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Let X be a connected metric space and U is a subset of X

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1. Connectedness of a metric space

A metric (topological) space X is disconnected if it is the union of two disjoint nonempty open subsets. Otherwise, X is connected.

Theorem 1.1. Let X be a connected metric space and U is a subset of X. Assume that (1) U is nonempty.

(2) U is closed.

(3) U is open.

Then U = X.

Proof. Suppose U 6= X. Then V = X \ U is nonempty. Since U is closed, V = X \ U is open. Moreover, U ∩ V = ∅ and X = U ∪ V is a disjoint union of nonempty open subsets of X. This implies that X is disconnected which leads to contradiction to the assumption

that X is connected. 

Let X be a metric (topological) space and A be a subset of X. Then A is also a metric (topological) space of A. Every open set in A is of the form U ∩ A for some open set U of X.

We say that A is a (dis)connected subset of X if A is a (dis)connected metric (topological) space.

Theorem 1.2. Let X and A be as above. Then A is disconnected if and only if there exist open sets U, V in X so that

(1) U ∩ V ∩ A = ∅ (2) A ∩ U 6= ∅ (3) A ∩ V 6= ∅ (4) A ⊆ U ∩ V.

Proof. Suppose A is disconnected. Then A = U0 ∪ V0, where U0, V0 are disjoint open nonempty subsets in A. Since A is a subspace of X, there are open sets U, V so that U0 = U ∩ A and V0 = V ∩ A. Since A = U0∪ V0, A ⊆ U ∪ V. Since U0 and V0 are nonempty, U ∩ A and V ∩ A are nonempty. Moreover. U ∩ V ∩ A = U0∩ V0 is empty.

Conversely, suppose there are open subsets U, V obeying (1)-(4). Take U0 = U ∩ A and V0 = V ∩ A. Then U0, V0 are nonempty disjoint open subsets of A so that A = U0 ∪ V0. Hence A is disconnected.



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