Connectedness is not weakly hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., connected space) not satisfying a topological space metaproperty (i.e., weakly hereditary property of topological spaces).
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Statement

It is possible to have a nonempty topological space $X$ and a nonempty closed subset $A$ of $X$ such that:

• $X$ is a connected space
• $A$ is not a connected space under the subspace topology from $X$.

Related facts

• Connectedness is not hereditary includes more motivating discussion, the examples here are a subset of the examples on that page.

Proof

Example using finite topological spaces

For a counterexample, $A$ must have at least two points, because the unique one-point space is connected. Therefore, $X$ must have at least three points. Below is one such example:

$X=\{-1,0,1\}$

with the topology defined as follows: the open subsets are:

$\{\},\{0\},\{-1,0\},\{0,1\},\{-1,0,1\}$

Or equivalently, the closed subsets are:

$\{\},\{-1\},\{1\},\{-1,1\},\{-1,0,1\}$

Clearly, $X$ is connected: the only nonempty closed subset containing $0$ is all of $X$, and therefore $X$ cannot be expressed as a union of two disjoint nonempty open subsets. In fact, framed more strongly, $X$ is an irreducible space.

Consider $A$ to be the subset $\{-1,1\}$ of $X$ with the subspace topology. $A$ has a discrete topology, and in particular, is a union of disjoint closed subsets $\{-1\}$ and $\{1\}$. Therefore, it is not connected.

Basically, the point $0$ serves the role of connecting the space, and removing it disconnects the space.

Example using the real line and a finite subset

Consider the following example:

• $X=\mathbb {R}$ is the set of real numbers endowed with the usual Euclidean topology.
• $A=\{-1,1\}$ is a subset of size two.

$X$ is connected. $A$ is discrete in the subspace topology. Explicitly, for instance, $\{-1\}$ is the intersection of $A$ with the open subset $(-\infty ,0)$ of $X$, hence is open in $A$, and similarly $\{1\}=A\cap (0,\infty )$ and hence is open in $A$.