Let $f : \mathbb{R} \to \mathbb{R}$ be a continous map then which of the following cannot be the image of
$[0,1)$ under $f$ ?
(a) $0$
(b) $(0,1)$
(c) $[0,1)$
(d) $[0,1]$
Now , I know the following theorem if $f$ is a continuous map iff inverse image of every closed set is a closed set.
So, (a) and (d) must be the correct choices,
But for option (a) I can easily set $f(x) = 0$, hence it must be incorrect.
But I don't really understand why isn't option (a) the correct choice, According to the theorem all the hypothesis are satisfied hence $0$ should be one of the answer.
Can someone please clear my doubt as why is option (a) Not the correct choice ?
Best Answer
I think the only good way to go about this really is to apply the property that $f$ maps compacts to compacts. So we see that $f([0,1])$ is compact, and $f([0,1])=f([0,1))\cup f(\{1\})$, but the latter is a single point, and adding a single point to $(0,1)$ can never make it closed (hence, not compact). Hence, $(0,1)$ must be wrong.
As for the other options:
$\{0\}=f([0,1))$ if $f(x)=0$ for all $x$.
$[0,1)=f([0,1))$ if $f(x)=x$
$[0,1]=f([0,1))$ if $f(x)=2\cdot|\frac{1}{2}-x|$