• Dec 4th 2005, 01:41 PM
ummmm
Let Nk= {1,2,…,k} be a finite set with k elements. This problems shows that if n, r elements of N with n > r, then there is no one-to-one function from Nn to Nr. We prove this by induction on n ≥ 2 for the following statement.
(Pn): For all r with 1 ≤ r < n, there is no one-to-one function from Nn to Nr.
a. The first step covers the general case when r=1 and any n>r.
Prove that if n > r=1, then there is no one-to-one function from Nn to Nr.
b. Show that the base case (P2) is true.
c. Assume the induction hypothesis (Pn) is true and prove that (Pn+1) is true.
Hint: If r ≥ 2 and f: Nn+1 to Nr were one-to-one, then the restriction of f to Nn into Nr \ {f(n+1)} would be one-to-one. You may also use the fact that there is a bijection from Nr \{f(n+1)} to Nr-1.
d. Conclude that (Pn) is true for all n≥ 2.
e. Prove that if n,r is an element of N with n not = r, then Nn does not have the same cardinality as Nr.
• Dec 4th 2005, 07:24 PM
CaptainBlack
This is fairly easy, as you are lead by the hand through the proof:

Quote:

(Pn): For all r with 1 ≤ r < n, there is no one-to-one function from Nn to Nr.

Let Nk= {1,2,…,k} be a finite set with k elements. This problems shows that if n, r elements of N with n > r, then there is no one-to-one function from Nn to Nr. We prove this by induction on n = 2 for the following statement.
(Pn): For all r with 1 = r < n, there is no one-to-one function from Nn to Nr.

a. The first step covers the general case when r=1 and any n>r.
Prove that if n > r=1, then there is no one-to-one function from Nn to Nr.
We will take:

(Pn): For all $\displaystyle n$ and $\displaystyle r$ in $\displaystyle \mathbb{N}$ with $\displaystyle 1 \leq r < n$, there is no one-to-one function
from $\displaystyle N_n$ to $\displaystyle N_r$.

Suppose there is a one-to-one function $\displaystyle f_{n,r}$ from $\displaystyle N_n$ to $\displaystyle N_r$ where $\displaystyle r=1$ and $\displaystyle n>r$.
Then $\displaystyle 1$ belongs to $\displaystyle N_n$ as does $\displaystyle 2$. Then by definition of a one-to-one function $\displaystyle f_{n,r}(1) \neq f_{n,r}(2)$,
but $\displaystyle f_{n,r}(1) = 1$, as does $\displaystyle f_{n,r}(2)$ as this is the only element in $\displaystyle N_r$, a contradiction, so there can be no
such function.

Quote:

b. Show that the base case (P2) is true.
Part a shows that for all $\displaystyle n>1$, there is no one-to-one function between
$\displaystyle N_n$ and $\displaystyle N_1$. Let $\displaystyle n=2$ and this is a proof that $\displaystyle P(2)$ is true.

Quote:

c. Assume the induction hypothesis (Pn) is true and prove that (Pn+1) is true.
Hint: If r = 2 and f: Nn+1 to Nr were one-to-one, then the restriction of f to Nn into Nr \ {f(n+1)} would be one-to-one. You may also use the fact that there is a bijection from Nr \{f(n+1)} to Nr-1.
Suppose that for some $\displaystyle n \geq 2$ P(n) is true and that P(n+1) is false.
Let $\displaystyle f_{n+1,r}$ be a one-to-one function between $\displaystyle N_{n+1}$ and $\displaystyle N_r, r<n$, which
exists by supposition. Then the restriction of $\displaystyle f_{n+1,r}$ to $\displaystyle N_n$ is a one-to-one
function from $\displaystyle N_n$ to $\displaystyle N_r$, but this contradict the supposition that P(n) is true.
Hence if P(n) is true so is P(n+1).

Quote:

d. Conclude that (Pn) is true for all n= 2.
By b P(2) is true, by c if P(n) is true so is P(n+1), hence
by the principle of induction P(n) is true for all $\displaystyle n \geq 2$.

Quote:

e. Prove that if n,r is an element of N with n not = r, then Nn does not have the same cardinality as Nr.
Two sets $\displaystyle A$ and $\displaystyle B$ have the same cardinality if there exits a one-to-one
function from $\displaystyle A$ to $\displaystyle B$ and a one-to-one function from $\displaystyle B$ to $\displaystyle A$.

If $\displaystyle n \neq r$ without loss of generality we may assume $\displaystyle n>r$,
and so there is no one-to-one function from $\displaystyle N_n$ to $\displaystyle N_r$
and so they do not have the same cardinality.

(We may assume $\displaystyle n>r$ because either $\displaystyle n>r$ or
$\displaystyle r>n$, and in the latter case the same argument goes through
but with $\displaystyle n$ and $\displaystyle r$ interchanged.)

RonL