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

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    Dear Sam, Thanks for your correction to my post, with which I agree; what was written was simply wrong! Best wishes,2012-06-04
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    A small addendum: The group of homeomorphisms of a compact space $X$ is complete (and separable) with respect to the bi-invariant metric $$d(f,g) = \sup_{x \in X} d_X(f(x),g(x)) + \sup_{x\in X} d_X(f^{-1}(x),g^{-1}(x)).$$ This makes this answer interesting: the group is complete with respect to the bilateral uniformity but it isn't with respect to the right uniformity as your example shows.2012-06-04
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    @t.b.: Quite right. It is also a example of a topological group where left and right uniformities don't coincide: The left-uniform structure corresponds to the metric $d(f,g) = \|f^{-1} - g^{-1}\|_\infty$ whereas the right-uniform structure corresponds to $d(f,g) = \|f-g\|_\infty$.2012-06-04