Interior, closure, boundary কী?

Example 12.12

$\overline{\intoo{0}{1}} = \intcc{0}{1}$; $\mathring{\intcc{0}{1}} = \intoo{0}{1}$; $\partial\intoo{0}{1} = \{0, 1\}$. For $A = \{\frac 1n : n \in \N^*\}$: $\overline A = A \cup \{0\}$, $\mathring A = \emptyset$, $\partial A = A \cup \{0\}$. For $\Q$: by density (Theorem 10.14) every real is adherent to $\Q$, so $\overline{\Q} = \R$ while $\mathring{\Q} = \emptyset$ (every interval contains irrationals): the boundary of $\Q$ is all of $\R$.

Example 12.13 (A full anatomy)

Let $A = \intoc{0}{1} \,\cup\, \bigl(\Q \cap \intoo{2}{3}\bigr) \,\cup\, \{4\}$. We compute the three sets of Definition 12.10, piece by piece.

Interior. A point of $\intoo{0}{1}$ has a whole interval inside $A$: interior. The point $1$: every interval around it leaks right of $1$, where $A$ has nothing until $2$: not interior. No point of $\Q \cap \intoo{2}{3}$ is interior (every interval contains irrationals, Theorem 10.14); neither is the isolated $4$. So $\mathring A = \intoo{0}{1}$.

Closure. Limits of points of $A$: all of $\intcc{0}{1}$ ($0 = \lim \frac1n$ with $\frac 1n \in A$); all of $\intcc{2}{3}$ (every real there is a limit of rationals of the interval, density again); and $4$. Nothing else: a point outside $\intcc{0}{1} \cup \intcc{2}{3} \cup \{4\}$ has positive distance to that closed set. So $\overline A = \intcc{0}{1} \cup \intcc{2}{3} \cup \{4\}$.

Boundary. $\partial A = \overline A \setminus \mathring A = \{0, 1\} \cup \intcc{2}{3} \cup \{4\}$.

The closing insight: the three operations act locally — each piece of $A$ contributes according to its own nature (a solid interval keeps its inside, a dense-but-porous piece turns entirely into boundary, an isolated point is pure boundary), and a two-line drawing of $A$ predicts every answer before any proof is written.

Example 12.15 (A closure computed exactly)

Let $G = \bigl\{\frac1m + \frac1n : m, n \in \N^*\bigr\}$ (from Example 12.7). Claim:

($\supseteq$) $\frac1m = \lim_n \bigl(\frac1m + \frac1n\bigr)$ and $0 = \lim_n \frac2n$: adherent by the sequential characterization. ($\subseteq$) Let $x = \lim_k \bigl( \frac{1}{m_k} + \frac{1}{n_k}\bigr)$; order each pair so that $m_k \leq n_k$. If $(m_k)$ is unbounded, a subsequence has $m_k \to \infty$, hence $n_k \to \infty$ too and $x = 0$. Otherwise $(m_k)$ takes finitely many values, one of them, say $m$, infinitely often; along that subsequence $\frac{1}{n_k} \to x - \frac1m$: if $(n_k)$ is bounded it takes some value $n$ infinitely often and $x = \frac1m + \frac1n \in G$; if not, $x = \frac1m$. Every case lands in the announced set. The closing insight: computing a closure is a compactness-style case analysis on indices — bounded index means finitely many values (pigeonhole), unbounded index means a limit escapes — and the answer displays the typical two-layer structure of limit points: the set, its first-generation limits, and their limit $0$.