If the set $ \{\emptyset,\{\emptyset\}\}$ has just one element that is $\{\emptyset\}$ and is empty otherwise, shouldn't it be equivalent to $\{\{\emptyset\}\}$?


closed as unclear what you're asking by Simply Beautiful Art, user21820, José Carlos Santos, Mars Plastic, Don Thousand Sep 27 at 16:35

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.

  • 12
    $\begingroup$ "If the set {Ф,{Ф}} has just one element": it doesn't $\endgroup$ – Gregory J. Puleo Sep 23 at 23:26
  • 1
    $\begingroup$ @Gregory J. Puleo, so its cardinality is 2 then? $\endgroup$ – Satyajit Sen Sep 23 at 23:28
  • 1
    $\begingroup$ $1 \neq \{ n\in\mathbb{N} : n^2=n\} = \{1\}$ but $1 \in \{1\}$. $\endgroup$ – hal4math Sep 23 at 23:28
  • 6
    $\begingroup$ The set $\{ \Phi,\{\Phi\}\}$ contains two elements, namely $\Phi$ and $\{\Phi\}$, while the set $\{\{\Phi\}\}$ contains one single element, $\{\Phi\}$. $\endgroup$ – Azif00 Sep 23 at 23:34
  • 3
    $\begingroup$ I live in a world were $0 \in \mathbb{N}_{0}$ but $0 \not\in \mathbb{N}$ :). $\endgroup$ – hal4math Sep 23 at 23:38

Assuming your "$\Phi$" is the empty set $\{\}$ (usually denoted "$\emptyset$," LaTeX code "\$\emptyset\$"), the important point is that $\emptyset$ is not nothing. It contains nothing, but that's not the same thing, any more than an empty bag is the same as no bag at all.

It's important to say at this point that you shouldn't push the bag metaphor too far, but in some contexts - including this one - I think it is useful.

In particular, $\{\emptyset,\{\emptyset\}\}$ has two elements - we can't ignore the by-itself $\emptyset$. After all, if we could the whole thing would evaporate: highlighting in red the bits that we erase at each step we'd get $$\mbox{$\{\color{red}{\emptyset},\{\color{red}{\emptyset}\}\}=\{\color{red}{\{\}}\}=\color{red}{\{\}}=\quad$ .}$$ I'm not even sure what that last thing is!

It's worth mentioning at this point that the usual framework of set theory builds everything up from the emptyset alone. So far from being a silly hair-splitting, caution around the emptyset is quite serious mathematics.

  • 2
    $\begingroup$ This question seems to come up in some form or another quite often, and the bag analogy is by far the best explanation for this that I have seen. Massive +1 from me. $\endgroup$ – DreamConspiracy Sep 24 at 0:34
  • $\begingroup$ I have used the bag analogy myself. It really helps describing the diffference between $\varnothing$ and $\{\varnothing\}$ and so on. $\endgroup$ – Arthur Sep 24 at 8:54
  • $\begingroup$ @DreamConspiracy: I don't see how anyone can successfully teach basic set theory without using an analogy to bags or containers of some kind. But this is the first time I have seen evaporating bags! =) $\endgroup$ – user21820 Sep 24 at 10:33
  • $\begingroup$ Just for the record, I have edited the post to write the sets with MathJax, making the same assumption as in the first sentence of this answer. $\endgroup$ – Arnaud D. Sep 24 at 10:50

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