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 $\sigma(E)$.)

  • $\begingroup$ 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

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.


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