Examine if there exists an $1-1$ and onto function $f:\mathbb{Q} \rightarrow \mathbb{Q} \setminus \{0\}$ that preserves order in $\mathbb{Q}$?
Solution
Yes, there exists. Reason being that any two sets that are totally arranged , countable, dense and not bounded sets are similar (identical) sets. Therefore , this function , as it is quite reasonable , must be the function of similarity.
Let us construct the function. We enumerate $\mathbb{Q}$ ($a_n$ sequence) and $\mathbb{Q}^*$ ($b_n$ sequence) . Having defined all $f(a_k), \; f^{-1} (b_k) , \; \; k \leq n$ we then appropriately define $f(a_{n+1}), \; f^{-1}(b_{n+1})$ in such way that we have the onto and the $1-1$. This can be done because $\mathbb{Q}$ is dense.
Uniting all functions we get an $1-1$ function and onto $\mathbb{Q}$ and we are done.
Solution
Yes, there exists. Reason being that any two sets that are totally arranged , countable, dense and not bounded sets are similar (identical) sets. Therefore , this function , as it is quite reasonable , must be the function of similarity.
Let us construct the function. We enumerate $\mathbb{Q}$ ($a_n$ sequence) and $\mathbb{Q}^*$ ($b_n$ sequence) . Having defined all $f(a_k), \; f^{-1} (b_k) , \; \; k \leq n$ we then appropriately define $f(a_{n+1}), \; f^{-1}(b_{n+1})$ in such way that we have the onto and the $1-1$. This can be done because $\mathbb{Q}$ is dense.
Uniting all functions we get an $1-1$ function and onto $\mathbb{Q}$ and we are done.
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