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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 $\color{red}{\cup\{\emptyset\}}$?

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  • $\begingroup$ A way of understanding this is to write $\varnothing$ as $\{\}$ and $\{\varnothing\}$ as $\{\{\}\}$. $\endgroup$
    – Trebor
    Apr 17 '19 at 9:06
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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 $\emptyset\in\{1,\emptyset\}$

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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 $.

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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).

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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 $a\in\emptyset.$

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