0
$\begingroup$

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

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

$\endgroup$

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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. $\endgroup$ – Stefan Mesken Oct 30 '16 at 22:50
  • 1
    $\begingroup$ 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 $I$. $\endgroup$ – ಠ_ಠ Oct 30 '16 at 22:52
  • 1
    $\begingroup$ I guess it's union of all elements of the set $\endgroup$ – Djura Marinkov Oct 30 '16 at 22:53
  • 3
    $\begingroup$ @ಠ_ಠ its not an uncommon notation. I mean it is explicitly part of the most common axiomatisation see Axiom of Union. $\endgroup$ – quid Oct 30 '16 at 22:53
  • $\begingroup$ @ಠ_ಠ I saw from definiton of open set. $\endgroup$ – PozcuKushimotoStreet Oct 30 '16 at 23:02
4
$\begingroup$

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 $\bigcup\{\mathbb{R}\}=\mathbb{R}$.

$\endgroup$
  • $\begingroup$ So, $\bigcup\mathbb{R}$ is equal to $\mathbb{R}$ , isn't it? $\endgroup$ – PozcuKushimotoStreet Oct 30 '16 at 23:46
  • 1
    $\begingroup$ @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. $\endgroup$ – Noah Schweber Oct 31 '16 at 0:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.