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Let $M$ be a non-empty set. Show that $M$ has as many subsets with an odd number of elements as subsets with an even number of elements.

I already found a solution which said to use the identity $\sum_{k = 0}^{n} (-1)^{k} {n \choose k} = 0$, which clearly can be seen that the sign alternates for odd and even $k$. I also found a prove of the identity here. The point is that, I want to know whether this can be proved in any other method apart from that identity, and if yes, how?

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Since the set is non-empty we can choose $m\in M$. Then there is a perfect pairing between the set of subsets containing $m$ and the set of subsets which don't contain $m$ (just add $m$ if it isn't there, and delete it if it is). But in each such pair one subset has an even number of elements and the other has an odd number of elements.

As an explicit example, suppose your original set has three elements $\{A,B,C\}$. Say we choose the element $A$. Then The pairing is $$\left[\emptyset, \{A\}\right], \left[\{B\}, \{A,B\}\right],\left[\{C\}, \{A,C\}\right],\left[\{B,C\}, \{A,B,C\}\right]$$

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  • $\begingroup$ What do you mean exactly with "perfect pairing". Can you explain it a bit? $\endgroup$ – user72151 Nov 8 '15 at 15:16
  • $\begingroup$ Just meant a pairing. The "perfect" part was just meant to emphasize that there are no "fixed points", that is every subset is paired with a subset not equal to itself. That's clear in this case, since one member of the pair contains $m$ and the other does not so the subsets can not coincide. $\endgroup$ – lulu Nov 8 '15 at 15:18
  • $\begingroup$ I did this drawing, assuming that my set $M = \{1,2,3\}$, and I chose $m = 2$. So, I listed the subsets that have $m$ in one side, and the ones that don't have it on the other side, and I did some kind of bipartite matching. I don't know whether this is the correct idea, but if it is, then $\{1,3\}$ matches with $\{2,3\}$, and both of them are of even length. Anyway, here is the graph: prntscr.com/90fw0e $\endgroup$ – user72151 Nov 8 '15 at 15:38
  • $\begingroup$ $\{1,2,3\}$ is paired with $\{1,3\}$. The full pairing is $$\[\emptyset,(2)\],\[(1),(1,2)\],\[(3),(2,3)\],\[(1,3),(1,2,3)\]$$ $\endgroup$ – lulu Nov 8 '15 at 15:40
  • $\begingroup$ And what happens to $\{2,3\}$, we can randomly pair as we want? $\endgroup$ – user72151 Nov 8 '15 at 15:44
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Let $M$ be a non-empty set, $O$ the set of odd subsets and $E$ the set of even subsets. As $M$ is non-empty, we may pick an element $m\in M$. Then $X\mapsto X\mathbin\Delta\{m\}$ is a bijection $O\to E$ (as well as $E\to O$), hence both must have equal cardinality. (Recall that $A\mathbin\Delta B:=(A\cup B)\setminus(A\cap B)$, i.e., we "toggle" membership of $m$)

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  • $\begingroup$ And what's XX in our case? Just a bijective function (or mapping)? Because I understood, what you meant with toggle membership of mm, I guess something like @lulu's answer above. But, I still don't see it mathematically, Simply, if $X$ includes $m$, then $XΔ\{m\}$, will be just $X$, and if it doesn't include we will have $X∪\{m\}$. $\endgroup$ – user72151 Nov 9 '15 at 10:43
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The OP knows the relevance of the identity, but here's a quick summary for any other reader. Give each subset with exactly $k$ elements a score of $(-1)^k$. The identity then states that the total score of all subsets is 0. However, while subsets with an even number of elements score $1$, those with an odd number of elements score $-1$. Therefore, the total score is the number of even-size subsets minus the number of odd-size subsets. Since this total is 0, there are equally many subsets of the two types.

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