Show that the topology on $\mathbb{R}$ generated by semi-closed intervals of type (a,b] contains the natural topology I'm a little confused why this is true. How could the natural topology be a subset of 
  $\bigcap_{i = 1}^{n}A_{\lambda }$ where {$A_{\lambda }$}$_{\lambda \in \Lambda  }$ is a family of subsets of X . 
How could the natural topology be a subset of a semi-closed interval? 
Any help would be appreciated. Thanks.
 A: The semi-closed subsets are used to generate a new topology on $\mathbb{R}$. In other words a set is open in this "semi-closed" topology if it is a union of semi-closed intervals. Just like an open set in the usual topology on $\mathbb{R}$ is a union of open intervals.
To see that the semi-closed topology contains all of the old open sets you can show that an open interval is the union of semi-closed intervals. Then the new topology also generates the old topology.
To see this notice that the union of $(a,b-\epsilon]$ where $0<\epsilon<b-a$ is exactly $(a,b)$. Thus $(a,b)$ is open in the new topology. Hence unions of open intervals are open hence all "old" open sets are still open.
A: $(a,b)=\bigcup_{i\in\mathbb{N}} (a,b-1/n]$, so the topology generated by the $(a,b]$'s contains the usual topology, which is the smallest topology containing all the $(a,b)$'s.
I am not sure I understand the question "How could the natural topology be a subset of $⋂_{i=1}^n A_λ$ where $\{A_λ\}_{λ∈Λ}$ is a family of subsets of $X$?"  A topology is a set of subsets of $X$, meeting certain conditions. It is thus not a subset of $X$ itself.
