Linear mapping between two different countably infinite sets

Take the natural numbers $$\mathbb{N}_0 = \{0, 1,2,3,\dots\}$$ and the integers $$\mathbb{Z} = \{\dots, -1, 0, 1, \dots\}$$.

Can we come up with a bijective linear mapping $$T: \mathbb{N} \rightarrow \mathbb{Z}$$?

Consider the mapping $$0 \mapsto 0$$, $$1 \mapsto 1$$, $$2\mapsto -1$$, $$3\mapsto 2$$, $$4 \mapsto -2$$, etc. This can be summarized by the function $$T(x) = \begin{cases} -\frac{x}{2} \text{ if } x \text{ even}, \\ \frac{x+1}{2} \text{ if } x \text{ odd}. \end{cases}$$

$$T$$ is not linear: if $$x$$ is even and $$y$$ is odd, then $$x+y$$ is odd and so $$T(x+y) = \frac{x+y+1}{2}$$, but $$T(x)+T(y) = \frac{-x+y+1}{2}$$.

Is it possible to come up with a linear mapping? Intuitively, a linear mapping should stretch, rotate, or translate a set, rather than "zig-zagging". I don't know if there is a way to do this to turn $$\mathbb{N}$$ into $$\mathbb{Z}$$?

In general: is there some explanation of when a linear mapping exists between two countably infinite sets, and when it doesn't exist?

Edit: it seems from some comments that a linear mapping exists between two sets when those sets are both vector spaces. Is this the answer?

• What about the linear map $T(x)=x$? Commented Aug 6 at 17:04
• Oh sorry, I meant to specify that it should be a bijection Commented Aug 6 at 17:04
• The bijections here are not linear, I think. Commented Aug 6 at 17:06
• @DietrichBurde thank you. I added a sentence about what I think a linear mapping should intuitively look like. Since every bijection between $\mathbb{N}$ and $\mathbb{Z}$ is "zigzagging", it seems like no linear map between them exists? Commented Aug 6 at 17:08
• Suppose f(x)>0 is a member of N then f(-x)=-f(x) for linearity but then f(-x)<0 which is a contradiction. f(x)=0 for all x in Z should probably satisfy linearity but I'm not sure. Commented Aug 7 at 5:04

Suppose $$T(0)=a$$ and $$T(1)=b$$. Then $$T^{-1}(a)=0$$ and $$T^{-1}(b)=1$$, so by linearity $$T^{-1}(a-b)=-1$$, which is not in $$\Bbb N$$.

• thank you! are you using the fact that there can be an arithmetic progression in $\mathbb{Z}$ whose image is not an arithmetic progression in $\mathbb{N}$ ? How do you show this? Commented Aug 6 at 17:13
• @900edges I’m not sure if my reasoning is correct here, but $\mathbb{N}$ is not a vector space since it doesn’t posses additive inverses to each element. And by definition, linear maps are defined on vector spaces. Commented Aug 6 at 17:28
• @PaulAsh Thanks for pointing that out. Is the answer to the general version of the question "linear mappings only exist from vector space to vector space"? Commented Aug 6 at 22:24
• @900edges I believe so, yes. (Btw, $\mathbb{Z}$ is also not a vector space). Commented Aug 6 at 23:01
• @PaulAsh: Well, I suppose that's right: a linear mapping is a mapping between vector spaces. But the axioms are just (i) $f(x)+f(y)=f(x+y)$ and (ii) $f(ax)=af(x)$. And these axioms make sense in wider contexts. Here, however, that $a$ is problematic: is it $a\in\Bbb Z$, or $a\in\Bbb N$? So perhaps the question is simply badly posed. Commented Aug 6 at 23:48

The term "linear mapping" is generally defined on vector spaces, and neither $$\mathbb{N}$$ nor $$\mathbb{Z}$$ is a vector space, so I'm going to try to rephrase your question to avoid that term:

Rephrased Question: Does there exist a bijective function $$T:\mathbb{N} \rightarrow \mathbb{Z}$$ such that for all $$x,y \in \mathbb{N}$$, $$T(x+y) = T(x) + T(y)$$?

The answer to this question is No. To prove this, we'll prove an even stronger statement:

Claim: Any $$T:\mathbb{N} \rightarrow \mathbb{Z}$$ such that $$T(x+y) = T(x) + T(y)$$ is of the form $$T(n) = nz$$, for some fixed $$z \in \mathbb{Z}$$.

Proof of Claim: First we see that $$T(0) = T(0+0) = T(0) + T(0)$$, so $$T(0) = 2T(0)$$. $$0$$ is the only integer equal to its double, therefore $$T(0) = 0$$.

Next, for any $$n > 0 \in \mathbb{N}$$, we can express it as the sum of $$n$$ copies of $$1$$. Thus:

$$T(n) = T(1 + \dots+1) = T(1) + \dots + T(1) = nT(1)$$.

Setting $$z = T(1)$$ proves the claim.

The claim implies the image of $$T$$ is either completely nonnegative or completely nonpositive, meaning it cannot be bijective.

• awesome thanks :) Commented Aug 9 at 14:16