If the set $ \{\emptyset,\{\emptyset\}\}$ has just one element that is $\{\emptyset\}$ and is empty otherwise, shouldn't it be equivalent to $\{\{\emptyset\}\}$?
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12"If the set {Ф,{Ф}} has just one element": it doesn't – Gregory J. Puleo Sep 23 '19 at 23:26
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1@Gregory J. Puleo, so its cardinality is 2 then? – Satyajit Sen Sep 23 '19 at 23:28
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1$1 \neq { n\in\mathbb{N} : n^2=n} = {1}$ but $1 \in {1}$. – hal4math Sep 23 '19 at 23:28
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Possible duplicate of Why is belonging not transitive? – Simply Beautiful Art Sep 23 '19 at 23:33
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6The set ${ \Phi,{\Phi}}$ contains two elements, namely $\Phi$ and ${\Phi}$, while the set ${{\Phi}}$ contains one single element, ${\Phi}$. – azif00 Sep 23 '19 at 23:34
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1@hal4math: and $0$? – Bernard Sep 23 '19 at 23:37
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3I live in a world were $0 \in \mathbb{N}_{0}$ but $0 \not\in \mathbb{N}$ :). – hal4math Sep 23 '19 at 23:38
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1And refer for that to Peano. Original work. – hal4math Sep 23 '19 at 23:45
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Technically a duplicate of this question, but this case might be more confusing. This is also related (and is linked to a few questions itself). – Arnaud D. Sep 24 '19 at 10:48
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@hal4math Excluding the cardinality of the empty set and the additive identity from $\mathbb{N}$? That sounds like an ugly world :) – user76284 Sep 28 '19 at 04:01
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@hal4math Not to mention the zeroth power in the binomial theorem or in a general power series. – user76284 Sep 28 '19 at 04:05
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@user76284 Well, the set of continuous function from the reals into reals which are differentiable at at least one point, is a meager set in the space of continuous function. So, not sure how much "uglier" everything gets here, especially in light of me just adding one subscript to a symbol to magically get all the things you mentioned. :) – hal4math Sep 28 '19 at 11:03
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@user76284 I of course get you. I just learned it differently in the beginning, actually the first proper thing I learned in math. So emotionally it stayed with me. :) – hal4math Sep 28 '19 at 11:19
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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.
Noah Schweber
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2This 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. – DreamConspiracy Sep 24 '19 at 00:34
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I have used the bag analogy myself. It really helps describing the diffference between $\varnothing$ and ${\varnothing}$ and so on. – Arthur Sep 24 '19 at 08:54
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@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! =) – user21820 Sep 24 '19 at 10:33
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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. – Arnaud D. Sep 24 '19 at 10:50