# proof of Tychonoff’s theorem in finite case

(The finite case of Tychonoff^{}’s Theorem is of course a subset of the infinite^{} case,
but the proof is substantially easier, so that is why it is presented here.)

To prove that ${X}_{1}\times \mathrm{\cdots}\times {X}_{n}$ is compact^{}
if the ${X}_{i}$ are compact, it suffices (by induction^{}) to prove that $X\times Y$ is compact
when $X$ and $Y$ are. It also suffices to prove that
a finite subcover can be extracted from every open cover of $X\times Y$
by only the *basis sets* of the form $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$.

###### Proof.

The proof is by the straightforward strategy of composing a finite subcover from a lower-dimensional subcover. Let the open cover $\mathcal{C}$ of $X\times Y$ by basis sets be given.

The set $X\times \{y\}$ is compact, because it is the image of a
continuous^{} embedding^{} of the compact set $X$.
Hence $X\times \{y\}$ has a finite subcover in $\mathcal{C}$: label the subcover
by ${\mathcal{S}}^{y}=\{{U}_{1}^{y}\times {V}_{1}^{y},\mathrm{\dots},{U}_{{k}_{y}}^{y}\times {V}_{{k}_{y}}^{y}\}$.
Do this for each $y\in Y$.

To get the desired subcover of $X\times Y$, we need to pick a finite number
of $y\in Y$. Consider ${V}^{y}={\bigcap}_{i=1}^{{k}_{y}}{V}_{i}^{y}$.
This is a finite intersection^{} of open sets, so ${V}^{y}$ is open in $Y$.
The collection^{} $\{{V}^{y}:y\in Y\}$ is an open covering of $Y$, so pick
a finite subcover ${V}^{{y}_{1}},\mathrm{\dots},{V}^{{y}_{l}}$.
Then ${\bigcup}_{j=1}^{l}{\mathcal{S}}^{{y}_{j}}$ is a finite subcover
of $X\times Y$.
∎

Title | proof of Tychonoff’s theorem in finite case |
---|---|

Canonical name | ProofOfTychonoffsTheoremInFiniteCase |

Date of creation | 2013-03-22 15:26:27 |

Last modified on | 2013-03-22 15:26:27 |

Owner | stevecheng (10074) |

Last modified by | stevecheng (10074) |

Numerical id | 4 |

Author | stevecheng (10074) |

Entry type | Proof |

Classification | msc 54D30 |