Let $f \colon X \rightarrow X$ continuous. Show that if $X = [0,1]$, then there exists a point $x$ such that $f(x) = x$. The point $x$ is called the fixed point of $f$. What happens if $X$ is the space $[0,1)$ or the space $(0,1)$?
The first part is already done. But I have not been able to know what happens if $ X = [0,1) $ or $ X = (0,1) $. Could you give me any suggestion?
I have noticed that if I define $ g(x) = f (x)-x $, the intermediate value theorem might not be applicable to $g $.
When $ X = [0,1) $ I have the counterexample $ f (x) = \frac{1}{2}x + \frac{1}{2}$. But when X = (0,1) the case will be the same?
Best Answer
Suppose $X=[0, 1)$. Then $X$ is homeomorphic to $Y:=[0, \infty)$. So if we can find a continuous function $Y\to Y$ which does not have a fixed point, then we can find such a function for $X$ too. But coming up with such a function for $Y$ is easy. Just take $f:Y\to Y$ as $f(y) = 1+y$ and we are done.
Similarly, when $X=(0, 1)$ then $X$ is homeomorphic to $\mathbb R$.