Why isn't $\{\{\emptyset\}\} = \{\emptyset,\{\emptyset\}\}$? [closed]

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 ThousandSep 27 at 16:35

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• "If the set {Ф,{Ф}} has just one element": it doesn't – Gregory J. Puleo Sep 23 at 23:26
• @Gregory J. Puleo, so its cardinality is 2 then? – Satyajit Sen Sep 23 at 23:28
• $1 \neq \{ n\in\mathbb{N} : n^2=n\} = \{1\}$ but $1 \in \{1\}$. – hal4math Sep 23 at 23:28
• The set $\{ \Phi,\{\Phi\}\}$ contains two elements, namely $\Phi$ and $\{\Phi\}$, while the set $\{\{\Phi\}\}$ contains one single element, $\{\Phi\}$. – Azif00 Sep 23 at 23:34
• I live in a world were $0 \in \mathbb{N}_{0}$ but $0 \not\in \mathbb{N}$ :). – 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.
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!
• I have used the bag analogy myself. It really helps describing the diffference between $\varnothing$ and $\{\varnothing\}$ and so on. – Arthur Sep 24 at 8:54