Connectedness is not weakly hereditary

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

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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.