Our "total" space is $[0,1]$. Now, denote $\mathcal{A}$ to be the collection all finite disjoint union of closed intervals of the form $[a,b]\subset [0,1]$.
I want to show that
$\mathcal{A}$ is an algebra. That is, $[0,1]\in \mathcal{A}$ and for any $E,F\in\mathcal{A}$, we have $E\setminus F\in\mathcal{A}$.
$[0,1]\in\mathcal{A}$ is for sure. However, I do not know how to prove the rest. For instance, if we have $$E:=\bigcup_{j=1}^{N}[a_{j}, b_{j}]\ \ \text{and}\ \ F:=\bigcup_{i=1}^{M}[c_{i}, d_{i}],$$ what could I do to analyze $E\setminus F$?
I tried to use a detour by showing that the collection $\mathcal{E}$ of all closed intervals $[a,b]$ and $\varnothing$ forms a semi-ring. That is, $\varnothing\in\mathcal{E}$, for any $E,F\mathcal{E}$, $E\cap F\in\mathcal{E}$, and for any $E\in\mathcal{E}$, $[0,1]\setminus E$ can be written as a finite disjoin unions of elements in $\mathcal{E}$.
If I can show this, then a theorem can directly give us $\mathcal{A}$ is an algebra.
However, it is not true that the complement of $[a,b]$ in $[0,1]$ can be written as a finite disjoint union of closed interval. For example, $[0,1]\setminus [a,b]=[0,a)\cup (b,1]$ which is only a disjoint union of half-closed interval.
To be honest, I am not really sure if the claim I am trying to proof is even correct.
Is there any way to prove it or disprove it? Thank you!
Best Answer
The claim is wrong.
$[\frac{1}{3},\frac{2}{3}] \backslash [\frac{1}{3},\frac{1}{2}]=(\frac{1}{2}, \frac{2}{3}] \notin \mathcal A$. This is because any finite union of closed intervals is compact and $(\frac{1}{2}, \frac{2}{3}]$ is not compact.