MHB Is f a Contraction Mapping on [1,∞)?

Poirot1
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f:[1,infinity)->[1,infinity)

$f(x)=x^{0.5}+x^{-0.5}$

I thought about using MVT but it doesn't work and I've tried showing it conventially but i can't reduce it to k|x-y|
 
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Poirot said:
f:[1,infinity)->[1,infinity) $f(x)=x^{0.5}+x^{-0.5}$ I thought about using MVT but it doesn't work and I've tried showing it conventially but i can't reduce it to k|x-y|

For $1\leq x <y<+\infty$ you'll get

$\left|f(y)-f(x)\right|=\dfrac{1}{2\sqrt{c}}\left(1-\dfrac{1}{c}\right)|y-x|$

Now, use that a global maximum for $F(c)=\dfrac{1}{2\sqrt{c}}\left(1-\dfrac{1}{c}\right)$ in $[1,+\infty)$ is $K=\dfrac{1}{3\sqrt{3}}<1$
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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