Is $[1,\frac{3}{2})$ open? 
Is $[1,\frac{3}{2})$ open in the subspace topology of $A=[1,2]\cup \{\frac{n-1}{n} : n \in \mathbb{N}\}$ inherited from $\mathbb{R}$ ?

I think it not open, I tried to write it as a union of open sets, but couldn't find any candidate.
 A: No, the set is not open. Remember the definition of "open" in the subspace topology: a set $B$ is open in $A$ if there exists an open subset $\mathcal{U} \subseteq \mathbb{R}$ (open in terms of the Euclidean topology on $\mathbb{R}$) such that $B = A \cap \mathcal{U}$. To show that $[1, \frac{3}{2})$ is not open, we must show that no such open set exists.
If such a $\mathcal{U}$ existed, it would have to include the point $1$. By the definition of the Euclidean topology, there must exist some $\varepsilon > 0$ such that $(1 - \varepsilon, 1 + \varepsilon) \subseteq \mathcal{U}$. But then, since $\frac{n - 1}{n} \to 1$ and $\frac{n - 1}{n} < 1$, there must exist some $N$ such that
$$n \ge N \implies 1 - \varepsilon < \frac{n - 1}{n} < 1 \implies \frac{n - 1}{n} \in \mathcal{U} \cap A.$$
But, $\frac{n - 1}{n} \notin [1, \frac{3}{2}) = \mathcal{U} \cap A$ by assumption. This is a contradiction, so the set isn't open.
A: Hint:
After your edit, note $[1, \tfrac 32 )^c = \{\tfrac{n-1}{n} \mid n \in \mathbb{N}\} \cup [\tfrac 32, 2].$
Recall a set is open if and only if its complement is closed. Does the complement contain all of its limit points?
