T1 is hereditary

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This article gives the statement, and possibly proof, of a topological space property (i.e., T1 space) satisfying a topological space metaproperty (i.e., subspace-hereditary property of topological spaces)
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Statement

Any subspace of a T1 space, endowed with the subspace topology, is again a T_1-space.

Definitions used

T1 space

Further information: T1 space

A topological space X is termed T_1 if it satisfies the following equivalent conditions:

  1. For any two distinct points x,y \in X, there exists an open subset U of X such that x \in U, y \notin U
  2. For every point x \in X, the singleton subset \{ x \} is closed in X
  3. For every point x \in X, the intersection of all open subsets of X containing x, is \{ x \}

Subspace topology

Further information: Subspace topology

The subspace topology on a subset A of X is defined in the following equivalent ways:

  1. A subset U of A is open in A iff there exists an open subset V of X such that V \cap A = U.
  2. A subset C of A is closed in A iff there exists a closed subset D of X such that D \cap A = C.

Proof

Proof in terms of two-point definition of T1

Given: A T_1-space X, a subset A

To prove: A is a T_1-space when endowed with the subspace topology

Proof: We need to show that if x \ne y are both points of A, then there exists an open subset of A containing x and not containing y.

Since x,y are distinct points of A, they are also distinct points of X. Since X is a T_1-space, there exists an open subset V of X such that x \in V and y \notin V. Now consider the set U = V \cap A. Then, by definition of subspace topology, U is open in A. Further, since x \in A and x \in V, we have x \in U. Since y \notin V, we have y \notin U. Thus, U is an open subset of X containing x and not containing y.

Proof in terms of points-are-closed definition of T1

Proof in terms of third definition of T1