Fixed points Property in discrete and indiscrete space Let $X$ be a set with at least two elements. Prove that the discrete space $(X,\tau)$ and the indiscrete space $(X,\tau')$ do not have the fixed point property. 
For the indiscrete space, I think like this. Since $(X,\tau')$ is an indiscrete space, so $\tau'={(\phi,X)}$. For any $x \in \tau'$, $f(\phi)=\phi$ or $f(x)=X$ for $x \in X$ . Note that $X$ is a set with at least two points, so $f(x)=X \neq x$. Hence indiscrete space has no fixed point. 
Now for the discrete space, i think that it will has the fixed point property. If $(X,\tau)$ is the discrete space, then $\tau$ consists of all the subsets of $X$, including all the singletons $\{x\}$, for $x\in X$. Then for $x\in X$, there is always a singleton such that $f(x)=\{x\}$. We can find the fixed point. What's wrong with my idea?
 A: Your first argument makes no sense: you seem to be confusing points of $X$ with subsets of $X$. The sets $\varnothing$ and $X$ aren't even in the domain of $f$: that domain is the set $X$, not the set $\tau'$. You have to show that there is a continuous function $f$ from $X$ to $X$ that has no fixed point. 
If $X$ is a two-point set, this is easy. Let $X=\{0,1\}$ with either the discrete or the indiscrete topology, and let $f:X\to X$ be defined by $f(0)=1$ and $f(1)=0$. Then $f$ is a continuous bijection with no fixed point. 
Now try to adapt this idea to arbitrary sets with at least two elements.
A: The basic observation is that any $f: X \rightarrow Y$ from a discrete space $X$ is continuous ($f^{-1}[O]$ is open in $X$ regardless of $f$ or $O \subseteq Y$), so any function from $X$ to $X$, where $X$ is discrete also is. And for a set with no structure but at least two points you can always find a function without a fixed point, see this question, and its answers.
The same holds for any function $f:X \rightarrow Y$ to an indiscrete space $Y$ ($f^{-1}[\emptyset] = \emptyset, f^{-1}[Y] = X$, which are always open in $X$), and so we can also choose any fixed point free function on an indiscrete space.
A: In both cases  it will be fail to have fixed point property
$1.$ For discrete space
Let $V \subset Y$ be a nonempty open subset of $Y$  with $a \in V$ and $b \in Y - V$, and let $U \subset X$ be any subset of $X$. Since $(X, \tau)$ is discrete topology so function $f: X \to Y$ defined by $$f(x) = \begin{cases} a & \text{ if } x \in U \\ b& \text{ if } x \notin U\end{cases}$$
is open .Here $f^{-1} (a)$ and $f^{-1}(b)$ both  are open in $U\subset X$
Therefore $f$ is continious but $f$ doesnot satisfied the property $f(x)= x$
$2.$ for indiscrete space
Consider $X = \{0, 1\}$ with the indiscrete topology i.e. the only two open sets are $\emptyset$ and $X = \{0, 1\}$ itself.
let $f:\{0, 1\} \to \{0, 1\}$ be defined by $f(x)=1-x$, then  $f$ will not have not have fixed point property
