# subspace topology in a metric space

###### Theorem 1.

Suppose $X$ is a topological space^{} whose topology is induced by a
metric $d$, and suppose $Y\mathrm{\subseteq}X$ is a subset.
Then the subspace topology in $Y$ is the same as the metric topology^{}
when by $d$ restricted to $Y$.

Let ${d}^{\prime}:Y:Y\to \mathbb{R}$ be the restriction of $d$ to $Y$, and let

${B}_{r}(x)$ | $=$ | $$ | ||

${B}_{r}^{\prime}(x)$ | $=$ | $$ |

The proof rests on the identity

$${B}_{r}^{\prime}(x)=Y\cap {B}_{r}(x),x\in Y,r>0.$$ |

Suppose $A\subseteq Y$ is open in the subspace topology of $Y$, then $A=Y\cap V$ for some open $V\subseteq X$. Since $V$ is open in $X$,

$$V=\cup \{{B}_{{r}_{i}}({x}_{i}):i=1,2,\mathrm{\dots}\}$$ |

for some ${r}_{i}>0$, ${x}_{i}\in X$, and

$A$ | $=$ | $\cup \{Y\cap {B}_{{r}_{i}}({x}_{i}):i=1,2,\mathrm{\dots}\}$ | ||

$=$ | $\cup \{{B}_{{r}_{i}}^{\prime}({x}_{i}):i=1,2,\mathrm{\dots}\}.$ |

Thus $A$ is open also in the metric topology of ${d}^{\prime}$. The converse direction is proven similarly.

Title | subspace topology in a metric space |
---|---|

Canonical name | SubspaceTopologyInAMetricSpace |

Date of creation | 2013-03-22 15:17:44 |

Last modified on | 2013-03-22 15:17:44 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 5 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 54B05 |