Let $f:[0,1)\cup[2,3]\to\mathbb{R}$ be given by $$f(x) = \begin{cases} x & x\in[0,1)\\ x-1 & x\in[2,3] \end{cases}$$ Show that $f$ is continuous, one-to-one, and $f([0,1)\cup[2,3]) = [0,2]$. It then follows that $f$ is invertible (where defined). Show then that $f^{-1}:[0,2]\to\mathbb{R}$ is not continuous.
Conitnuity
Let $x_0\in [0,1)$, and suppose that $0 < |x - x_0| < \delta$.
Then if $\delta = \epsilon$, $0 < |f(x) - f(x_0)| = |x - x_0| < \delta = \epsilon$.
So $f$ is continuous on the interval $[0,1)$.
Let $x_0\in [2,3]$, and suppose that $0 < |x - x_0| < \delta$.
Then if $\delta = \epsilon$, $0 < |f(x) - f(x_0)| = |x - 1 - x_0 + 1| = |x - x_0| < \delta = \epsilon$.
So $f$ is continuous on the interval $[2,3]$.
The above also implies that the limit of $f$ at $x\to 1$ from the left is $1$, and the limit of $f$ as $x\to 2$ from the right is $1$.
Therefore $f$ is continuous.
I feel like I'm messing something up with my claim about limits. It feels hand-wavy and I can't figure out why.
One-to-one
Suppose that $f(a) = f(b)$, where $a,b\in [0,1).$
Then $f(a) = a = b = f(b)$. Therefore $a=b$.
Now suppose that $f(a) = f(b)$, where $a,b\in [2,3].$
Then $f(a) = a-1 = b-1 = f(b)$. Therefore $a=b$.
Because $f([0,1)) = [0,1)$, and $f([2,3]) = [1,2]$, no points in the first set are in the second set, therefore all points have been considered, and $f$ is one-to-one.
I don't feel confident about this last step either
$\pmb{f([0,1)\cup[2,3]) = [0,2]}$
As shown in the last part, since $f([0,1)) = [0,1)$ and $f([2,3]) = [1,2]$, $f([0,1)\cup[2,3]) = [0,2].$
Is this legit?
$\pmb{f^{-1}:[0,2]\to\mathbb{R}}$ is not continuous
$$f^{-1}(x) = \begin{cases} x & x\in [0,1) \\ x+1 & x\in [1,2] \end{cases} $$ As shown in the first section, the limit of $f^{-1}(x)$ as $x\to 1$ from the left is $1$, and the limit of $f^{-1}(x)$ as $x \to 1$ from the right is $2$.
Therefore $f^{-1}(x)$ is not continuous.
Not sure if I need to show what the inverse is