FALSE: a totally disconnected space carries the discrete topology

False claim: if every connected component of a topological space is a single point (Connected components, quasicomponents, and totally disconnected spaces) then the space carries the discrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

The implication holds in the other direction — a discrete space is totally disconnected, as Connected components, quasicomponents, and totally disconnected spaces shows — and it is that true statement which the false one attempts to reverse.

Witness. The set $\mathbb{Q}$ of rationals inside $\mathbb{R}$, as a subspace of the usual topology (Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable, 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, 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, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Every component of $\mathbb{Q}$ is a single point, and no singleton is open in $\mathbb{Q}$, so the topology is not discrete.

Here $\mathbb{Q}$ denotes the copy ${\mathbb{Q}}_{\mathbb{R}}$ of the rationals inside $\mathbb{R}$ (Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable).

• What separates the two notions. Total disconnectedness forbids large connected pieces; discreteness demands that each point be isolated. In $\mathbb{Q}$ every point is a limit of other points, so nothing is isolated, and yet no two points can be joined inside $\mathbb{Q}$ by an order-convex set, because the irrationals block every interval. Both facts hold simultaneously, and step 1.3 and step 2.1 are exactly the two of them.

• The witness is not exotic. No construction is needed: $\mathbb{Q}$ with its usual topology is the standard example, and the only inputs are the density of the rationals and of the irrationals ([A3]) together with the classification of the connected subsets of the line ([A1]).

• A second, non-countable witness exists. The set of irrationals, and the Cantor set, are totally disconnected and not discrete for the same reason. The Cantor set case is already recorded elsewhere in the library as The Cantor set contains no interval of positive length yet has no isolated point, so every connected subset of it is a single point, which proves that every connected subset of it is a single point while no point of it is isolated.