Let $C$ be a compact totally disconnected infinite subset of $\big\{(x,y)\in \Bbb R^2: x\geq 0\big\}$. So, $C$ is homeomorphic to a subspace of Cantor Space. Let $C^\dagger:=\big\{(-x,y)\in \Bbb R^2:(x,y)\in C\big\} $.
Is $\Bbb R^2\backslash C$ homeomorphic to $\Bbb R^2\backslash (C\cup C^\dagger)$?
When $C\subseteq[0,1]$ itself is the Cantor Set then I guess $\Bbb R^2\backslash C\cong \Bbb R^2\backslash (C\cup C^\dagger)$.
$\textbf{Context:}$ Note that the space of ends of $\Bbb R^2\backslash C$ is $C\sqcup \{\infty\}$, so the question is equivalent to saying that doubling the ends doesn't alter the space.
If the above question has a negative answer in general, then under what extra conditions on $C$ the same question has an affirmative answer?
If $C = \{1\} \times \bigl(\{0\} \cup \{1/n \mid n \in \mathbb N\} \bigr)$ then $C \cup C^\dagger$ is not homeomorphic to $C$, but if two spaces are homeomorphic then their end spaces are homeomorphic as well.
In general, your question is equivalent to asking when $C \cup C^\dagger$ is homeomorphic to $C$. This is true when $C$ is the whole Cantor set, and I think it is generally false when $C$ is homeomorphic to a countable ordinal $\alpha$ for which the $\alpha ^{\text{th}}$ derived set is empty, such as the counterexample I gave above: the least such $\alpha$ will be a successor ordinal, and the $\alpha-1^\text{st}$ derived set will be finite; the cardinality of that set is doubled when you replace $C$ by $C \cup C^\dagger$. There's probably some more true examples, e.g. when the first derived set of $C$ is a Cantor set.