Is the discrete uniformity the only complete uniformity on the discrete topology? Here's a simple statement about uniformities that seems like it has to be true but which I can't quite seem to prove at the moment.  Say $X$ is a uniform space, and suppose 1. that $X$ is complete and that 2. considered as a topological space, $X$ is discrete.  Does $X$ have to be discrete as a uniform space?
Thank you all!
Edit: As per Mike F's answer, the statement is actually false.
 A: A metric space is:


*

*topologically discrete if, for all $x$, there exists $\epsilon_x > 0$ such that $d(x,y)<\epsilon_x$ implies $x=y$. Equivalently, there is no accumulation point in the space.

*uniformly discrete if there exists a global $\epsilon > 0$ such that $d(x,y)<\epsilon$ implies $x=y$. Turning this around, a metric space is not uniformly discrete if and only if the distance function takes on arbitrarily small positive values.


We want a metric space which is (1) complete, (2) topologically discrete, (3) not uniformly discrete. Let's try to achieve this with a subspace $X \subseteq \mathbb{R}$, with the standard metric structure. Let $A$ denote the set of accumulation points of $X$ in $\mathbb{R}$.


*

*$X$ is complete if and only if it is closed in $\mathbb{R}$, i.e. $A \subseteq X$.

*$X$ is topologically discrete if and only if has no accumulation point in $X$, i.e. $A \cap X = \varnothing$. 

*Combining the above, $X$ is complete and topologically discrete if and only if $A = \varnothing$. 


So our goal is:

Find a subset $X$ of $\mathbb{R}$ which has no accumulation points in $\mathbb{R}$ and arbitrarily small positive distances between points. 

Now that we know what we're looking for, it's not hard to think of examples. Here's one possiblity.

Take $X = \mathbb{N} \cup \{ n + \frac{1}{n} : n \in \mathbb{N}\}$, in the standard absolute value metric. Then $X$ is a complete metric space which is topologically discrete but not uniformly discrete.

