# For continuous $f:X\rightarrow [0,1]$ is it true that $a<b\implies \overline{f^{-1}[0,a)}\subset f^{-1}[0,b)$?

Let $f:X\rightarrow [0,1]$ be continuous function. Is it true that $a<b\implies \overline{f^{-1}[0,a)}\subset f^{-1}[0,b)$? I'm asking because I can't think of a counterexample. (This is not homework.)

## 1 Answer

If $a < b$ then $[0,a] \subset [0,b)$ and as $f$ is continuous you have that $f^{-1}([0,a])$ is closed giving

$$\overline{f^{-1}([0,a))} \subseteq \overline{f^{-1}([0,a])} = f^{-1}([0,a]) \subseteq f^{-1}([0,b))$$