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:

• $X = \R$ 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 $A$.