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Am i right that $\sigma$ - algebra $\Rightarrow$ topology?
The defintion of $\sigma$-algebra is:
The family of set $\mathcal{M}$ is $\sigma$ -algebra on X if:

  1. $\emptyset \in X $
  2. $\forall A \in \mathcal{M} :X \backslash A \in \mathcal{M}$
  3. $\forall A_j \in \mathcal{M},j \in \mathbb{N}:\bigcup_{j=1}^{\infty}Aj \in \mathcal{M} $

The family $\mathcal{\tau} $ is topology on X if:

  1. $\emptyset ,X \in \tau $
  2. If family $\mathcal{N} \subset \tau $ then $\bigcup \mathcal{N} \in \tau$
  3. $U,V \in \tau \Rightarrow U \cap V\in \tau$
MIkiti
  • 25

2 Answers2

1

If $\mathcal{M}$ is a $\sigma$-algebra on $X$ such that $\{x\}\in\mathcal{M}$ for all $x\in X$, then if $X$ were a topololgy, since every subset $A\subseteq X$ is the union $\bigcup\{\{x\} : x\in A\}$, it would follow that $A\in \mathcal{M}$ for every $A\subseteq X$. Thus $\mathcal{M} = \mathcal{P}(X)$ is the family of all subsets of $X$.

The existence of subsets of $\mathbb{R}$ which are not Lebesgue measurable (say, Vitali set), implies that the $\sigma$-algebra of Lebesgue measurable sets is not a topology.

Alternatively, let $X$ be uncountable and let $\mathcal{M}$ be the $\sigma$-algebra of countable and cocountable sets. There exists a subset $A\subseteq X$ such that neither $A$ nor $A^c$ is countable. Thus $\mathcal{M}$ is not a topology.

Jakobian
  • 10,247
1

No.

There are $\sigma$-algebra's that are not a topology. For instance the collection of subsets of $\mathbb R$ that are countable or co-countable.

This is a $\sigma$-algebra but is not closed under unions. Note that interval $(0,1)$ is not countable and is not co-countable. Nevertheless it can be written as a union of countable sets:$$(0,1)=\bigcup_{x\in(0,1)}\{x\}$$

drhab
  • 151,093