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To prove equality we need to prove a. $\emptyset\subseteq\{\emptyset\}$ and b. $\{\emptyset\}\subseteq \emptyset$

For a. because the empty set is a subset of every set, in particular it is a subset of $\{\emptyset\}$ therefore $\emptyset\subseteq\{\emptyset\}$

As for b. the empty set is empty therefore it can not include elements or be a subset (else from the empty set itself), is this claim concludes that $\{\emptyset\}\nsubseteq \emptyset$ or is it vacuous truth?

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    $\begingroup$ The empty set is empty. But $\{\varnothing\}$ isn't. $\endgroup$ – MathematicsStudent1122 Jun 30 '16 at 13:08
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    $\begingroup$ The difference is made clear notationally remembering that $\emptyset=\{\,\}$ is another way to indicate the empty set. Therefore, id clear that $$ \{\,\}\subsetneq \{\emptyset\}$$ $\endgroup$ – guestDiego Jun 30 '16 at 13:19
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    $\begingroup$ The best analogy i've ever heard is "an empty box is not the same as a box containing an empty box" $\endgroup$ – Elliot G Jun 30 '16 at 13:22
  • $\begingroup$ $\emptyset=\{\emptyset\}\rightarrow \emptyset\in\emptyset$. $\endgroup$ – user170039 Jun 30 '16 at 13:47
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    $\begingroup$ Some related posts: math.stackexchange.com/questions/941062/… or math.stackexchange.com/questions/51752/… $\endgroup$ – Martin Sleziak Jun 30 '16 at 14:24
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You could just note that the set $\{\emptyset\}$ has one element (namely $\emptyset$) while $\emptyset$ does not contain any elements. So, they are not equal.

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  • $\begingroup$ So if I wanted it to be true, is should have been $\emptyset=\{\}$ $\endgroup$ – gbox Jun 30 '16 at 13:10
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    $\begingroup$ @gbox: That depends on what you mean by $\{\}$. If your definition of this expression is the empty set, then yes they would indeed be equal. $\endgroup$ – Thomas Jun 30 '16 at 13:11
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As others have stated, $\emptyset$ is a set with no elements and $\{\emptyset\}$ is a set with one element. Therefore these two sets can't be equal.

To address your specific question:

As for b. the empty set is empty therefore it can not include elements or be a subset (else from the empty set itself), is this claim concludes that $\{\emptyset\}\nsubseteq \emptyset$ or is it vacuous truth?

I think by "be a subset" you meant "have a subset," in which case, yes, that claim concludes that $\{\emptyset\} \nsubseteq \emptyset$. The reason I think you meant "have a subset" is because "be a subset" doesn't make sense in the context. The "subset" in question in this direction of the proof is $\{\emptyset\}$, which is the set containing the empty set. Although this set can potentially be a subset of another set, it can't be a subset of $\emptyset$ because $\emptyset$ has no elements. Therefore $\{\emptyset\}$ is not a subset of $\emptyset$.

To put it more concisely:

In general, $\{a\} \subseteq A$ if and only if $a \in A$.

So in our case, $\{\emptyset\} \subseteq \emptyset$ if and only if $\emptyset \in \emptyset$.

But of course $\emptyset \notin \emptyset$, because $\emptyset$ is empty.

Therefore $\{\emptyset\} \nsubseteq \emptyset$, and so $\{\emptyset\} \neq \emptyset$.

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  • $\begingroup$ "In general, $\{a\} \subseteq A$ if and only if $a \in A$." But In the case of $2\in{1,2,3} \not \implies {2}\subseteq {1,2,3}$ $\endgroup$ – gbox Jun 30 '16 at 13:23
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    $\begingroup$ @gbox, not sure I'm interpreting your comment correctly because I don't see any curly braces in your example (as of the time I write this) but my response would be this: The example is incorrect because $2 \in \{1, 2, 3\}$ actually does imply $\{2\} \subseteq \{1,2,3\}$. $\endgroup$ – user307169 Jun 30 '16 at 13:26
  • $\begingroup$ @gbox, viewing this on the app now and I can see your unrendered TeX. It looks like I did interpret you correctly. Sounds like you're trying to say that $2\in \{1,2,3\}$ does not imply that $\{2\} \subseteq \{1,2,3\}$, but that's just not correct because the implication is indeed true. Clarifying for you and for anyone else who comes here wondering. $\endgroup$ – user307169 Jul 1 '16 at 1:33
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$\emptyset$ is the empty set. It contains no elements.

$\{\emptyset\}$ is the set containing the empty set. It contains one element: the empty set.

Since $\{\emptyset\}$ contains one element and $\emptyset$ does not contain any elements, they cannot be the same.

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$\#\emptyset=0$ while $\#\{\emptyset\}=1$.

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