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Note: Actually there's no error in the book and the manual. I actually misread it. The answer is of a different question :
True or False: {0} ⊂ {0}

This question is from Discrete Math Book by Rosen.

{{∅}} ⊂ {∅,{∅}}

Answer in the manual is:

This is false. For one set to be a proper subset of another, the two sets cannot be equal.

How is {{∅}} is equal to the set {∅,{∅}} ? I know the two sets have different cardinal numbers. So they cannot be equal.

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    $\begingroup$ @Vim Are you sure? $\endgroup$
    – user261263
    Aug 24, 2015 at 16:48
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    $\begingroup$ @Vim No, is not true $\endgroup$
    – user261263
    Aug 24, 2015 at 16:51
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    $\begingroup$ This is a silly question, but it has happened to me many times before. Are you sure that you're looking at the right answer on the answers key? $\endgroup$
    – Asaf Karagila
    Aug 24, 2015 at 16:57
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    $\begingroup$ FTR in my university classes $\subset$ did not mean "proper." You would need $\subsetneq$ for that. $\endgroup$
    – djechlin
    Aug 24, 2015 at 18:53
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    $\begingroup$ @djechlin: I agree that $\subsetneq$ or even $\subsetneqq$ is better for proper inclusion. But the fact is that a lot of people use $\subset$ for proper inclusion. To make matters worse, there are plenty of people who use $\subset$ for improper inclusion, which creates quite a clash. This is why it is always best to use $\subseteq$ and $\subsetneq$ which are entirely unambiguous. $\endgroup$
    – Asaf Karagila
    Aug 25, 2015 at 0:52

1 Answer 1

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$\begingroup$

Is $\{ \{ \emptyset \} \}$ a subset of $\{ \emptyset, \{ \emptyset \} \}$?

Well, it's equivalent to asking whether $\{ \emptyset \} \in \{ \emptyset, \{ \emptyset \} \}$.

And it obviously is.

The answer in the manual is extremely wrong, if you've quoted it correctly. $\{ \{ \emptyset \} \}$ is not equal to $\{ \emptyset, \{ \emptyset \} \}$. (More easily read, $\{ 5 \}$ is not equal to $\{ \emptyset, 5 \}$.) It is possible there is a typographical error in the book, or that you've mis-read it, or misunderstood it.

Notice that $\{ \{ \emptyset \} \}$ is equal to $\{ \{ \emptyset \}, \{ \emptyset \} \}$, because sets must have distinct elements and we discard duplicates: $\{1, 1 \} = \{ 1 \}$. This is one possible way you might have misread the book, or that the book might have been printed incorrectly.

Note: But make sure what the definition of $\subset$ is! Some use $A \subset B$ to mean any subset (i.e., include $A = B$); others use $A \subseteq B$ for this, in which case they use $A \subset B$ if $A \subseteq B$ but $A \ne B$.

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  • $\begingroup$ I have just pasted the ans from the manual. $\endgroup$ Aug 24, 2015 at 17:05
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    $\begingroup$ @ShashiDwivedi: Out of curiosity I checked the book (K. H. Rosen, Discrete mathematics and its applications, 6th ed.). In that edition, the one you mention is exercise 8f) from section 2.1: given that the book gives the answers of the odd-numbered exercises only, are you sure to have checked the right answer? $\endgroup$ Aug 24, 2015 at 18:36
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    $\begingroup$ @MassimoOrtolano Bear in mind that exercise numbering can vary wildly between editions and international versions. $\endgroup$
    – mrp
    Aug 24, 2015 at 19:02
  • $\begingroup$ @MassimoOrtolano Yup.. I was wrong. But I was having doubts on this question. I was confused if I were reasoning correctly for this question. I got it here. Thanks! $\endgroup$ Aug 24, 2015 at 20:03
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    $\begingroup$ ShashiDwivedi: In case you didn't state the question from the book correctly, you should add a note which clarifies this circumstance. The reader of your question shouldn't get the wrong impression, that the book has a flaw. Regards, $\endgroup$
    – epi163sqrt
    Aug 25, 2015 at 5:00

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