Decide whether the following sets have the fixed-point property. My definition for the FPP: A topological space $X$ has the FPP if for every continuous function $f: X \to X,$ there exists an $x_0 \in X$ such that $f(x_0)=x_0.$
Do the $A$ and $B$ have the fixed-point property?

$ A = \{(x,0):-1 \leqslant x \leqslant 1 \} \cup \{(0,y):0 \leqslant y \leqslant 1 \}$

Suppose $a \in A.$ Then either $a = (x,0)$ where $-1 \leqslant x \leqslant 1$ or $a = (0,y)$ where $0 \leqslant y \leqslant 1.$ 

$B = \{(x,y):x^2+y^2=1 \} \cup \{(x,0):1 \leqslant x \leqslant 2 \}$

Suppose $b \in B.$ Then either $b =(x,y)$ where all ordered pairs $(x,y)$ lie on the unit circle centered at $(0,0)$ or $b =(x,0)$ with $1 \leqslant x \leqslant 2.$ 
I'm having trouble coming up with functions without fixed points here. Is there a function that doesn't have fixed points in these sets? Or do these sets have the FPP? 
 A: The case for $B$ is more or less solved in the comment. 
The case for $A$ is harder (if you do not use the general theorem) because it does satisfy the fixed point property. To see this, let $f : A\to A$ be continuous. Let $p = (0,0)$. Note that if $f(A)$ does not contain $p$, then $f(A)$ must stay inside one of the three intervals ($p$ is the common intersection point in $A = "\perp"$, taking away $p$ gives you three connected components). For example if $f(A) \subset I_1 =\{(x, 0): -1\le x\le 0\}$, then 
$$f|_{I_1} :I_1 \to I_1$$
is continuous and has a fixed point (I assumed you know how to show this using intermediate value theorem)
So we assume $p \in f(A)$. If $f(p) = p$, we are done. If not, then $f(p)$ must lie in one of the three intervals. Assume $f(p) \in \{(x, 0) : -1\le x<0\}$. Then by continuity, there is an open set $U$ containing $p$ so that $f(U) \subset \{ (x,0) : -1\le 0 <1\}$. In particular, there is a closed interval 
$$I = \{(x, 0) : x_0 \le x \le 0\}$$
so that $f(I)\subset I_1$. Now if $f(I_1)$ does not contain $p$, then $f(I_1) \subset I_1$ and so has a fixed point. If $f(I_1)$ contains $p$, let $x_0 <0$ so that $f((x_0, 0)) = p$ and $f((x, 0)) \in I_1$ for all $x \in [x_0, 0]$. Write $f((x, 0)) = (h(x), 0)$ for all $x\in [x_0, 0]$. Define
$$g : [x_0, 0] \to \mathbb R$$
by $g(x) = h(x) -x.$ Then $g(x_0) = 0 - x_0 >0$ and $g(0) = h(0) <0$ (as $f(p) \in \{(x,0) :  -1\le x<0 \}$. Thus by the intermediate value theorem again, there is $y\in [x_0, 0]$ so that $g(y) = 0$. That is $y = h(y)$. This implies 
$$f((y, 0)) = (h(y), 0) = (y, 0)$$
and so $(y, 0)$ is a fixed point of $f$. 
Remark: One can also prove the fixed point property of $A$ using Lefschetz's fixed point theorem.
