Proof: By Induction

(related to Proposition: Number of Subsets of a Finite Set)

If \(X\) is a finite set with \(|X|=n\), then following the definition of binomial coefficients the number of its subsets \(X_i\subseteq X\) must be

\[|\{X_i,~X_i\subseteq X\}|=\sum_{k=0}^n \binom nk~~~~~~~~~~~~~~~~~~~~( * ).\]

Moreover, we observe some special cases.

Special case \(n=0\).

Since \(X=\emptyset\), there is only one subset \(X_i\subseteq\emptyset\), namely \(X_i=\emptyset\), thus \(|\{X_i,X_i\subseteq \emptyset\}|=1=2^0\).

Special case \(n=1\).

Let \(S=\{a\}\). Then there are two subsets \(X_i\subseteq\{a\}\), namely \(\{\emptyset\}\) and \(\{a\}\), thus \(|\{X_i,X_i\subseteq \{a\}\}|=2=2^1\).

Special case \(n=2\).

Let \(X=\{a,b\}\). Then there are four subsets \(X_i\subseteq\{a,b\}\), namely \(\{\emptyset\}\), \(\{a\}\), \(\{b\}\), and \(\{a,b\}\), thus \(|\{X_i,X_i\subseteq \{a,b\}\}|=4=2^2\).

Special case \(n=3\).

Let \(X=\{a,b,c\}\). Then there are eight subsets \(X_i\subseteq\{a,b,c\}\), namely \(\{\emptyset\}\), \(\{a\}\), \(\{b\}\), \(\{c\}\), \(\{a,b\}\), \(\{b,c\}\), \(\{a,c\}\), and \(\{a,b,c\}\), thus \(|\{X_i,X_i\subseteq \{a,b,c\}\}|=8=2^3\).

The above special cases suggest that the general formula is \[|\{X_i,X_i\subseteq X\}|=2^{|X|}.\] Comparing this with the formula \(( * )\), we conjecture for \(|X|=n\), \(n\ge 0\) that
\[|\{X_i,X_i\subseteq X\}|=\sum_{k=0}^n \binom nk=2^n~~~~~~~~~~~~~~~~~~~~( * * ),\] and prove the formula \(( * * )\) by induction.

Base Case (see also special case \(n=0\) above)

\[\sum_{k=0}^0\binom nk=\binom 00=2^0=1.\]

Induction step: Let \(( * * )\) be proven for all \(n\ge 0\).


\[\begin{array}{ccl} \sum_{k=0}^{n+1}\binom {n+1}k&=&\binom {n+1}0+\sum_{k=1}^{n+1}\binom {n+1}k~~~~~~~~~~~~~~~~~( *** )\\ &=&1+\sum_{k=1}^{n}\left(\binom nk+\binom n{k-1}\right)~~~~~~~~~~~~~~~~~( **** )\\ &=&1+\left(-1+\sum_{k=0}^{n}\binom nk\right)+\left(\sum_{j=0}^{n}\binom n{j}\right)~~~~~~~~~~~~~~~~~( ***** )\\ &=&2^n+2^n=2\cdot 2^n=2^{n+1}~~~~~~~~~~~~~~~~~( ****** ).\\ \end{array}\]

Thereby, in \( ( *** ) \) we have split the sum into the first term \(\binom{n+1}0\) from the rest of the sum, which equals \(1\), since there is only one subset with the cardinality \(0\) (i.e. \(\emptyset\)) of any set \(X\) with the cardinality \(n+1\). In \( ( **** ) \) we have applied the recursive formula for binomial coefficients. In \( ( ***** ) \) we have included in the first sum the term \(\binom n0\) which equals \(1\). By doing so, we corrected this inclusion by \(-1\), which cancels with the \(+1\) included in the previous step. We also have shifted the index of the second, replacing \(k-1\) by \(j=k-1\). In \( ( ****** ) \) we have performed the induction step.

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  1. Aigner, Martin: "Diskrete Mathematik", vieweg studium, 1993