Does there exist a surjection from the integers to the naturals?

[docnet]

Homework Statement

a) Does there exist a surjection from the integers to the natural numbers?

b) Does there exist a surjection from the real numbers to the natural numbers?

Relevant Equations

$\mathbb{R},\quad \mathbb{N},\quad\mathbb{Z}$

a)
Yes.
One surjection from $\mathbb{Z}$ to $\mathbb{N}$ is the double cover of $\mathbb{N}$ induced by $f:\mathbb{Z}\longmapsto\mathbb{N}$ with
$$f(z)=\begin{cases}
-z & ,\forall z<0\\
z+1 & ,\forall 0\leq z
\end{cases}$$
b)
Yes.
One surjection from $\mathbb{R}$ to $\mathbb{N}$ is the the projection $p:\mathbb{R}\longmapsto\mathbb{N}$ with $f(r)=r$ for all $r\in\mathbb{N}\subset\mathbb{R}$.

 

[Office_Shredder]

Does $\mathbb{N}$ not include 0 for you?

(b) feels incomplete. What is the relationship between p and f, and what are you supposed to to with real numbers that are not natural numbers?

 

[docnet]

Does $\mathbb{N}$ not include 0 for you?

If $\mathbb{N}$ includes $0$, a surjection from $\mathbb{Z}$ to $\mathbb{N}$ is given by
$$f(z)=\begin{cases}
-z & ,\forall z<0\\
z & ,\forall 0\leq z
\end{cases}$$

(b) feels incomplete. What is the relationship between p and f, and what are you supposed to to with real numbers that are not natural numbers?

I agree with you. They were all meant to be $p$'s, but I hit $f$ by force of habit. The "complete" projection function is $p:\mathbb{R}\longmapsto\mathbb{N}$ with
$$p(r)=\begin{cases}r& ,\forall r\in\mathbb{N}\\
\text{undefined}& ,\forall r\notin\mathbb{N}
\end{cases}$$

 

[Office_Shredder]

p(r) can't be undefined, by definition a function is always defined. But notice it doesn't matter much what value you pick - you already covered the surjection part, so you could make it e.g.. always 0 for the rest of the space

If $\mathbb{N}$ includes $0$, a surjection from $\mathbb{Z}$ to $\mathbb{N}$ is given by
$$f(z)=\begin{cases}
-z & ,\forall z<0\\
z & ,\forall 0\leq z
\end{cases}$$

I agree that works. But which one does your class/textbook use? I was just surprised because they usually include 0, but if it didn't then your first answer is fine.

 

[docnet]

p(r) can't be undefined, by definition a function is always defined. But notice it doesn't matter much what value you pick - you already covered the surjection part, so you could make it e.g.. always 0 for the rest of the space

Thank you. I understand, and I can't believe I missed that.
Then let's make it $p:\mathbb{R}\longmapsto\mathbb{N}$ with
$$p(r)=\begin{cases}r& ,\forall r\in\mathbb{N}\\
0 & ,\forall r\notin\mathbb{N}
\end{cases}$$

I agree that works. But which one does your class/textbook use? I was just surprised because they usually include 0, but if it didn't then your first answer is fine.

Those questions are taken from a midterm review sheet that I stumbled upon. The class is named "Introduction to higher maths" and it is a proof-based course I have not taken, so I don't know what the professor wants to do.

 

[PeroK]

In general if $A \subseteq B$ are non-empty sets and $a \in A$, then $f:B \rightarrow A$ defined by:
$$f(b)=\begin{cases}b& \ \ (b \in A)\\
a & \ \ (b \notin A)
\end{cases}$$ is a surjection.

 

[PeroK]

Does $\mathbb{N}$ not include 0 for you?

There is no definitive standard about this.

https://en.wikipedia.org/wiki/Natural_number

 

[docnet]

p(r) can't be undefined, by definition a function is always defined. But notice it doesn't matter much what value you pick - you already covered the surjection part, so you could make it e.g.. always 0 for the rest of the space

Question: Does this imply that any function that has an infinity, like $f(x)=\frac{1}{x}$ does at $x=0$, and like $f(x)=\tan(x)$ does at $x=\frac{\pi}{2}+\pi n$ is not a proper function over all of $\mathbb{R}$? (really, curious.)

 

[PeroK]

Question: Does this imply that any function that has an infinity, like $f(x)=\frac{1}{x}$ does at $x=0$, and like $f(x)=\tan(x)$ does at $x=\frac{\pi}{2}+\pi n$ is not a proper function over all of $\mathbb{R}$? (really, curious.)

Yes. The domain of the function $\frac 1 x$ is $\mathbb R - \{0\}$.

Note that technically this function is continuous at every point in its domain. That said, it's common to talk about a discontinuity at $x = 0$. What that really means is that we cannot extend the function $\frac 1 x$ to be continuous on $\mathbb R$.

We can compare this to the function $\frac{\sin x}{x}$, which likewise is not defined at $x = 0$, but which can be extended to a continuous function on $\mathbb R$ by defining the function to be $1$ at $x = 0$. This is called a removable discontinuity. Although, again, it's not technically a discontinuity in the function, but a disconnectedness of the domain that is being removed.