Closed set কী?

Example 12.7 (The sequential test, both ways)

Closed: $F = \Z \cup \bigl\{n + \frac1n : n \geq 2\bigr\}$. Let $u_k \in F$ with $u_k \to \ell$. The window $\intcc{\ell - 1}{\ell + 1}$ contains only finitely many points of $F$ (finitely many integers, finitely many $n + \frac1n$), and beyond some rank all $u_k$ lie in it: the sequence then takes finitely many values, and, converging, is eventually constant (as in Exercise 12.3): $\ell \in F$. Closed — even though $F$ contains pairs of points at distance $\frac1n$, arbitrarily close.

Not closed: $G = \bigl\{\frac1m + \frac1n : m, n \in \N^*\bigr\}$. The sequence $\frac1n + \frac1n \in G$ tends to $0$, and $0 \notin G$ (a sum of two positive terms): the sequential test fails, $G$ is not closed. Interestingly, each $\frac1m$ does belong to $\overline G \cap G$: indeed $\frac1m = \frac{1}{m+1} + \frac{1}{m(m+1)} \in G$. The closing insight: to prove closedness, control all convergent sequences at once (usually via a local finiteness or a closed-formula argument); to disprove it, one well-chosen escaping sequence suffices — the asymmetry makes the negative direction the easy one, and the counterexamples of this chapter all have this one-line shape.

Example 12.8

Segments $\intcc{a}{b}$, half-lines $\intco{a}{+\infty}$, finite sets, $\Z$ (a convergent sequence of integers is eventually constant) are closed. $\intoc{0}{1}$ is neither open (fails at $1$) nor closed ($\frac 1n \to 0 \notin$ the set): most sets are neither. $\R$ and $\emptyset$ are both open and closed — and they are the only such subsets of $\R$ (Exercise 12.9).

Example 12.9 (An open set assembled from infinitely many pieces)

$\R \setminus \Z = \bigcup_{n \in \Z} \intoo{n}{n+1}$: an infinite union of open intervals, open by Proposition 12.3 — so $\Z$ is closed with no sequential argument needed. Note the division of labor in the stability rules: unions of open sets may be arbitrary (each point only needs its own certificate, supplied by the one set containing it), while intersections must stay finite (certificates must be intersected, and infinitely many radii can shrink to nothing). Exercise 12.10 will show this example is the general shape: every open subset of $\R$ is a countable disjoint union of open intervals.