Is $\operatorname{Homeo}([0,1])$ Weil-Complete?

Solution 1:

The answer is no. That's simply because a uniform limit of homeomorphisms need not be a homeomorphism (for notational convenience, I shall consider $[0,2]$ instead of $[0,1]$):

Let $$\varphi_n(x) = \begin{cases} \frac xn & \text{for }x\in \left[0,1\right] \\ 2\left(1-\frac{1}n \right)(x-1)+\frac1 n & \text{for }x\in [1,2] \end{cases}$$

and $$\varphi(x) = \begin{cases} 0 & \text{for }x\in \left[0,1\right] \\ 2(x-1) & \text{for }x\in [1,2] \end{cases}$$

Then $\varphi_n \in \text{Homeo}([0,2])$ for all $n$ and $\|\varphi_n - \varphi\|_\infty \le \frac 1n \to 0$. So $\varphi_n$ is Cauchy, but since $\varphi\notin \text{Homeo}([0,2])$, the sequence cannot have a limit in $\text{Homeo}([0,2])$.