# closed set in a subspace

In the following, let $X$ be a topological space^{}.

###### Theorem 1.

Suppose $Y\mathrm{\subseteq}X$ is equipped with the subspace topology,
and $A\mathrm{\subseteq}Y$.
Then $A$ is closed (http://planetmath.org/ClosedSet) in $Y$ if and only if
$A\mathrm{=}Y\mathrm{\cap}J$ for some closed set^{} $J\mathrm{\subseteq}X$.

###### Proof.

If $A$ is closed in $Y$,
then $Y\setminus A$ is open (http://planetmath.org/OpenSet) in $Y$,
and by the definition of the subspace topology,
$Y\setminus A=Y\cap U$ for some open $U\subseteq X$.
Using properties of the set difference^{} (http://planetmath.org/SetDifference),
we obtain

$A$ | $=$ | $Y\setminus (Y\setminus A)$ | ||

$=$ | $Y\setminus (Y\cap U)$ | |||

$=$ | $Y\setminus U$ | |||

$=$ | $Y\cap {U}^{\mathrm{\complement}}.$ |

On the other hand, if $A=Y\cap J$ for some closed $J\subseteq X$, then $Y\setminus A=Y\setminus (Y\cap J)=Y\cap {J}^{\mathrm{\complement}}$, and so $Y\setminus A$ is open in $Y$, and therefore $A$ is closed in $Y$. ∎

###### Theorem 2.

Suppose $X$ is a topological space, $C\mathrm{\subseteq}X$ is a closed set equipped with the subspace topology, and $A\mathrm{\subseteq}C$ is closed in $C$. Then $A$ is closed in $X$.

###### Proof.

This follows from the previous theorem: since $A$ is closed in $C$, we have $A=C\cap J$ for some closed set $J\subseteq X$, and $A$ is closed in $X$. ∎

Title | closed set in a subspace |
---|---|

Canonical name | ClosedSetInASubspace |

Date of creation | 2013-03-22 15:33:32 |

Last modified on | 2013-03-22 15:33:32 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 9 |

Author | yark (2760) |

Entry type | Theorem |

Classification | msc 54B05 |