Solution: We need to show that the collection $\mathcal{T} $ is a topology on $\mathbb{R} $. Note that $\emptyset = I_\infty \in \mathcal{T} $ and $\mathbb{R} = I _{-\infty } \in \mathcal{T} $. Now we will show that the collection is closed under arbitrary union and finite intersection.

Let $I$ be an index set and for each $i \in I$, $I_{x_i}\in \mathcal{T} $. We claim that $ \bigcup_{i\in I} U_{x_i} \in \mathcal{T} $. Note that if $I_{-\infty } \in \mathcal{T} $, then the union will be $I_{-\infty }$ and hence we are done. So let us assume that $I_{-\infty } \notin \mathcal{T} $. Then the set $\{ x_i : i \in I \} $ is bounded below and let $x = \inf \{ x_i: I \in I \} $. Then we claim that \[ \bigcup_{i \in I} I_{x_i} = I_x. \] Since $x$ is the infimum, so $x \leq x_i$ for any $i \in I$ and hence $I_{x_i} \subseteq I_x$. Therefore, the union is also a subset of $I_x$. For other inequality, note that for any $y \in I_x$, $x < y$ and hence $y \in \{ x_i : i \in I \} $ and therefore, there exists $j \in I$ such that $x_j < y$. Therefore, $y \in I_{x_j}$ and hence $I_x \subseteq \cup I_{x_i}$. This proves the equality and hence $\mathcal{T} $ is closed under arbitrary union.

To show that it is closed under finite intersection, let $I_{x_i} \in \mathcal{T} $ for $i = 1,2, \dots, n$. Then \[ \bigcap_{i=1}^{n} I_{x_i} = I_x, \] where $x = \max \{ x_1, \dots, x_n \} $.