Cardinality of Infinite Sigma-Algebra is at Least Cardinality of Continuum

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Theorem

Let $\mathcal M$ be an infinite $\sigma$-algebra on a set $X$.


Then $\mathcal M$ is has cardinality at least that of the cardinality of the continuum $\mathfrak c$:

$\operatorname{card}\left({\mathcal M }\right) \ge \mathfrak c$


Corollary

Let $\mathcal M$ be an infinite $\sigma$-algebra on a set $X$.


Then $\mathcal M$ is uncountable.


Proof

We first show that $X$ is infinite.

By the definition of a $\sigma$-algebra, $\mathcal M$ is a subset of $\mathcal P(X)$.

Were $X$ finite, by Cardinality of Power Set of Finite Set, the cardinality of $\mathcal M$ would be at most $2^{\operatorname{card}(X)}$

As $2^{\operatorname{card}(X)}$ is finite if $X$ is finite, $X$ must be infinite by the assumption that $\mathcal M$ is infinite.

By the definition of $\sigma$-algebra, $X \in \mathcal M$.

Also, by Sigma-Algebra Contains Empty Set, $\varnothing \in \mathcal M$.

Construct a countable collection of sets $\left\langle {F_1, F_2, F_3, \ldots } \right\rangle_{\N}$ as follows:

$F_1 = \varnothing$
$F_2 = X$

We can continue this construction using the axiom of choice:

$F_3 = \text{ any set in } \mathcal M \setminus \left \{ { \varnothing, X } \right \}$
$\ \ \vdots$
$F_n = \text{ any set in } \mathcal M \setminus \left \{ {F_1, F_2, \ldots, F_{n-1} } \right \}$
$\ \ \vdots$

Consider an arbitrary $S \in \bigcup_{k \mathop \in \N} F_k$

Then $S \in F_k$ for some $F_k$.

By the well-ordering principle, there is a smallest such $k$.

Then for any $j < k$, $S \notin F_j$

Thus the sets in $\langle F_i \rangle$ are disjoint.

Recall $\mathcal M$ is infinite.

Then by Relative Difference between Infinite Set and Finite Set is Infinite, $\mathcal M \setminus \left\{ {F_1, F_2, \cdots , F_{k-1}} \right\}$ is infinite.

Thus this process can continue indefinitely, choosing an arbitary set in $\mathcal M$ that hasn't already been chosen for an earlier $F_k$.

By the definition of a $\sigma$-algebra, $\displaystyle \bigsqcup_{i \mathop \in \N} F_i$ is measurable.

By the definition of an indexed family, $\langle F_i \rangle_{i \mathop \in \N}$ corresponds to a mapping $\iota: \N \hookrightarrow \bigsqcup_{i \mathop \in \N} F_i$

Such a mapping $\iota$ is injective because:

  • each $F_i$ is disjoint from every other,
  • each $F_i$ contains a distinct $x \in X$,
  • $X$ has infinitely many elements

Define:

$\iota^*: \mathcal P \left({\N}\right) \to \mathcal M$:
$\iota^*\left({N}\right) = \bigsqcup_{i \mathop \in N} F_i$

That is, for every $N \subseteq \N$, $\iota^*\left({N}\right)$ corresponds to a way to select a countable union of the sets in $\langle F_i \rangle$.

Because any distinct $F_i, F_j$ are disjoint, any two distinct ways to create a union $S \mapsto \bigsqcup_{i \mathop \in S} F_i$ will result in a different union $\iota^*(S)$.

Thus $\iota^*$ is an injection into $\mathcal M$.

Then the cardinality of $\mathcal M$ is at least $\mathcal P \left({\N}\right)$.

From Continuum equals Cardinality of Power Set of Naturals, $\R \sim \mathcal P \left({\N}\right)$.

Thus $\mathcal M$ is uncountable and $\operatorname{card}\left({\mathcal M}\right) \ge \mathfrak c$


$\blacksquare$


Proof of Corollary

Follows from the main result, as the real numbers are uncountable.

$\blacksquare$


Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.


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