1
$\begingroup$

If $E$ is the collection of all open intervals $(a,b)$ in $X=[0,1]$, how do I know that the $\sigma(E)$ contains all closed intervals $[a,b] \subset X$, in particular closed intervals involving the endpoints? (Regular closed intervals I can represent as countable intersections of open intervals, hence they are in $\sigma(E)$.)

$\endgroup$
  • $\begingroup$ For a closed interval $[a,b]$ take the intersection of the sequence $(a-\frac{1}{n},b+\frac{1}{n})$. $\endgroup$ – Pedro Oct 1 '14 at 14:42
  • $\begingroup$ @Pedro: That won't work if $a=0$ or $b=1$ since in either case $(a-\frac1n,b+\frac1n)$ is not a subset of $[0,1]$ $\endgroup$ – MPW Oct 1 '14 at 14:45
  • $\begingroup$ Maybe you really mean for $E$ to comprise intervals that are open subsets of $X$? In the subspace topology, that would include intervals of the form $[0,b)$ and $(a,1]$. I'm not sure you can even generate $[0,1]$ using your proposed $E$. $\endgroup$ – MPW Oct 1 '14 at 14:48
  • $\begingroup$ Since $\sigma(E)$ is an algebra, it contains complements and finite unions. Try to use that to construct a sequence that converges to $[0, 1]$? So, we know $(1/4, 1/2) \in \sigma(E)$, so $[0, 1/4] \cup [1/2, 1] \in \sigma(E)$ and now union that with say $(1/8, 3/4)$. $\endgroup$ – Robert Cardona Oct 1 '14 at 14:49
  • $\begingroup$ @MPW you are right, for $[0,1]$ take the complement of the intesection of the sequence $(\frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac{1}{n})$ and then union with $(0,1)$. $\endgroup$ – Pedro Oct 1 '14 at 14:51
1
$\begingroup$

Let $X = [0, 1]$.

If we define $E = \{(a, b) \cap X : a, b \in \mathbb R$ and $a < b\}$, then

  • $(-1, a) \cap X = [0, a) \in E \subseteq \sigma(E)$ and
  • $(b, 2) \cap X = (b, 1] \in E \subseteq \sigma(E)$.

So if we want $[a, b]$ in general, we observe that $(b, 1], [0, a) \in E \subseteq \sigma(E)$ and so are there complements: $[0, b], [a, 1]$.

Taking their intersection, we get $[a, b] \in \sigma(E)$ since $\sigma(E)$ is closed under intersections.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.