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Please help me with this exercise, I try to apply the definitions and I can not prove it.

  • Give an inductive set $A \neq \mathbb{N} $
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closed as off-topic by Andrés E. Caicedo, Lord Shark the Unknown, miracle173, mlc, draks ... Jun 12 '17 at 8:14

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  • $\begingroup$ I think we are willing to help. But what did you try? How do you show that two sets are equal, in general? What is the problem with the direction from right to left? $\endgroup$ – martin.koeberl Apr 8 '17 at 23:57
  • $\begingroup$ Strictly speaking, notation $\left\lbrace {a} \right\rbrace = \left\lbrace {b, c} \right\rbrace$ is meaningless: a set with one element cannot be identified with a set with 2 elements. Besides, don't exaggerate with the number of "pleases". $\endgroup$ – Jean Marie Apr 9 '17 at 0:04
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    $\begingroup$ @JeanMarie the intention of the exercise is of course to imply that if a set has repeated elements, such as $\{1,1\}$, that it should be thought of as the set without those elements repeated, in this case the set $\{1,1\}$ being trimmed down to the more canonical representation $\{1\}$. If that notation bothers you so much, then the question could be better worded perhaps as $\{x~\mid~x=a\}=\{x~\mid~x=b~\text{or}~x=c\}$ $\endgroup$ – JMoravitz Apr 9 '17 at 0:10
  • $\begingroup$ @JMoravitz I agree, it's surely the convention taken in this homework. But, on the other hand, it is a rather accepted convention that a set defined in extension has no repeated elements. $\endgroup$ – Jean Marie Apr 9 '17 at 0:21
  • $\begingroup$ @JeanMarie Maybe the OP means for the "please"s to be a set - in the same vein that $\{b,c\}$ is a set. In which case there is only a single "please". $\endgroup$ – Χpẘ Apr 9 '17 at 0:47
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Outline:

Supposing that $\{a\}=\{b,c\}$. Then by definition of set equality, this means that simultaneously:

  • $\{a\}\subseteq \{b,c\}$

  • $\{a\}\supseteq \{b,c\}$

Now... by definition of being a subset (or superset) this implies...


In the other direction, suppose that $a=b=c$. Then we need to check the validity of the two statements: $\{a\}\subseteq \{b,c\}$ and $\{a\}\supseteq \{b,c\}$.

Let $x\in \{a\}$. Then $x=a$ but since $a=b$ it follows that $x\in\{b,c\}$ so...

Similarly...

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