The limit set of an unbounded path/ray Given an infinite path $\gamma:[0,+\infty) \to \mathbb{R}^n$, the limit set $L$ is the set of limit points of sequences $\gamma (t_k)$ for a sequence $t_k \to \infty$.
If $\gamma$ is bounded (or more generally a path in a compact Hausdorff space), then the limit set is non-empty, compact and connected, by a variant of Cantor's intersection theorem.
Question: How disconnected can the limit set $L$ be, if $\gamma$ is unbounded? Can $L$ have infinitely/uncountably many connected components? Can $L$ be totally disconnected (while having at least two points)?
I am almost sure that the answer is "no" to the last question.
What I know:
If $L$ contains at least two points $p\neq p'$, then $L$ is uncountable and has no isolated points. In fact for any $0<\varepsilon<d(p,p')$, I can find a point in $L \cap \partial B_{\varepsilon}(p)$.
But how "badly" can these points behave?
Secondary question: Can I find at least some (bounded?) open neighbourhood $U$ of $p$, such that there is a (path-)connected subset of $L$ containing $p$ and a point in $\partial U$?
It seems like it should follow from:
In the one-point compactification $\dot{\mathbb{R}}^n$ of $\mathbb{R}^n$, the limit set of an unbounded $\gamma$ is $L \cup \{\infty\}$ and this set is connected.
This means every point in $L$ is "connected to $\infty$".
 A: I think I can find a path in ${\mathbb R}^2$ such that $L=C\times [0, \infty )$,  where $C$ is Cantor's set, and hence $L$ has uncountably
many connected components.
Just take a path that attempts to "surround" this set closer
and closer in successive passes.
I hope this is a clear enough description but I can provide more details upon request.

EDIT:
Regarding the second part of the first question (does this post have too many questions?),  the answer is again no.
Since $L\cup \{\infty \}$ is compact and connected,  the question may be restated as:

Can a Hausdorff,  compact,  connected space $X$ with two or more points become totally disconnected upon removal  of a single point?

To see why not suppose otherwise, so there is $x_0$ in $X$ such that $Y:= X\setminus\{x_0\}$ is totally disconnected.  Since $Y$
is open in $X$, it is locally compact, and hence zero-dimensional, that is, its topology has a basis of clopen sets.
Pick any point $y$ in $Y$ and let $K$ be a compact neighborhood of
$y$ (relative to the topology of $Y$).   Using that $Y$ is zero dimensional, choose a clopen neighborhood $C$ of
$y$, with $C\subseteq K$, so that $C$ is compact and open.
Consequently $C$ is compact and open as a subset of $X$,  as well.  However $C\neq X$, because $x_0\notin C$, and $C\neq \emptyset$,
because $y\in C$. Therefore $X$ is disconnected, contradicting the hypothesis.
