# find $σ(E)⊂P(X)$ with $X=\left \{ 1,2,3,4 \right \}$ and $E=\left \{ \left \{ 1,2,3\right \} \left \{ 3,4\right \}\right \}$ [closed]

I have this small exercise that I don't know how to solve.

Can someone help me?

Let $$X=\left \{ 1,2,3,4 \right \}$$ and $$E=\left \{ \left \{1,2,3\right \} \left \{ 3,4\right \}\right \}$$.

I want to find $$S= σ(E)⊂P(X)$$

Can someone tell me how to do it?

My thoughts are:

I wanted to use the following steps:

1.Start with an empty set $$S=\emptyset$$. 2.Iterate through each subset $$e \in E$$. 3.For each element $$x$$ in the subset $$e$$, add it to the set $$S$$ if it is not already in the set. 4.repeat step 2 and 3 for all subsets in E. 5.The final set $$S$$ will be the output. For the example provided, $$X=\left { 1,2,3,4 \right }$$ and $$E=\left { \left { 1,2,3\right } ,\left { 3,4\right }\right }$$

$$S=\emptyset$$

For the first subset $${ 1,2,3}$$, we add the elements 1,2,3 to S resulting in $$S=\left {1,2,3 \right }$$. For the second subset $${3,4}$$, we add the element 4 to S resulting in $$S=\left {1,2,3,4 \right }$$.

Am I right?

• It would be good if you could add your thoughts, your ideas and your attempts at solving it. Simply saying "please solve this for me" is usually not well received. Having said this, I'll try to answer your question. Commented Nov 17, 2022 at 11:20
• You have not defined what $\sigma(E)$ means. Until you do no one can answer. Commented Nov 17, 2022 at 11:27
• If it's about generating $\sigma$-algebras then have a look at this answer. Commented Nov 17, 2022 at 11:45

The usual way of computing such small $$\sigma$$-algebras is to simply do all the intersections, unions and complements and add them, until the set is closed under these operations. Thus, as $$\lbrace 3\rbrace = \lbrace 1,2,3\rbrace\cap \lbrace 3,4\rbrace$$, $$\lbrace 4\rbrace = \lbrace 1,2,3\rbrace^c$$, $$\lbrace 1,2 \rbrace =\lbrace 3,4\rbrace^c$$ and $$\lbrace 1,2,4\rbrace=\lbrace4\rbrace\cup\lbrace 1,2\rbrace$$, we know that these sets must lie in $$\sigma(E)$$, along with $$\emptyset$$ and $$X$$ of course.

We now claim that $$\sigma(E)=\mathcal{A}:=\lbrace \emptyset,\lbrace 1,2\rbrace,\lbrace 3\rbrace,\lbrace 4\rbrace,\lbrace 1,2,3\rbrace,\lbrace 1,2,4\rbrace,\lbrace 3,4\rbrace,X\rbrace$$. Clearly $$E\subset \mathcal{A}$$, and by our calculations above, $$\mathcal{A}\subset\sigma(E)$$, so showing that $$\mathcal{A}$$ is a $$\sigma$$-algebra, we would be done.

There are several ways to do this. The simplest is to check that it contains all unions, intersections and complements. This is what I recommend for you to check.

A more elegant, but slightly deeper, way to do it is the following. I write this just for you to see more refined arguments - if you do not understand this, ignore it.

Instead of checking directly it is a $$\sigma$$-algebra, we observe directly from $$E$$ that $$1$$ and $$2$$ can never be seperated using the operations of a $$\sigma$$-algebra, so that e.g. $$\{1\}\notin \sigma(E)$$. Thus $$\sigma(E)\neq P(X)$$. Then, as every finite $$\sigma$$-algebra has cardinality $$2^n$$ and $$\mathcal{A}\subset \sigma(E)$$ but $$\mathcal{A}$$ has the maximal possible cardinality, $$8$$, we must have equality.

The easiest way to do the proof is to show that $$\{1,2\},\{3\},\{4\}\in\sigma(E)$$, and quickly note that these are all the atoms. Then $$\sigma(E)$$ is the $$\sigma$$-algebra generated just by unions of these, i.e. $$\mathcal{A}$$ as defined above.

• Sorry but I didn't undertstand this steps: "$\lbrace 3\rbrace = \lbrace 1,2,3\rbrace\cap \lbrace 3,4\rbrace$, $\lbrace 4\rbrace = \lbrace 1,2,3\rbrace^c$, $\lbrace 1,2 \rbrace =\lbrace 3,4\rbrace^c$ and $\lbrace 1,2,4\rbrace=\lbrace4\rbrace\cup\lbrace 1,2\rbrace$" . Can you explain them in more details? Commented Nov 17, 2022 at 12:36
• We know that $E\subset \sigma(E)$. We also know that $\sigma(E)$ is closed under complement, (countable) union and intersection. Thus, $\sigma(E)$ must contain any such operation applied to the elements of $E$. Further, the same logic holds for the new sets I generate, as they lie in $\sigma(E)$. Thus, all these sets I have produced must lie in $\sigma(E)$. This shows that $\mathcal{A}\subset\sigma(E)$. Commented Nov 17, 2022 at 14:39