Let $X = [0,1) \cup \{2\} \subset \mathbb{R}$. Consider two topologies on X. The subspace topology on X induced by the standard topology on $\mathbb{R}$, and the order topology on X induced by the natural order on $\mathbb{R}$. Is 1 a limit point of X with either topology?
I'm struggling with this question because 1 is not contained in X, and therefore is not an element of any of the open sets. I'm not sure if the fact that 1 is not in X disqualifies it from being a limit point. Any pointers are appreciated.
The question with $1$ is nonsense, as $1 \notin X$ as you rightly say. But for $x=2$ the question is relevant: in the subspace topology $2$ is an isolated point, as $\{2\} = (1,3) \cap X$ is open in $X$, while in the order topology $2$ has special neighbourhoods of the form $\langle a,2]$, where $a \in X$ because $2 = \max X$. And as $a$, the left endpoint, must be a point of $X$ itself it must be one of $[0,1)$ but then $a < 1$ and we have that the order-interval $\langle a, 2] = \{x \in X\mid x < a\}$ contains points from $(a,1)$ and so $2$ is indeed a limit point of $X$. In the order topology $X$ is just homeomorphic, even order-isomorphic, to $[0,1]$ using the map $f(x)=x$ for $x< 1$ and $f(1)=2$ as an order isomorphism.
This is probably the intended idea for this problem.