Measurable space of nonempty compact intervals, $\sigma ( CI_{\mathbb R}) \neq \mathcal I_{\mathbb R}$ If we define
$$CI_{\mathbb R} = \{ [a,b] \subset \mathbb R : (a,b \in \mathbb R \ \text{ and } \ a \leq b ) \}$$
$$ \mathcal I_{\mathbb R} = \sigma_{CI_{\mathbb R}}  \left ( \{ \{ A \in CI_{\mathbb R} : \inf( A) < c \} : c \in \mathbb R \}  \cup \{ A \in CI_{\mathbb R} : \sup( A) < c \} : c \in \mathbb R \} \right )  $$
The idea is to construct a measurable space for confidence intervals. The space $( CI_{\mathbb R}, \mathcal I_{\mathbb R} ) $ is nice. However, I am trying to understand how come these two sets are not equal :
$$\sigma ( CI_{\mathbb R}) \neq \mathcal I_{\mathbb R}$$
LHS should be bigger, do you have an example of an element in LHS that is not in RHS?

An answer given by my lecturer :
On the LHS, we have the power set of $CI_{\mathbb R}$ (which is the largest \sigma-algebra on $CI_{\mathbb R}$).
Let $X$ be an uncountable set (in our case $X = CI_{\mathbb R}$), and let $Y = \sigma_{X} \{ Y_j, j \in J \}$ (in our case $Y = I_{\mathbb R}$),  where $J$ is some (uncountable) set, and $Y_j \subset 2^{X}$ (i.e., power set of $X$).
Note that in our case every $Y_j$ is uncountable, its complement is uncountable, and $Y_j \cap Y_i$ is either uncountable or empty set.
Hence, one can show that every element of $Y$ is uncountable. However, $2^{X}$ also includes countable sets (e.g., singletons). Hence, there are uncountable many sets which are contained in $2^X$ but not in $Y$.
