# What is difference between $\cup\emptyset =\emptyset$ and $\cup$ {$\emptyset$}$=\emptyset$. [closed]

What is difference between $\cup\emptyset =\emptyset$ and $\cup$ {$\emptyset$}$=\emptyset$.

Also What is $\cup$ {$\mathbb{R}$}$=\mathbb{R}$ mean?.

## closed as unclear what you're asking by Stefan Mesken, user223391, Siong Thye Goh, E. Joseph, ShaileshOct 31 '16 at 0:38

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• The first line lists two ways to obtain the empty set and the 2nd line is true by the mere definition of $\bigcup$. Since this can't possibly be the answer you are looking for, I have no idea what your actual question is. – Stefan Mesken Oct 30 '16 at 22:50
• Where did you see this notation? Usually you would write something like $A \cup B$ for a binary union, or $\bigcup_{i \in I} A_i$ for a union of sets $A_i$ indexed by some set $I$. – ಠ_ಠ Oct 30 '16 at 22:52
• I guess it's union of all elements of the set – Djura Marinkov Oct 30 '16 at 22:53
• @ಠ_ಠ its not an uncommon notation. I mean it is explicitly part of the most common axiomatisation see Axiom of Union. – quid Oct 30 '16 at 22:53
• @ಠ_ಠ I saw from definiton of open set. – PozcuKushimotoStreet Oct 30 '16 at 23:02

Remember that the notation "$\bigcup\mathcal{A}$" means "The set of all things which are an element of an element of $\mathcal{A}$."
• If $\mathcal{A}$ is empty, it has no elements, so it certainly has no elements-of-elements; so $\bigcup \emptyset=\emptyset$.
• What if $\mathcal{A}=\{\emptyset\}$? Then $\mathcal{A}$ does indeed have an element . . . but that element has no elements. So there are still no "elements of elements of $\mathcal{A}$", so $\bigcup\{\emptyset\}=\emptyset$.
• Now $\bigcup\{\mathbb{R}\}$ means "the set of all things which are elements of $\mathbb{R}$." This is just $\mathbb{R}$! So indeed we have $\bigcup\{\mathbb{R}\}=\mathbb{R}$.
• So, $\bigcup\mathbb{R}$ is equal to $\mathbb{R}$ , isn't it? – PozcuKushimotoStreet Oct 30 '16 at 23:46
• @Kahler No, it is not. $\bigcup\mathbb{R}$ is the set of elements of elements of $\mathbb{R}$. So: what are the elements of $2$? of $\pi$? etc. At this point it depends how exactly we treat "$\mathbb{R}$" in set theory (a set of equivalence classes of Cauchy sequences? a set of Dedekind cuts?), but in every formalization I'm aware of arbitrary real numbers are not elements of real numbers. – Noah Schweber Oct 31 '16 at 0:07