Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 9469

Contraction-like mapping without fixed point?

$
0
0

If $(X,d)$ is a complete metric space and $\xi:\;X\to X$ satisfies:

$$d(x,y)<n+1\Rightarrow d(\xi(x),\xi(y))<n$$$$d(x,y)<1/n\Rightarrow d(\xi(x),\xi(y))<1/(n+1)$$for all $n= 1,2,\dots$, does $\xi$ necessarily have a fixed point?

I am not sure what to do with this question; I would guess the answer is no, but I am having trouble constructing a counter-example. If someone could offer a hint I would be thankful.


Viewing all articles
Browse latest Browse all 9469

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>