The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$"

Give $\mathbb{R}$ its usual topology, the metric topology of $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and let $E\subseteq \mathbb{R}$ carry the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then $E$ is a connected subset of $\mathbb{R}$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) if and only if $E$ is order-convex (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, The order topology of a linearly ordered set, with the open rays as a subbasis; order-convex sets, order-density, the least upper bound property, and linear continua), that is

$x,z\in E\text{ and }x\le w\le z⟹w\in E\mathrm{.}$

In particular each of the nine interval forms of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length is connected, and so are $\mathrm{\varnothing }$ and every singleton.

What has to be checked, and it is not the mathematics. The characterisation itself is the published A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval, which is stated for the connectedness of Separated sets, disconnection, and connected subset of $\mathbb{R}$ — a condition phrased with the open sets of Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen and the closure of Interior, closure, boundary and exterior of a subset of $\mathbb{R}$. The present corollary says the same thing for the connectedness of Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets in the topological space $\mathbb{R}$. What licenses the transport is that the two descriptions of "open in $\mathbb{R}$" are the same condition word for word, which is unfolded in the proof rather than quoted.

• Nothing here re-proves the hard direction. The mathematical content — that order-convexity is exactly connectedness on the line — is A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval, whose proof uses the least upper bound property. This corollary only checks that the vocabulary of the general definition and the vocabulary of the real-line definition denote the same conditions, so that the published theorem may be quoted afterwards without a translation step each time.

• "Interval" is read as "order-convex" throughout. The published theorem records that the converse classification — that every order-convex subset of $\mathbb{R}$ is empty or one of the nine written forms — is not proved, and Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length records the same omission. The statement above is therefore written with order-convexity and not with a list of forms.

• The identification is one sentence and is deliberately not routed through a conventions remark. A dependency edge onto a remark that itself points at material developed further on would mark every consequence of this corollary as resting on later material, which would be false of everything on this page. The computation $B(x,r)=(x-r,x+r)=N_r(x)$ is short enough to carry in the open.