The set of homeomorphism $h:[a,b]\rightarrow [a,b]$ has empty interior

Question

Let $C$ be the set of all continuous functions $f:[a,b]\rightarrow [a,b]$, with the metric $d(f,g)=\sup_{x \in X}|f(x)-g(x)|$. Let $X\subset C$ be the set of all homeomorphism $h:[a,b]\rightarrow [a,b]$. Show that $X$ has empty interior.

Some ideas: To suppose not, and take $h\in int(X)$. And somehow using the fact that $h$ is homemorphism to show that any $g\in B(f;r)\subset X$ must satisfy:$$b-a = d(f,g)<r$$

and this would imply contradiction, because $b-a$ is exactly the lenght of the interval $[a,b]$ which contains the image of any $g$.

Any help? Suggestion?

score 1 · Accepted Answer · answered Sep 07 '17 at 01:34

1

Let $h$ be an homeomorphism, for every $c>0$, consider $U=f^{-1}(f(b)-c/2,f(b)+c/2)$. It is an open neighborhood which contains $b$, there exists $a<d<b$ such that $(d,¸b]\subset U$. Define $g$ such that $g(x)=f(x), x<d, g(x)=f(d), x\geq d$, $g\in B(f,c)$ and $g$ is not invertible.

answered Sep 07 '17 at 01:34

Tsemo Aristide

87,475

How does this argument show that the interior is empty? for me it seems that it only shows that $X$ is not open. – user2345678 Sep 07 '17 at 01:54
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Take any ball $B(f,c), c>0$ it contains a non invertible element, so the interior is empty. – Tsemo Aristide Sep 07 '17 at 01:55

score 0 · Answer 2 · answered Sep 07 '17 at 01:34

if a function $f$ is a homeomorphism then it must have $f(a) =a$, $f(b)=b$ or $f(a) =b$ and $f(b)=a$.

Simply squeeze the function $f$ slightly so it maps to $[c,d] \subset [a,b]$ . The new function is not a homeomorphism but can be made arbitrarily close to $f$ by reducing the amount of squeezing. So $f$ is a limit of non-homeomorphisms so it is not an interior point.

The set of homeomorphism $h:[a,b]\rightarrow [a,b]$ has empty interior

2 Answers2