Converse of Banach fixed-point theorem

Question

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?

Can anyone give link to proof of this problem please?

I thought on this problem some hours but I haven't any ideas. I guess that it's really hard problem.

Possible duplicate of Contraction mapping in an incomplete metric space — Weaam, Oct 27 '15 at 20:51
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? — RFZ, Oct 28 '15 at 05:01
@Weaam, So what? Sorry but how we conclude from here that $A_j\cup {0}$ is convex and compact? — RFZ, Oct 28 '15 at 06:47
For a fixed $j$, can you show that $f_j(t) : [0,1] \to \mathbb{R}^2$ defined by $f_j(t) = (t,t/j)$ is continuous? Then it will follow that $f_j([0,1])$ is also compact. — Weaam, Oct 28 '15 at 06:55
@Weaam, Yes I can. Both $t$ and $t/j$ are continuous on $[0,1]$. Hence $(t,t/j)$ is also continuous on $[0,1]$. Maybe you mean that "$f_j([0,1])$" is also compact? — RFZ, Oct 28 '15 at 06:59
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 07:11

score 0 · Answer 1 · answered Oct 27 '15 at 20:32

0

Didn't going through the details, but I think you can find the answer in this article and article cited.

answered Oct 27 '15 at 20:32

iiivooo

1 Answers1