bijection between natural numbers and set of strictly growing finite rows

Question

I have to construct a bijection between $\Bbb N$ and $\Bbb S_+$ ,where $\Bbb S_+$ is the set of strictly growing finite rows: $\Bbb S_+ = \{ (n_0,n_1...n_k) \;|\; k \in \Bbb N,n_i \in \Bbb N, n_0 <n_1<n_2<...<n_k\}$ and $\Bbb N$ are the natural numbers.

I thought that if I represent the rows via numbers in binary : 43 = 101011 when I have 1 i will write the index as an element in my row, 43 ->101011 -> 0<1<3<5 this means that if I use the function $f(x)=x$ which is a bijection I compare the natural numbers with a row. Does this work?

Also how do I put this in a more formal way?

I don't understand at all how you get from $101011$ to $0,1,3,5$ nor how you are using $f(x)=x$ — Ross Millikan, Nov 13 '17 at 16:21
101011 from right to left from 0 to 5 right most "1" is on the 0th position and we get 0 so on for the rest — yolo expectz, Nov 13 '17 at 16:24

Ross Millikan · Answer 1 · 2017-11-13T16:35:01.417

1

Your approach works fine. A nice way to describe it is $$f: \Bbb S_+ \to \Bbb N\\s\to \sum_{i \in s}2^{i}$$ This assumes $0 \in \Bbb N$ as it appears in your example and assumes that you allow the empty sequence, which will be mapped to $0$.

edited Nov 13 '17 at 16:35

answered Nov 13 '17 at 16:28

Ross Millikan

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the "s" is an element of $\Bbb S_+$ right? – yolo expectz Nov 13 '17 at 16:39
That is right. It is one of your increasing sequences. – Ross Millikan Nov 13 '17 at 18:54

score 0 · Answer 2 · answered Nov 25 '17 at 06:49

We may assume ${\Bbb N}=\{1,2,3,...\}$. There is a natural bijection between $S_+$ and ${\Bbb N}^{k+1}$ given by: $$ \phi (n_0,n_1,...,n_k) = (n_0, n_1-n_0,...,n_k-n_{k-1}).$$ Compose this with a bijection between ${\Bbb N}^{k+1}$ and ${\Bbb N}$ and you are done. The latter you may construct first between ${\Bbb N}^2$ and ${\Bbb N}$ (zig-zag-path) and then use induction in $k$.

bijection between natural numbers and set of strictly growing finite rows

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