The closure of $X = [0,1) \cup (1,2] \cup \{3\}$ and the closure of its complement For $X = [0,1) \cup (1,2] \cup \{3\}$. I can say that $\overline{X} = [0,3]$ and the $\overline{X^c} = 1$?
 A: No.

*

*$\overline X = [0,2] \cup \{3\}$

*$X^c = (-\infty,0) \cup \{1\} \cup (2,3) \cup (3,\infty)$

*$\overline{X^c} = (-\infty,0) \cup \{1\} \cup (2,\infty)$
A good intuition for whether some element $s$ is in the closure $\overline S$ of a subset $S$ of a metric space is whether you can find a sequence $\{s_n\}_{n \in \Bbb N}$ of elements of $S$ such that $s_n \to s$. If such a sequence $\{s_n\}_{n \in \Bbb N} \subseteq S$ exists such that $s_n \to s$, then $s \in \overline S$.
(This follows readily from the definition of $\overline S$ as $S \cup S'$, where $S'$ denotes the set of limit points. Apply the definition of sequence convergence in a metric space. Note in particular that this means $S \subseteq \overline S$: just take constant sequences in $S$ to justify this or use the original definition.)
So to see why your answers are wrong:

*

*Can you find a sequence in $X$ converging to anything in $(2,3)$? Say, to $2.5$?

*Knowing what $X^c$ is, it should be not difficult to see why you have much more than just $\{1\}$ in the closure - and, again, in particular, $X^c \subseteq \overline{X^c}$. But you can also trivially get a sequence converging to $3$, say the sequence
$$s_n = 2 + 9 \sum_{k=1}^n \left( \frac{1}{10} \right)^k = 2.\underbrace{999\ldots 999}_{n \; \text{dots}}$$
which lets us include that too.

