# local dimension of a locally Euclidean space

Let $X$ be a locally Euclidean space. Recall that the local dimension^{} of $X$ in $y\in X$ is a natural number^{} $n\in \mathbb{N}$ such that there is an open neighbourhood $U\subseteq X$ of $y$ homeomorphic to ${\mathbb{R}}^{n}$. This number is well defined (please, see parent object for more details) and we will denote it by ${\mathrm{dim}}_{y}X$.

Proposition^{}. Function $f:X\to \mathbb{N}$ defined by $f(y)={\mathrm{dim}}_{y}X$ is continuous^{} (where on $\mathbb{N}$ we have discrete topology).

Proof. It is enough to show that preimage^{} of a point is open. Assume that $n\in \mathbb{N}$ and $y\in X$ is such that $f(y)=n$. Then there is an open neighbourhood $U\subseteq X$ of $y$ such that $U$ is homeomorphic to ${\mathbb{R}}^{n}$. Obviously for any $x\in U$ we have that $U$ is an open neighbourhood of $x$ homeomorphic to ${\mathbb{R}}^{n}$. Therefore $f(x)=n$, so $U\subseteq {f}^{-1}(n)$. Thus (since $y$ was arbitrary) we’ve shown that around every point in ${f}^{-1}(n)$ there is an open neighbourhood of that point contained in ${f}^{-1}(n)$. This shows that ${f}^{-1}(n)$ is open, which completes^{} the proof. $\mathrm{\square}$

Corollary. Assume that $X$ is a connected^{}, locally Euclidean space. Then local dimension is constant, i.e. there exists natural number $n\in \mathbb{N}$ such that for any $y\in X$ we have

$${\mathrm{dim}}_{y}X=n.$$ |

Proof. Consider the mapping $f:X\to \mathbb{N}$ such that $f(y)={\mathrm{dim}}_{y}X$. Proposition shows that $f$ is continuous. Therefore $f(X)$ is connected, because $X$ is. But $\mathbb{N}$ has discrete topology, so there are no other connected subsets then points. Thus there is $n\in \mathbb{N}$ such that $f(X)=\{n\}$, which completes the proof. $\mathrm{\square}$

Remark. Generally, local dimension need not be constant. For example consider ${X}_{1},{X}_{2}\subseteq {\mathbb{R}}^{3}$ such that

$${X}_{1}=\{(x,0,0)|x\in \mathbb{R}\}\mathit{\hspace{1em}\hspace{1em}}{X}_{2}=\{(x,y,1)|x,y\in \mathbb{R}\}.$$ |

One can easily show that $X={X}_{1}\cup {X}_{2}$ (with topology^{} inherited from ${\mathbb{R}}^{3}$) is locally Euclidean, but ${\mathrm{dim}}_{(0,0,0)}X=1$ and ${\mathrm{dim}}_{(1,1,1)}X=2$.

Title | local dimension of a locally Euclidean space |
---|---|

Canonical name | LocalDimensionOfALocallyEuclideanSpace |

Date of creation | 2013-03-22 18:55:34 |

Last modified on | 2013-03-22 18:55:34 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 7 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 53-00 |