-3

Can you find a set $A$ which satisfies $A = \{2,|A|\}$? Justify your answer.

Matthew Towers
  • 11,540
YGX
  • 7
  • 1
    @DietrichBurde : Not so. In that case, $A={2}$ so $|A|=1$, which gives also that $A={2,1}$ -- contradiction. – MPW Apr 14 '20 at 14:43
  • 1
    Why do you tag "measurable sets?" – Dietrich Burde Apr 14 '20 at 14:45
  • 1
    @MPW I see, this is called multiset, if an element is contained twice. – Dietrich Burde Apr 14 '20 at 14:46
  • @DietrichBurde : Yes, that allows repeated occurrences of an element. You could also use an ordered set, such as $(2,2)$. – MPW Apr 14 '20 at 14:49
  • I think if one allows multisets, then one needs to specify that. Otherwise, sets do not have multiple elements, although the may be represented with individual elements presented mulitple times (but actually only be one elements). Example: ${c,o,n,s,t,r,u,c,t,i,o,n} = {c,i,n,o,r,s,t,u}$. – fleablood Apr 14 '20 at 15:40

3 Answers3

-1

Hint: Note that as written, your set could have either one or two distinct elements, depending on whether $2$ and $|A|$ are equal or not. Try considering these cases separately.

MPW
  • 44,860
  • 2
  • 36
  • 83
-1

3)If we do not allow multisets and do not allow sloppy notation, then whenever we have $M = \{x,y\}$ we know that $x \ne y$ and $|M| = 2$.

So if $A = \{2,|A|\}$ we know that $|A|\ne 2$ and $|A| = 2$. Which is a contradictions. So no such set exists.

I do not think this should be the answer because I believe $A= \{2,2\} =\{2\}$ and $|A|=1$ should be legitimate things to write.

.....

2)If we do not allow multisets (which we shouldn't) but allow sloppy notation (which in my opinion we should) and allow multiple representation of a single element-- for example; if $K=\{x^2\% 10|x\in \mathbb Z\}$ then $K=\{... (-3)^2\%10, (-2)^2\%10, (-1)^2\%10,0^2\%10, 1^2\%10, 2^2\%10, 3^2\%10, ....\} = \{..... 9,4,1,0,1,4,9,6,5,6,9,4,1,0,..\} = \{0,1,4,5,6,9\}$ and $|K|=6$..or for example $M=\{$letters in the work "construction"$\}=\{c,o,n,s,t,r,u,c,t,i,o,n\}=\{c,o,n,s,t,r,u,i\}$ and $|M|=8$---

Then if $M= \{x,y\}$ then $|M| = 2$ if $x \ne y$ and $|M| = 1$ if $x=y$.

So if $A= \{2,|A|\}$ we would have $|A| =1$ if $|A| = 2$ or we would have $|A|=2$ if $|A|\ne 2$. Both are contradictions and no such set can exists.

In my opinion this should be the answer.

......

1) If we allow the word "set" to include "multisets" (which in my opinion we should not),

then $|\{2,2\}| = 2$. And also $|\{x,y|\}|= 2$ regardless as to whether $x=y$ or $x \ne y$. So if $A = \{2,|A|\}$ then $|A| = 2$ and thus we have $A = \{2,|A|\} = \{2,2\}$.

[Ordered sets are a form of multisets, so the $A= (2,2)$, the ordered pair, will also by such that $|(x,y)| = 2$ so $A=(2,2) = (2,|A|)$, will work]

In my opinion, these should not be the answers as "multisets" are not sets, and sets can not have multiple representations of elements. The set $\{2, |A|\}$, in my opinion, can be a set with two different elements, or a set with one distinct element that can be represent by being written twice. I think if you allow multisets and consider that a set with one element written twice, or in a distinct order to be a different object than a set with the one element written once, you are obligated to specify that.

.....

fleablood
  • 130,341
-2

If $A=2$, then $A=\{2\}$, a contradiction.

If $A\neq 2$, then $|A| = 2$, which means that in fact $A=\{2\}$ and so $|A|=1$, another contradiction.

Luke Collins
  • 8,838