# topological condition for a set to be uncountable

###### Theorem.

A nonempty compact^{} Hausdorff space with no isolated points^{} is uncountable.

###### Proof.

Let $X$ be a nonempty compact Hausdorff space with no isolated points. To each finite $0,1$-sequence $\alpha $ associate a point ${x}_{\alpha}$ and an open neighbourhood ${U}_{\alpha}$ as follows. First, since $X$ is nonempty, let ${x}_{0}$ be a point of $X$. Second, since ${x}_{0}$ is not isolated, let ${x}_{1}$ be another point of $X$. The fact that $X$ is Hausdorff implies that ${x}_{0}$ and ${x}_{1}$ can be separated by open sets. So let ${U}_{0}$ and ${U}_{1}$ be disjoint open neighborhoods of ${x}_{0}$ and ${x}_{1}$ respectively.

Now suppose for induction^{} that ${x}_{\alpha}$ and a neighbourhood ${U}_{\alpha}$ of ${x}_{\alpha}$ have been constructed for all $\alpha $ of length less than $n$. A $0,1$-sequence of length $n$ has the form $(\alpha ,0)$ or $(\alpha ,1)$ for some $\alpha $ of length $n-1$. Define ${x}_{(\alpha ,0)}={x}_{\alpha}$. Since ${x}_{(\alpha ,0)}$ is not isolated, there is a point in ${U}_{\alpha}$ besides ${x}_{(\alpha ,0)}$; call that point ${x}_{(\alpha ,1)}$. Now apply the Hausdorff property to find disjoint open neighbourhoods ${U}_{(\alpha ,0)}$ and ${U}_{(\alpha ,1)}$ of ${x}_{(\alpha ,0)}$ and ${x}_{(\alpha ,1)}$ respectively. The neighbourhoods ${U}_{(\alpha ,0)}$ and ${U}_{(\alpha ,1)}$ can be chosen to be proper subsets^{} of ${U}_{\alpha}$. Proceed by induction to find an ${x}_{\alpha}\in {U}_{\alpha}$ for each finite $0,1$-sequence $\alpha $.

Now define a function $f:{2}^{\omega}\to X$ as follows. If $\alpha $ is
eventually zero, put $f(\alpha )={x}_{\alpha}$. Otherwise, consider the sequence
$({x}_{({\alpha}_{0})},{x}_{({\alpha}_{0},{\alpha}_{1})},{x}_{({\alpha}_{0},{\alpha}_{1},{\alpha}_{2})},\mathrm{\dots})$ of points in $X$. Since $X$ is compact and Hausdorff, it is closed and limit point compact, so the sequence has a limit point^{} in $X$. Let $f(\alpha )$ be such a limit point. Observe that for each finite prefix $({\alpha}_{0},\mathrm{\dots},{\alpha}_{n})$ of $\alpha $, the point $f(\alpha )$ is in ${U}_{({\alpha}_{0},\mathrm{\dots},{\alpha}_{n})}$.

Suppose $\alpha $ and $\beta $ are distinct sequences in ${2}^{\omega}$. Let $n$ be the first position where ${\alpha}_{n}\ne {\beta}_{n}$. Then $f(\alpha )\in {U}_{({\alpha}_{0},\mathrm{\dots},{\alpha}_{n})}$ and $f(\beta )\in {U}_{({\beta}_{0},\mathrm{\dots},{\beta}_{n})}$, and by construction ${U}_{({\alpha}_{0},\mathrm{\dots},{\alpha}_{n})}$ and ${U}_{({\beta}_{0},\mathrm{\dots},{\beta}_{n})}$ are disjoint. Hence $f(\alpha )\ne f(\beta )$, implying that $f$ is an injective function. Since the set ${2}^{\omega}$ is uncountable and $f$ is an injective function from ${2}^{\omega}$ into $X$, $X$ is also uncountable. ∎

###### Corollary.

The set $\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{]}$ is uncountable.

###### Proof.

Being closed and bounded^{}, $[0,1]$ is compact by the Heine-Borel Theorem; because $[0,1]$ is a
subspace^{} of the Hausdorff space $\mathbb{R}$, it too is Hausdorff; finally, since $[0,1]$ has no isolated points, the preceding theorem implies that it is uncountable.
∎

Title | topological condition for a set to be uncountable |
---|---|

Canonical name | TopologicalConditionForASetToBeUncountable |

Date of creation | 2013-03-22 16:15:15 |

Last modified on | 2013-03-22 16:15:15 |

Owner | mps (409) |

Last modified by | mps (409) |

Numerical id | 15 |

Author | mps (409) |

Entry type | Theorem |

Classification | msc 54D10 |

Classification | msc 54A25 |

Classification | msc 54D30 |