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So i will start by saying that i am using the discrete topology for $\mathbb{Z}$ and for $\mathbb{N}$

Defining the function.

$f: (\mathbb{Z},\tau_1)\longrightarrow (\mathbb{N},\tau_2)$

$f(x) = \begin{cases} 2x &\quad\text{if}\ge0, & x \in \mathbb{Z} \\ -2x-1 &\quad\text{if}<0 \\ \end{cases} $

the inverse would be:

$f^{-1}(x) \begin{cases} y/2 &\quad\text{if y even}, & x \in \mathbb{N} \\ \frac{y+1}{2} &\quad\text{if y odd} \\ \end{cases}$

im trying to use the fact - that in the discrete topologies all the subsets of $\mathbb{Z} $ and $\mathbb{N} $ are opened - to prove that $f$ is continuous. Can i do that? And can i prove that this function is bijective on this topology and about the inverse ?

sorry for any mistake, and any help or tip would be greatly appreciated

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  • $\begingroup$ You are correct: as far as the domain has discrete topology, the function is continuous. And the function you created is one to one. $\endgroup$ – Miles Zhou Oct 9 '18 at 2:53
  • $\begingroup$ thanks a lot!!, still trying to do the intermediate steps that are only in my head hahaah $\endgroup$ – Saiten Oct 9 '18 at 2:54
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Your function $f$ works, and your definition of the inverse shows that $f$ is a bijection.

You're left to show that $f$ and $f^{-1}$ are continuous - or equivalently that $f$ is continuous and open.

For example, to show $f$ is continuous, you need to show that the preimage of any open set is open. But in the discrete topology every set is open...

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  • $\begingroup$ Thanks , now I'm finishing writing $\endgroup$ – Saiten Oct 9 '18 at 3:03

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