since $gamma(t)$ is a cut point of $L_{t+s}$ it cannot be that that $L_{t+s} setminus L_t$ did not grow ”along” $gamma(t)$. Growing say along the middle of the curve around $gamma(t/2)$ should contradict some combination of the notion of cut point and continuity of $gamma$ but I find it hard to write down precisely.

Thanks in advance!

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