# T1 is hereditary

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 $y$.