# Converse of contraction mapping theorem

Let $$(X,d)$$ be a metric space such that, if $$Y \subset X$$ is closed, then every contraction mapping on $$Y$$ has a fixed point. Show that $$X$$ is complete.

This problem appeared in an exam and I attempted a solution, but admittedly was not able to get very far. I started by letting $$X$$ not be complete and considering a Cauchy sequence in $$X$$, $$\lbrace y_n \rbrace \rightarrow y, y \notin X$$. Then, maybe we could construct a contraction mapping which would have fixed point as $$y$$, i.e $$f(y)=y$$, and arrive at a contradiction that way.

Suppose $$X$$ is not complete. Then, take a Cauchy sequence $$(x_n)$$ that does not converge in $$X$$. The set $$Y = \{x_n : n \in \mathbb{N} \}$$ is closed in $$X$$, so we will work on this subspace. To show $$Y$$ is closed, we show that the complement is open. So let $$w \in X \setminus Y$$ be any arbitrary point. If we can find an open ball center $$w$$ radious $$\epsilon$$, denote $$B(w, \epsilon)$$, that does not contain any points in $$Y$$, then there is nothing to do. Otherwise, if $$w$$ is chosen so that no such ball exists, then $$w$$ is a limit point of $$(x_n)$$, but this contradicts our choice of $$(x_n)$$! Hence, $$Y$$ is closed.

Now, we an always replace $$(x_n)$$ by a subsequence of itself. So WLOG, we can assume the elements of $$x_n$$ are all distinct, since we can always throw away duplicates. After that, lets replace it with this special subsequence $$(y_n)$$, which I will describe later. This special subsequence will be chosen so that the map $$f(y_n) = y_{n+1}$$ is a contraction map. Then by hypothesis, $$f$$ must have a fixed point, and it can't be any of the $$y_n$$ (because then $$y_{n+1} = f(y_n) = y_n$$ contradicts the distinctness of the sequence). Thus, $$(y_n)$$ has no fixed point, which contradicts the hypothesis.

Now, it remains to find this subsequence $$(y_n)$$, and show that $$f$$ is contraction map. The idea is that we want consecutive terms of $$(y_n)$$ to not be close to each other. Take $$y_1 = x_1$$. Then let $$\epsilon_1 = \inf\{ |x_1 - x_m| : m > 1 \}$$. If $$\epsilon_1 = 0$$, then $$y_1$$ is the limit of $$x_n$$, so this Cauchy sequence has a limit. Otherwise, $$\epsilon_1 > 0$$. Then, by Cauchyness of $$(x_n)$$, you find $$M$$ sufficiently large so that for all $$n, m \ge M$$, $$|x_n - x_m| \le \epsilon_1 / 6$$. Then, discard all the elements $$x_i$$ with $$2 \le i \le M$$ from the sequence $$(x_n)$$. Now, with this modified $$(x_n)$$, we take $$x_2$$, and repeat the construction to inductively create a sequence, which we denote $$(y_n)$$.

Intuitively, what we have done is to modify $$(x_n)$$ so that $$x_1$$ is 'isolated', the remaining elements of $$(x_n)$$ are all clustered in some ball of radius at most $$\epsilon_1 / 3$$, and $$x_1$$ is at least distance $$\epsilon_1$$ away from this cluster. Try to draw a picture to see this.

Then now $$f(y_n) = y_{n+1}$$ is contraction, because for any $$n, m$$, $$|x_{n+1} - x_{m+1}| < \epsilon_n / 6 < \epsilon_n / 2 < \frac{1}{2} |x_n - x_m|$$ The first inequality is because $$x_{n+1}$$ and $$x_{m+1}$$ are contained in the ball constructed while choosing $$x_n$$. And so their distance is at most $$\epsilon_n / 6$$.

• I don't understand why $Y$ is closed. If $Y= \lbrace x_n \rbrace$ is closed, wouldn't that imply that $\lbrace x_n \rbrace$ is convergent to a point in $X$? – Avijit Dikey Sep 18 '20 at 7:17
• That implication does not hold. Consider $X$ the rational numbers, with the subspace topology from $\mathbb{R}$. Let $(y_n)$ be a sequence of rationals converging to $\sqrt 2$. Then $\{ y_n \}$ is a closed subset of $X$, but does not converge to any point in $X$. – eatfood Sep 18 '20 at 7:38
• Also, your comment made me realize that more justification for why $Y$ is closed might be needed. – eatfood Sep 18 '20 at 7:40