# 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

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 :$$S=\{(a,b] ; a,b\in \mathbb R,a appear $$\color{red}{\cup\{\emptyset\}}$$?

• A way of understanding this is to write $\varnothing$ as $\{\}$ and $\{\varnothing\}$ as $\{\{\}\}$. Commented Apr 17, 2019 at 9:06

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 $$\emptyset\in\{1,\emptyset\}$$
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 $$S$$.
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).
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 $$a\in\emptyset.$$