Thread:

1. Measure Theory

Quick question: Does the sigma-algebra generated by a countable number of sets have a countable number of elements?

2. It can be shown that any infinite sigma algebra has cardinality greater than or equal to that of the continuum.

3. Thanks, but how? I tried proving that you can't have a bijection from the natural numbers to the sigma-algebra, with no sucess...

4. Given a countably infinite $\sigma$-algebra $\Sigma$ over an infinite set X, let $\{E_n\}_{n\in\mathbb{N}}$ be an enumeration of $\Sigma$. Wlog, we can take these to be disjoint. Then, for every binary sequence, we may produce another set in $\Sigma$. Namely, consider a sequence $b=b_1b_2b_3\ldots$, and let $F_b=\bigcup_{b_i=1}E_i$.

5. Originally Posted by Tikoloshe
Given a countably infinite $\sigma$-algebra $\Sigma$ over an infinite set X, let $\{E_n\}_{n\in\mathbb{N}}$ be an enumeration of $\Sigma$. Wlog, we can take these to be disjoint. Then, for every binary sequence, we may produce another set in $\Sigma$. Namely, consider a sequence $b=b_1b_2b_3\ldots$, and let $F_b=\bigcup_{b_i=1}E_i$.
This is not immediate. I don't think it works quite that simply, there is no guarantee that two sequences will give different sets.

You probably need to use the axiom of choice. Let X be a set and $\mathcal{F}$ an infinite sigma algebra over X. Then define the equivalence relation ~ as x~y if and only if $x \in A \in \mathcal{F} \iff y \in A$. Then each equivalence class E(x) is measurable, as it is the countable intersection of sets in the sigma-algebra containing x.

Let I be the set of representatives from each equivalence class (Axiom of Choice here). Then $X= \bigcup_{x \in I} E(x)$ is a disjoint union. Hence if we suppose that $\mathcal{F}$ is countable then I must either be finite or countable. Thus we have an injection from $\{0,1\}^I$ to $\mathcal{F}$ as we can pick an element of $\mathcal{F}$ by taking $\bigcup_{x\in S \subset I}E(x)$ (as they are equivalence classes, you do have a guarantee of getting distinct sets).

This is impossible as $|\{0,1\}^I|> |I|$ if I is countable or finite.

6. Originally Posted by Focus
This is not immediate. I don't think it works quite that simply, there is no guarantee that two sequences will give different sets.
I don’t understand your objection. If $\{E_i\}_{i\in\mathbb{N}}$ is a countable sequence of disjoint, nonempty sets, then using different index sets, the unions over the $E_i$ will be distinct sets. I.e., if $A,B\subset\mathbb{N}$ where $A\neq B$, then $\bigcup_{i\in A}E_i\neq\bigcup_{i\in B}E_i$. Given an arbitrary countable sequence in a $\sigma$-algebra, you can of course make a new sequence of pairwise disjoint, nonempty sets (without AC).

Originally Posted by Focus
You probably need to use the axiom of choice.
I don’t think AC is necessary for the proof (c.f. mine above). Also, here is another proof which takes a different approach.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 29, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9290536642074585, "perplexity_flag": "head"}