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After learning about uniformities on topological groups, we were given several sources to read. I came across the term "Weil-complete." A topological group is Weil-complete if it is complete with respect to the left (or right) uniformity.

We learned that the sets $\{(x,y) \in G \times G : x^{-1}y \in U \}$, where $U$ is neighborhood of the neutral element, form a base of entourages for the left uniformity (similarly the sets $\{(x,y) \in G \times G : yx^{-1} \in U \}$ form a base of entourages for the right uniformity).

If $G$ is the group $\operatorname{Homeo}([0,1])$ of all self-homeomorphisms of $[0,1]$, and it is equipped with the topology of uniform convergence, would $G$ be Weil-complete?

I've been looking at this for quite some time, but haven't been able to make any progress. Any help would be greatly appreciated!

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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 $\text{Homeo}([0,2])$.