# proof of delta system lemma

Since there are only ${\mathrm{\aleph}}_{0}$ possible cardinalities for any element of $S$, there must be some $n$ such that there are an uncountable number of elements of $S$ with cardinality $n$. Let ${S}^{*}=\{a\in S\mid |a|=n\}$ for this $n$. By induction^{}, the lemma holds:

If $n=1$ then there each element of ${S}^{*}$ is distinct, and has no intersection^{} with the others, so $X=\mathrm{\varnothing}$ and ${S}^{\prime}={S}^{*}$.

Suppose $n>1$. If there is some $x$ which is in an uncountable number of elements of ${S}^{*}$ then take ${S}^{**}=\{a\setminus \{x\}\mid x\in a\in {S}^{*}\}$. Obviously this is uncountable and every element has $n-1$ elements, so by the induction hypothesis there is some ${S}^{\prime}\subseteq {S}^{**}$ of uncountable cardinality such that the intersection of any two elements is $X$. Obviously $\{a\cup \{x\}\mid a\in {S}^{\prime}\}$ satisfies the lemma, since the intersection of any two elements is $X\cup \{x\}$.

On the other hand, if there is no such $x$ then we can construct a sequence^{} $$ such that each ${a}_{i}\in {S}^{*}$ and for any $i\ne j$, ${a}_{i}\cap {a}_{j}=\mathrm{\varnothing}$ by induction. Take any element for ${a}_{0}$, and given $$, since $\alpha $ is countable^{}, $$ is countable. Obviously each element of $A$ is in only a countable number of elements of ${S}^{*}$, so there are an uncountable number of elements of ${S}^{*}$ which are candidates for ${a}_{\alpha}$. Then this sequence satisfies the lemma, since the intersection of any two elements is $\mathrm{\varnothing}$.

Title | proof of delta system lemma |
---|---|

Canonical name | ProofOfDeltaSystemLemma |

Date of creation | 2013-03-22 12:55:03 |

Last modified on | 2013-03-22 12:55:03 |

Owner | Henry (455) |

Last modified by | Henry (455) |

Numerical id | 5 |

Author | Henry (455) |

Entry type | Proof |

Classification | msc 03E99 |