If the empty set is a subset of every set, why write ... $\cup \{∅\}$? I met the notation $ S=\{(a,b] ; a,b\in \mathbb R,a<b\}\cup\{\emptyset\} $
I know $S$ is a family of subsets ,a set of intervals, and from set theory $\emptyset$ is a subsets of every set then why in the notation :$ S=\{(a,b] ; a,b\in \mathbb R,a<b\}\cup\{\emptyset\} $ appear $\color{red}{\cup\{\emptyset\}}$?
 A: The answer is: the given definition uses $\cup\,\{\emptyset\} $, not $\cup\,\emptyset $, so it adds the empty set as an element, not a subset of $S $.
A: Because the empty set $(\emptyset)$ is one thing, but what you have there is $\{\emptyset\}$, which is a different thing: it's a set with a single element (which happens to be the empty set).
A: It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $\emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $\emptyset$ is a subinterval. The reason for using $\{\emptyset\}$ is show you can write out the collection of all such subintervals in a nice form. 
As for the empty set is a subset of every set, well that's a vacuous truth. For all $a\in\emptyset$ if $X$ is a set it follows that $a\in X.$ This is true, because there are no $a\in\emptyset.$ 
A: It is because the emptyset $\emptyset$ is a subset of every set, but not an element of every set.
It is $\emptyset\in S$ and you might want that to show, that the elements of $S$ define a topology.
Or to be more clear it is $\{1\}\neq\{1,\emptyset\}$. The set on the left has one element, the set on the right has two elements, with $\emptyset\in\{1,\emptyset\}$
