Does $R \subsetneq \mathbb{N}$ contradict '$f:\mathbb{N}\rightarrow R$ is one-to-one and onto' There is a paradox I can't resolve. If we take the set of natural numbers $\mathbb{N}^0$. We could take the element $0$ as our starting position and reference to all other elements by defining a successor function $S(n)=n+1$ (i.e. $1$ is $S(0)$, 2 is $S(S(0))$ and so on).

If we take the set of all references it would look like $R=\{S, SS, SSS, SSSS, ...\}$. Now it seems obvious that for every element in $\mathbb{N}$ (without 0) there is a corresponding element in $R$ ($\forall x \in \mathbb{N}:x \in R$). Or in other terms, there is a function $f:\mathbb{N}\rightarrow R$ that is one-to-one and onto.

However, one can rewrite the set $R$ by (notationally) replacing $S$ with $1$. $R$ would then look like $R=\{1, 11, 111, 1111, ...\}$. In this case $R$ is the set of all natural numbers only composed of a single digit. Of course this is only a subset of all natural numbers ($R \subsetneq \mathbb{N}$).

Does $R \subsetneq \mathbb{N}$ contradict '$f:\mathbb{N}\rightarrow R$ is one-to-one and onto'? Where is the mistake in the above argument?
I'm not entirely sure if this answers your question, but you can indeed form a bijection $f: \mathbb{N} \to \{1,\ 11,\ 111, \ \dots\}$ via:
$$
f(n) = 10^{n-1} + 10^{n-2} + \ \cdots \ + 1 = \sum_{i= 0}^{n-1}10^i
$$

This is an injective mapping, because $f$ is monotone, and surjective because for each number $l =\sum_{i= 0}^M10^i \in R$, $f(M+1) = l$.  

Edit: also, as someone else has said, in general $A \subsetneq B$ does not imply $\#A &lt; \#B$. For example, $\mathbb{Z}$ is in a bijective correspondence with $\mathbb{Q}$.
There is no contradiction because $\Bbb N$ is infinite and infinite sets can biject with its proper subset (assuming axiom of choice).