17
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ @Vim Are you sure? $\endgroup$ – user261263 Aug 24 '15 at 16:48
  • 3
    $\begingroup$ @Vim No, is not true $\endgroup$ – user261263 Aug 24 '15 at 16:51
  • 13
    $\begingroup$ 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 '15 at 16:57
  • 5
    $\begingroup$ FTR in my university classes $\subset$ did not mean "proper." You would need $\subsetneq$ for that. $\endgroup$ – djechlin Aug 24 '15 at 18:53
  • 2
    $\begingroup$ @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 '15 at 0:52
29
$\begingroup$

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 $A \ne B$.

$\endgroup$
  • $\begingroup$ I have just pasted the ans from the manual. $\endgroup$ – Shashi Dwivedi Aug 24 '15 at 17:05
  • 3
    $\begingroup$ @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? $\endgroup$ – Massimo Ortolano Aug 24 '15 at 18:36
  • 2
    $\begingroup$ @MassimoOrtolano Bear in mind that exercise numbering can vary wildly between editions and international versions. $\endgroup$ – mrp Aug 24 '15 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! $\endgroup$ – Shashi Dwivedi Aug 24 '15 at 20:03
  • 4
    $\begingroup$ 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, $\endgroup$ – Markus Scheuer Aug 25 '15 at 5:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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