Is $\operatorname{Homeo}([0,1])$ Weil-Complete? 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!
 A: 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 $\text{Homeo}([0,2])$.
A: This is to provide a visual tool based on Sam's earlier answer, as well as point out that the distance function t.b. provides in the comments is not bi-invariant (if by bi-invariant invariance under left and right compositions is meant); indeed it's neither left nor right invariant.
In my opinion a geometric/soft approach to global matters related to homeomorphisms of the unit interval is often fruitful; for such an approach one identifies homeomorphisms with their graphs. One particular class of useful homeomorphisms seems to be piecewise linear ones, especially for the purposes of explicit calculations, and indeed the example Sam provided is using piecewise linear homeomorphisms (although there seems to be a typo in Sam's formula).

First consider the interactive graph at https://www.desmos.com/calculator/uraldtcztn. As in Sam's answer, dragging one of the points vertically to a limit demonstrates that the group $\text{Homeo}([0,1])$ of homeomorphisms of $[0,1]$ is not a closed subset of the space $C^0([0,1];[0,1]])$ of continuous self-maps of $[0,1]$, endowed with the topology of uniform convergence.
Likewise taking the two draggable points to lie on the same horizontal line and dragging them to kiss demonstrates that $\text{Homeo}([0,1])$ fails also to be open in $C^0([0,1];[0,1]])$.

Secondly let's see that the distance function t.b. provides is neither left nor right invariant. For $f,g\in\text{Homeo}([0,1])$ let's put
$$\rho(f,g)=\sup_{x\in [0,1]} |f(x)-g(x)|=\lambda(f^{-1},g^{-1}),$$
$$\lambda(f,g)=\sup_{x\in [0,1]} |f^{-1}(x)-g^{-1}(x)|=\rho(f^{-1},g^{-1}),$$
$$d(f,g)=\rho(f,g)+\lambda(f,g).$$
It's clear that all three of $\rho,\lambda,d$ are distance functions on $\text{Homeo}([0,1])$ compatible with the topology of uniform convergence. It's also clear that $\rho$ is invariant under right compositions and $\lambda$ is invariant under left compositions:
$$\rho(f\circ h,g\circ h)=\rho(f,g),\,\, \lambda(h\circ f,h\circ g)=\lambda(f,g).$$
We claim that $d$ is invariant under neither left nor right compositions. By symmetry it suffices to check left composition. Suppose for any $f,g,h$ we have $d(h\circ f,h\circ g)=d(f,g)$. Then
$$\rho(h\circ f,h\circ g) 
= d(h\circ f,h\circ g) - \lambda(h\circ f,h\circ g) 
= d(f,g) - \lambda(f,g) = \rho(f,g),$$
that is, $\rho$ itself is bi-invariant. Let's abbreviate the phrase "orientation preserving piecewise linear homeomorphism of $[0,1]$" by "$\text{PL}_+$". Let $f$ be the $\text{PL}_+$ determined by $f(1/2)=3/4$ (thus the graph of $f$ consists of the line segments connecting $(0,0)$ to $(1/2,3/4)$ and this latter point to $(1,1)$). Next let $h$ be the $\text{PL}_+$ determined by $h(1/2)=1/2$ and $h(3/4)=7/8$. Then we have
\begin{align*}
\rho(h\circ f,h)
\geq |h(f(1/2))-h(1/2)|
= 3/4
> 1/4
=\rho(f,\text{id}_{[0,1]}),
\end{align*}
a contradiction.

Indeed one can prove more generally that there is no distance function on $\text{Homeo}([0,1])$ compatible with the topology of uniform convergence that is bi-invariant. For this one can use the following theorem (see e.g. Klee's paper "Invariant Metrics in Groups (Solution of a problem of Banach)"; compare Hewitt & Ross' Abstract Harmonic Analysis, Vol. 1 (2e), p.68, Thm.8.2, there the theorem is attributed to (Garrett) Birkhoff and Kakutani):

Theorem: Let $G$ be a Hausdorff topological group. Then $G$ admits a bi-invariant distance function compatible with its topology iff there is a complete countable system $N_1,N_2,...$ of neighborhoods such that for any $g\in G$ and $k\in\mathbb{Z}_{\geq1}$: $g N_k g^{-1}=N_k$.

Since $\text{Homeo}([0,1])$ admits a distance function it is Hausdorff. To see that it does not admit a bi-invariant distance function, one can take a  $\text{PL}_+$ arbitrarily close to $\text{id}_{[0,1]}$ (w/r/t the uniform topology) and conjugate it away from $\text{id}_{[0,1]}$ using another appropriately chosen $\text{PL}_+$; see the interactive graph https://www.desmos.com/calculator/xz7ehxjdmj.
