For which of the following subspaces $X$ of $\mathbb{R}$ does every continuous surjective map $f : X \to X$ have a fixed point?
choose the correct option
$a.$$[3,∞) $
$b. [1, 2]∪[3,∞)$
My attempt : I think option $a$ is correct and option $b$ is not correct. Take a continuous function
$$f:[3,\infty]\to[3,\infty]$$
and define
$$g:[3,\infty]\to[3,\infty]$$ $$g(x)=f(x)-x$$
It follows that $g(3)\geq 0$ and $g(\infty)\leq 0$. By the intermediate value theorem there is $c\in[3,\infty]$ such that $g(c)=0$. Thus $f(c)-c=0$ and so $f(c)=c$.
Therefore every continuous surjective map $f :[3,\infty) \to[3,\infty)$ have a fixed point
option $b$ is not correct because $[1,2]\cup [3,\infty]$ doesn't have the fixed Point Property. Take $$f(x)=\begin{cases} 1&\mbox{if }x\in[1,2] \\ 31&\mbox{otherwise} \end{cases} $$
First, recall that a function $f \colon X \to X$ is surjective if for any $y \in X,$ there exists $x \in X$ such that $f(x) = y.$ So, the second function you gave is not surjective because for $3 \in [1, 2] \cup [3, \infty),$ there is no corresponding $x \in [1, 2] \cup [3, \infty)$ such that $f(x) = 3.$
Furthermore, note that we are working in $\mathbb{R},$ so $\infty$ is not an element of $\mathbb{R}.$ So, it doesn't make sense to discuss the interval $[3, \infty],$ since $\infty$ can't be contained in any interval of the real numbers.
Addressing part (b) first, it turns out that any surjective continuous $f \colon [1, 2] \cup [3, \infty) \to [1, 2] \cup [3, \infty)$ must have a fixed point. You can find the details in this first answer to this question: Existence of fixed point for a continuous function on an infinite closed set.
Now, we will address part (a). Suppose that $f \colon [3, \infty) \to [3, \infty)$ is surjective and continuous. We claim that $f$ must have a fixed point. If $f(3) = 3,$ then $3$ is a fixed point for $f,$ and we are done. So, we can suppose $f(3) \neq 3.$
Since the range of $f$ must lie in $[3, \infty),$ it follows that $f(3) \in [3, \infty).$ Since $f(3) \neq 3$, we must have $f(3) > 3.$ On the other hand, since $f$ is surjective, there must exist some $x_{0} \in [3, \infty)$ such that $f(x_{0}) = 3.$ We know that $x_{0}$ can't be equal to $3$ (since $f(3) \neq 3$), so we must have $x_{0} > 3.$
Now, consider the function $g \colon [3, \infty) \to \mathbb{R},$ given by $$g(x) = f(x) -x.$$ Since $f$ is continuous, $g$ is also continuous. Note that $$g(3) = f(3) - 3 > 3 -3 =0,$$ and $$g(x_{0}) = f(x_{0}) - x_{0} = 3 - x_{0} < 3 - 3 = 0$$ (we have used the facts that $f(3) > 3$ and $x_{0} > 3$ here).
So, $g(3)$ is positive, and $g(x_{0})$ is negative. Since $g$ is continuous, by the intermediate value theorem, there exists $x_{1} \in (3, x_{0})$ such that $g(x_{1}) = 0.$
Then, we have $$0 = g(x_{1}) = f(x_{1}) - x_{1},$$ hence we have $$f(x_{1}) = x_{1}.$$ Since $x_{1} \in (3, x_{0}) \subseteq [3, \infty),$ we see that $x_{1}$ is a fixed point for $f \colon [3, \infty) \to [3, \infty).$ This proves that any continuous surjective $f \colon [3, \infty) \to [3, \infty)$ has a fixed point, as claimed.