# Does the set of all finite subsets of positive integers form a group under set intersection?

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.

• No, the set of all positive integers is not finite – J. W. Tanner Feb 13 at 5:18
• Yes $\emptyset$ $\in$ $A$. Its cardinality is $0$, hence it is a finite set. – Ufomammut Feb 13 at 5:25
• 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. – Lucas Corrêa Feb 13 at 5:30
• The sets $\{1\},\{2\},...,\{n\},...$ are subsets of $A$. – Lucas Corrêa Feb 13 at 5:33
• 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 – Alex J Best Feb 13 at 5:34

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.
• 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.) – Brian Tung Feb 13 at 6:37
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 $$J=J\cap I=J\cap\emptyset=\emptyset$$.