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 $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$.

• $U$ is open when for every $x\in U$ there is a real $\epsilon >0$ with $N_{\epsilon }(x)\subseteq U$.
• $F$ 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).

$\mathrm{\varnothing }$ and $\mathbb{R}$ are clopen. The condition defining openness quantifies over the elements of the set, so it holds vacuously for $\mathrm{\varnothing }$; 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 $y\in N_{\epsilon }(x)$ and put $\delta :=\epsilon -\mathrm{\mid }y-x\mathrm{\mid }$, which is $>0$ because $y\in N_{\epsilon }(x)$. The nesting property of The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$ gives $N_{\delta }(y)\subseteq N_{\epsilon }(x)$. So every point of $N_{\epsilon }(x)$ 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 $\mathrm{\mid }y-x\mathrm{\mid }<\delta$ then $y>x-\delta \ge x-(x-a)=a$ and $y<x+\delta \le x+(b-x)=b$, so $y\in (a,b)$; hence $N_{\delta }(x)\subseteq (a,b)$.
• $(a,\mathrm{\infty })$: for $x>a$ take $\delta :=x-a>0$; then $\mathrm{\mid }y-x\mathrm{\mid }<\delta$ gives $y>x-\delta =a$.
• $(-\mathrm{\infty },b)$: for $x<b$ take $\delta :=b-x>0$; then $\mathrm{\mid }y-x\mathrm{\mid }<\delta$ gives $y<x+\delta =b$.
• $(-\mathrm{\infty },\mathrm{\infty })=\mathbb{R}$: 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 $x\mathrm{\notin }[a,b]$ 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 $y\mathrm{\notin }[a,b]$. If $x>b$, take $\delta :=x-b>0$; every $y\in N_{\delta }(x)$ has $y>x-\delta =b$, hence $y\mathrm{\notin }[a,b]$. So $\mathbb{R}\setminus [a,b]$ is open.
• $[a,\mathrm{\infty })$: its complement is $(-\mathrm{\infty },a)$, which is open by the previous paragraph.
• $(-\mathrm{\infty },b]$: its complement is $(b,\mathrm{\infty })$, which is open.
• $(-\mathrm{\infty },\mathrm{\infty })=\mathbb{R}$: its complement is $\mathrm{\varnothing }$, 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: $\mathrm{\varnothing }$ 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: $\mathrm{\varnothing }$ 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 $\epsilon$ at each point, and that $\epsilon$ 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 $\epsilon$ 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.