Does there exist a function $f:\mathbb{R} \rightarrow \mathbb{R}$ with $f(\mathbb{R})=\mathbb{R}^*$ that is $1-1$ ?
Solution
Of course. There are plenty of examples for which the above holds. For example one such function would be $f(x)=x$ unless $x$ is natural. In naturals we define $f(0)=1, \; f(1)=2, \; f(2)=3$ etc. Another function would be:
$$f(x) = \begin{cases}
x &,&x <0 \\
1-x &, & 0< x \leq 1 \\
3-x &, &1< x \leq 2 \\
...\\
2n+1-x &, & n< x \leq n+1\\
...\\
\end{cases}$$
Of course such function cannot be continuous. We can see that from Bolzano. If $f$ takes the values $\pm 1$ then it shall also take the value $0$.
Solution
Of course. There are plenty of examples for which the above holds. For example one such function would be $f(x)=x$ unless $x$ is natural. In naturals we define $f(0)=1, \; f(1)=2, \; f(2)=3$ etc. Another function would be:
$$f(x) = \begin{cases}
x &,&x <0 \\
1-x &, & 0< x \leq 1 \\
3-x &, &1< x \leq 2 \\
...\\
2n+1-x &, & n< x \leq n+1\\
...\\
\end{cases}$$
Of course such function cannot be continuous. We can see that from Bolzano. If $f$ takes the values $\pm 1$ then it shall also take the value $0$.
No comments:
Post a Comment