Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen

Let $U, F \subseteq \mathbb{R}$, with neighbourhoods as in The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$.

• $U$ is open when for every $x \in U$ there is a real $\varepsilon > 0$ with $N_\varepsilon(x) \subseteq U$.
• $F$ is closed when its complement $\mathbb{R} \setminus F$ is open.
• A set is clopen when it is both open and closed.

The whole of the topology of $\mathbb{R}$ developed on this page rests on this one definition: closedness is defined as openness of the complement, and every other description of a closed set on this page is a theorem (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, A point lies in the closure of $A \subseteq \mathbb{R}$ iff some sequence in $A$ converges to it, so a subset of $\mathbb{R}$ is closed iff it is sequentially closed).

$\varnothing$ and $\mathbb{R}$ are clopen. The condition defining openness quantifies over the elements of the set, so it holds vacuously for $\varnothing$; and for $x \in \mathbb{R}$ one has $N_1(x) \subseteq \mathbb{R}$, so $\mathbb{R}$ is open. Since each of the two is the complement of the other, each is also closed.

Every neighbourhood is open. Let $y \in N_\varepsilon(x)$ and put $\delta := \varepsilon - |y - x|$, which is $> 0$ because $y \in N_\varepsilon(x)$. The nesting property of The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$ gives $N_\delta(y) \subseteq N_\varepsilon(x)$. So every point of $N_\varepsilon(x)$ has a neighbourhood inside it.

The four open forms of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length are open sets. Let $a, b \in \mathbb{R}$.

• $(a,b)$: for $x$ with $a < x < b$, both $x - a > 0$ and $b - x > 0$, so $\delta := \min\{x - a,\ b - x\}$ is a positive real (the minimum of a two-element set of reals exists, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set). If $|y - x| < \delta$ then $y > x - \delta \ge x - (x - a) = a$ and $y < x + \delta \le x + (b - x) = b$, so $y \in (a,b)$; hence $N_\delta(x) \subseteq (a,b)$.
• $(a,\infty)$: for $x > a$ take $\delta := x - a > 0$; then $|y - x| < \delta$ gives $y > x - \delta = a$.
• $(-\infty,b)$: for $x < b$ take $\delta := b - x > 0$; then $|y - x| < \delta$ gives $y < x + \delta = b$.
• $(-\infty,\infty) = \mathbb{R}$: already treated above.

The four closed forms of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length are closed sets. In each case the complement is shown open directly.

• $[a,b]$: if $x \notin [a,b]$ then $x < a$ or $x > b$ by trichotomy (Ordered field). If $x < a$, take $\delta := a - x > 0$; every $y \in N_\delta(x)$ has $y < x + \delta = a$, hence $y \notin [a,b]$. If $x > b$, take $\delta := x - b > 0$; every $y \in N_\delta(x)$ has $y > x - \delta = b$, hence $y \notin [a,b]$. So $\mathbb{R} \setminus [a,b]$ is open.
• $[a,\infty)$: its complement is $(-\infty,a)$, which is open by the previous paragraph.
• $(-\infty,b]$: its complement is $(b,\infty)$, which is open.
• $(-\infty,\infty) = \mathbb{R}$: its complement is $\varnothing$, which is open.

• Open and closed are not opposites, and not exhaustive. A set may be neither: the half-open interval $[0,1)$ is neither open nor closed (FALSE: every subset of $\mathbb{R}$ is either open or closed). A set may be both: $\varnothing$ and $\mathbb{R}$ are clopen. The words are inherited from the interval terminology of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, and the agreement between the two usages is exactly the two lists verified above: an interval called open there is an open set here, and an interval called closed there is a closed set here.

• A clopen set is a disconnection waiting to happen. If $A$ is clopen and both $A$ and $\mathbb{R} \setminus A$ are nonempty, then each of the two is its own closure, so the two are separated in the sense of Separated sets, disconnection, and connected subset of $\mathbb{R}$ and $\mathbb{R} = A \cup (\mathbb{R} \setminus A)$ is a disconnection. Since $\mathbb{R}$ is order-convex it is connected (A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval), so no such $A$ exists: $\varnothing$ and $\mathbb{R}$ are the only clopen subsets of $\mathbb{R}$.

• The half-open forms are the ones the two lists omit, and deliberately so: $[a,b)$ and $(a,b]$ with $a < b$ are neither open nor closed as subsets of $\mathbb{R}$.

• The radius depends on the point. Openness asks for some $\varepsilon$ at each point, and that $\varepsilon$ may shrink to nothing as the point approaches the edge of the set, as the computation for $(a,b)$ shows: there $\delta = \min\{x - a,\ b - x\}$ tends to $0$ as $x$ tends to either endpoint. Asking instead for a single $\varepsilon$ that works simultaneously at every point of the set is a strictly stronger condition, and it is not what is defined here; nothing on this page uses it.