MHB Smallest Sigma Algebra .... Axler, Example 2.28 ....

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I am reading Sheldon Axler's book: Measure, Integration & Real Analysis ... and I am focused on Chapter 2: Measures ...

I need help in order to make a meaningful start on verifying the first part of Axler, Example 28 ...

The relevant text reads as follows:
Axler - Borel Subsets of R ... including Example 2.28 .png


Can someone please help me to make a meaningful start on verifying Example 2,28 ... that is, to show that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is the set of all subsets $E$ of $X$ such that $E$ is countable or $X \setminus E$ is countable ... ...
Help will be much appreciated ...

Peter
 
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Peter said:
Can someone please help me to make a meaningful start on verifying Example 2,28 ... that is, to show that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is the set of all subsets $E$ of $X$ such that $E$ is countable or $X \setminus E$ is countable ... ...
Let $\mathcal{S}$ be be set of all subsets of $X$ that are either countable or co-countable (where "countable" is understood to include finite or empty, and "co-countable" means having a countable complement). Then $\mathcal{S}$ is a $\sigma$-algebra. To see that, notice that it certainly contains the empty set and is closed under complementation. Also every subset of a countable set is countable, and (by complementation) every superset of a co-countable set is co-countable. Now suppose that $E_1,E_2,\ldots$ is a sequence of sets in $\mathcal{S}$. If anyone of those sets is co-countable then so is the union $\bigcup E_n$. Alternatively, if they are all countable then so is the union. That shows that $\mathcal{S}$ is closed under countable unions and is therefore a $\sigma$-algebra.

But it is easy to see that any $\sigma$-algebra $\mathcal{S}$ that contains $\mathcal{A}$ must contain all the countable subsets of $X$ and their complements. So it contains $\mathcal{S}$. Therefore $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$.
 
Opalg said:
Let $\mathcal{S}$ be be set of all subsets of $X$ that are either countable or co-countable (where "countable" is understood to include finite or empty, and "co-countable" means having a countable complement). Then $\mathcal{S}$ is a $\sigma$-algebra. To see that, notice that it certainly contains the empty set and is closed under complementation. Also every subset of a countable set is countable, and (by complementation) every superset of a co-countable set is co-countable. Now suppose that $E_1,E_2,\ldots$ is a sequence of sets in $\mathcal{S}$. If anyone of those sets is co-countable then so is the union $\bigcup E_n$. Alternatively, if they are all countable then so is the union. That shows that $\mathcal{S}$ is closed under countable unions and is therefore a $\sigma$-algebra.

But it is easy to see that any $\sigma$-algebra $\mathcal{S}$ that contains $\mathcal{A}$ must contain all the countable subsets of $X$ and their complements. So it contains $\mathcal{S}$. Therefore $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$.
Thanks Opalg ...

Still reflecting on what you have written ...

Peter
 
Opalg said:
Let $\mathcal{S}$ be be set of all subsets of $X$ that are either countable or co-countable (where "countable" is understood to include finite or empty, and "co-countable" means having a countable complement). Then $\mathcal{S}$ is a $\sigma$-algebra. To see that, notice that it certainly contains the empty set and is closed under complementation. Also every subset of a countable set is countable, and (by complementation) every superset of a co-countable set is co-countable. Now suppose that $E_1,E_2,\ldots$ is a sequence of sets in $\mathcal{S}$. If anyone of those sets is co-countable then so is the union $\bigcup E_n$. Alternatively, if they are all countable then so is the union. That shows that $\mathcal{S}$ is closed under countable unions and is therefore a $\sigma$-algebra.

But it is easy to see that any $\sigma$-algebra $\mathcal{S}$ that contains $\mathcal{A}$ must contain all the countable subsets of $X$ and their complements. So it contains $\mathcal{S}$. Therefore $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$.
Thanks again, Opalg ...

You write:

" ... ... But it is easy to see that any $\sigma$-algebra $\mathcal{S}$ that contains $\mathcal{A}$ must contain all the countable subsets of $X$ and their complements. So it contains $\mathcal{S}$. Therefore $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$. ... "

I am finding it difficult to see exactly why $\mathcal{S}$ is the smallest $\sigma$-algebra that contains $\mathcal{A}$. ...

Can you elaborate/explain further ... ?

Peter
 
The structure of the argument consists of these two parts:
1) $\mathcal{S}$ is a $\sigma$-algebra containing $\mathcal{A}$;
2) Every $\sigma$-algebra that contains $\mathcal{A}$ must contain $\mathcal{S}$.
Those two facts together say that $\mathcal{S}$ is the smallest $\sigma$-algebra containing $\mathcal{A}$.
 
Thanks Opalg ...

Peter
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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