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Let A be a set of all finite subsets of positive integers. I have proved closure and associativity under intersection. I am kind of confused about existence of identity. Originally, I was thinking that the set of all positive integers also lives in A and when any set M in A intersects with that set, we will get M back. However, I am not sure if the set of all positive integers even lives in A cause it is not finite, is it?

Any help would be great.

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    $\begingroup$ No, the set of all positive integers is not finite $\endgroup$
    – J. W. Tanner
    Feb 13, 2019 at 5:18
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    $\begingroup$ Yes $\emptyset$ $\in$ $A$. Its cardinality is $0$, hence it is a finite set. $\endgroup$
    – Four Seasons
    Feb 13, 2019 at 5:25
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    $\begingroup$ If $I$ is identity, $I \cap \{n\} = \{n\}$ for all $n \in \mathbb{N}$. So, $n \in I$ for all $n \in \mathbb{N}$. Thus, $I$ is not finite. $\endgroup$
    – Lucas
    Feb 13, 2019 at 5:30
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    $\begingroup$ The sets $\{1\},\{2\},...,\{n\},...$ are subsets of $A$. $\endgroup$
    – Lucas
    Feb 13, 2019 at 5:33
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    $\begingroup$ I think they just mean the singleton set containing the integer $n$. The statement that $I \cap \{1,\ldots,n\} = \{1,\ldots, n\}$ is true for all $n$ so $\{1,\ldots , n\}\subset I$ for all $n$ which gives the same conclusion $\endgroup$
    – Alex J Best
    Feb 13, 2019 at 5:34

2 Answers 2

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You have already given yourself the answer.

Proof by contradiction: Assume there is an identity element $I\in A$ with the usual property. Since $I$ is a finite subset of integers there exists a maximum element $n=\max I$. Now consider the set $B=I \cup \{n+1\}$. $B$ is finite, thus an element of $A,$ but $B\cap I\neq B$. Hence, $I$ is not the identity element.

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    $\begingroup$ Unless finite subsets exclude the empty set, you cannot assume the existence of a maximum element $n$ of $I$. (To be sure, it doesn't take much work to fix this.) $\endgroup$
    – Brian Tung
    Feb 13, 2019 at 6:37
  • $\begingroup$ @Briang Tung you are completely right! $\endgroup$
    – maxmilgram
    Feb 13, 2019 at 10:07
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Another proof by contradiction which doesn't worry about infinity:

If $A$ does form a group then there is some identity $I$ and some $\emptyset^{-1}$ with $\emptyset\cap\emptyset^{-1}=I$, but $\emptyset\cap\emptyset^{-1}=\emptyset$ so $I=\emptyset$. For a contradiction take any nonempty $J\in A$ (say $J=\{1\}$), then $J=J\cap I=J\cap\emptyset=\emptyset$.

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