$\mathbb{Q}$ as a subspace of $\mathbb{R}$: every component is a single point, no point is isolated, and the space is not locally connected anywhere

Let $\mathbb{Q} = \mathbb{Q}_{\mathbb{R}}$ be the copy 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) with the subspace topology of the usual 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, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ 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). Then:

1. Every component is a single point: $C(q) = \{q\}$ for every $q \in \mathbb{Q}$, so $\mathbb{Q}$ is totally disconnected (Connected components, quasicomponents, and totally disconnected spaces).
2. No point is isolated: $\{q\}$ is not open in $\mathbb{Q}$, so the topology is not discrete (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).
3. $\mathbb{Q}$ is not locally connected at any of its points.
4. The components are closed and not open. They are closed by The components of a space are its maximal connected subsets, they partition it, and each of them is closed and not open by claim 2, so $\mathbb{Q}$ is a space in which every component fails to be clopen.

$\mathbb{Q}$ is countably infinite ($\mathbb{Q}$ is countably infinite) while its complement in $\mathbb{R}$ is uncountable (The irrationals are uncountable); it is the abundance of the complement, not the scarcity of $\mathbb{Q}$, that drives claim 1.

• Total disconnectedness and discreteness come apart here, and the two steps that separate them are step 2.1 and step 2.2. The first says the irrationals block every interval, so no connected set can span two rationals; the second says the rationals themselves are everywhere, so no rational is isolated. A space can be shredded into points without those points being separated.

• Local connectedness fails for a structural reason, not a delicate one. In any totally disconnected space the only candidates for a connected neighbourhood are singletons, so local connectedness at $x$ is equivalent to $\{x\}$ being open. Hence a totally disconnected space is locally connected exactly when it is discrete, and step 3.1 is that observation applied to $\mathbb{Q}$.

• The same argument applies to the irrationals. Nothing in steps 1.1 and 2.1 used countability of $\mathbb{Q}$; only that its complement meets every interval. The irrationals have that property too by [A3], so they are totally disconnected and not discrete as well, and they are uncountable (The irrationals are uncountable).