Let $X\subseteq \mathbb{R}^{n} $ be open and $$[x]_{\sim} = \{y \in X \mid \text{there exists a continuous path from }x \text{ to }y \text{ in X}\}.$$ I want to show that $[x]_{\sim}$ is closed in/wrt. $X$.
My attempt:
Suppose for the sake of contradiction there existed a sequence $(x_{n})_{n\in\mathbb{N}} \subseteq [x]_{\sim}$ s.t. $x_{n} \to x^{*} \not\in [x]_{\sim}$. Consider now the function \begin{align*} f\colon [0, 1] \to X, \quad f(t) = \begin{cases} x^{*}, & t = 0 \\ l_{n}(m_{n}(t)), & t \in \left[\frac{1}{n}, \frac{1}{n-1}\right] \end{cases} \end{align*} with \begin{align*} &l_{n}(t) = t \cdot x_{n} + (1 - t)\cdot x_{n-1}\quad \quad \text{(line segment joining }x_{n} \text{ and }x_{n-1}) \\[5pt] &m_{n}(t) = \left(t - \frac{1}{n}\right) \cdot (n^{2} - n) \quad \quad \text{(naturally map } [1 / n, 1 / (n-1)] \text{ to } [0, 1]) .\end{align*} Since $x_{1}\in [x]_{\sim}$ there exists a continous path from $x$ to $x_{1}$ which can be joined with the continous path $f$ from $x_{1}$ to $x^{*}$. However, this means that $x^{*} \in [x]_{\sim}$, a contradiction!
Firstly, does my attempt look correct? Secondly, what would be other ways to prove the above?
I assume that we are working in the Topology of the subset $A$:
if $[x]_∼ =A$ we are done. Suppose for the sake of contradiction exists $y\in A\backslash [x]_∼$ such that $B(y,1/n)\cap [x]_∼\neq \emptyset\;\; \forall n\in\mathbb N\;\;$ i.e $\;\;A\backslash[x]_∼$ isn't open. Hence exists $\{x_n\}\subset [x]_∼ $ which converge to $y$. Since $A$ is open, for $n_0$ big enough $B(y,1/n)\subset A, \forall n\geq n_0$. So the function: $$f:[0,1]\rightarrow B(y,1/n_0)\subset A\;\;\; \text{such that }\; f(t)= (1-t)x_{n_0}+ty$$ is well-defined and continuous in $A$. So the function we are looking for is the juxtaposition of the continuous function that connects $x$ to $x_0$ with $f$. (hence $y$ stay in $[x]_∼$ and this is a contradiction, so $A\backslash[x]_∼$ is open i.e. $[x]_∼$ is closed)