Converse of Banach fixed-point theorem

Banach fixed-point theorem states that:

If $(X,d)$ - complete metric space and $f:X\to X$ is contraction mapping. Then exists unique point $x_0$ such that $f(x_0)=x_0.$

And I have the following question:

Is there incomplete metric space in which every contraction mapping has fixed point?

• The example by Suzuki and Takahashi $S = \bigcup_{n \in \mathbb{N}} A_n \cup \{0\}$, where $A_n = \{(t, t/n) | t \in (0,1]\}$ for $n \in \mathbb{N}$. S is not complete and every continuous mapping on $S$ has a fixed point in S. – Weaam Oct 27 '15 at 21:14
• @Weaam, Can I ask you couple questions? 1) Why $S$ is not complete? 2) He proves that every continuous mapping has fixed point. Consequently every contraction mapping also has fixed point. Am I right? – ZFR Oct 28 '15 at 5:01
• Is convexity of $f_j([0,1])$ note clear? In any case, if $L_1=(t_1, t_1/j), L_2 = (t_2, t_2/j)$ in $A_j \cup \{0\}$, then $t_x = (1-x)t_1+t_2 \in [0,1]$ and thus $L_1 + x(L_2-L_1) = (t_x, t_x/j) \in A_j \cup \{0\}$. – Weaam Oct 28 '15 at 7:11