Natural injection $\binom{[n]}{k} \to \binom{[n]}{k+1}$ for $0\leq k < \frac n 2$.

Asked Nov 12 '22 at 16:13

Active Nov 12 '22 at 16:13

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Let $n$ and $k$ be integers with $0\leq k < \frac n 2$, and denote $[n]=\{1,2,\dots,n\}$.

There are a myriad ways one can show that $\binom{n}{k} \leq \binom{n}{k+1}$, but I can't think of any "natural" injection $f: \binom{[n]}{k} \to \binom{[n]}{k+1}$.

I could of course just construct such map by settting $\binom{[n]}{k} = \{A_i : 1\leq i \leq \binom n k\}$ and $\binom{[n]}{k+1} = \{ B_i : 1\leq i \leq \binom n {k+1} \}$ and $f(A_i)=B_i$ for all $1\leq i \leq \binom n k$ but that would feel "over-engineered". Such a simple statement should correspond to a simple and intuitive map, shouldn't it?

asked Nov 12 '22 at 16:13

Steve

@DavidK If $S$ is a set and $k$ is an integer, then $\binom{S}k$ is defined to be the set of size-$k$ subsets of $S$. This is a common convention. – Mike Earnest Nov 12 '22 at 16:34
$X \subset f(X)$ would be nice if possible, too. – aschepler Nov 12 '22 at 16:38
It seems you can just add one element to every set in $\binom{[n]}{k},$ but which element? – David K Nov 12 '22 at 16:40
1

@DavidK And in an invertible way: Given $f(X)$, it must be possible to figure out which element must be removed to get back to $X$. – aschepler Nov 12 '22 at 16:42
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I gave an injection in this answer, one that satisfies $X\subset f(X)$. Should I close as a duplicate, or reproduce that answer here? – Mike Earnest Nov 12 '22 at 17:36
@MikeEarnest excellent! Thank you! – Steve Nov 13 '22 at 16:09

Natural injection $\binom{[n]}{k} \to \binom{[n]}{k+1}$ for $0\leq k < \frac n 2$.

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