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.

enter image description here

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 $\epsilon_n / 6$.

  • $\begingroup$ 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 $X$? $\endgroup$ – Avijit Dikey Sep 18 '20 at 7:17
  • $\begingroup$ 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 $X$. $\endgroup$ – eatfood Sep 18 '20 at 7:38
  • $\begingroup$ Also, your comment made me realize that more justification for why $Y$ is closed might be needed. $\endgroup$ – eatfood Sep 18 '20 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.