# Existence of unique fixed point in compact Metric space [duplicate]

Let $(X,d)$ be compact. Show: for a map $f$ that when $\forall x, y \in X$ with $x\neq y$

$d(f(x),f(y))<d(x,y)$ is fulfilled.

Then $f$ has a unique fixed point.

## marked as duplicate by user99914, John B, cansomeonehelpmeout, samerivertwice, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 22 '18 at 17:37

Assume there are two fixed points, let $x$ and $y$.
As $f(x)=x$ and $f(y)=y$, then $$d(f(x),f(y))=d(x,y)<d(x,y),$$ a contradiction.
• The statement $f$ has a unique fixed point means $f$ has a fixed point and it does not have more than one fixed point. Where did you prove the existence of a fixed point? – Kavi Rama Murthy May 22 '18 at 8:58