Dieudonné complete and topologically complete are equivalent for every space $X$. How can we show that:
For every topological space $X$ the following conditions are equivalent:


*

*A space $X$ is topologically complete if $X$ is homeomorphic to a closed subspace of a product of metrizable spaces.


*A topological space $X$ is Dieudonne complete if there exists a complete uniformity
on the space $X$.

Tychonoffness is a necessary condition to proof?
Thanks.
 A: Every uniformizable space is completely regular, so a Dieudonné complete space is necessarily completely regular, as of course is any closed subspace of a product of metrizable spaces. Moreover, a closed subspace of a product of metrizable spaces is Hausdorff, so the spaces in question must be Tikhonov.
The key result is Theorems $39.11$ in Stephen Willard’s General Topology, which in our setting says that every Hausdorff uniform space is uniformly isomorphic to a subspace of a product of metric spaces. If $\mathscr{U}$ is a complete uniformity on $X$, and $e:X\to Y\subseteq\prod_{i\in I}X_i$ is a uniform isomorphism of $X$ to a subspace $Y$ of the product of the metric spaces $\langle X_i,d_i\rangle$, you can use the completeness of $X$ to show that $Y$ must be a closed subset of $\prod_{i\in I}X_i$.
In the other direction, show that every metric space is homeomorphic to a closed subspace of a product of completely metrizable spaces. (If $X$ is metric, with metric completion $\widehat X$, consider the product $\prod_{x\in\widehat X\setminus X}(\widehat X\setminus\{x\})$.) Thus, every topologically complete space is homeomorphic to a closed subspace of a product of complete metric spaces. Suppose that $X$ is a closed subset of $\prod_{i\in I}X_i$, where each $\langle X_i,d_i\rangle$ is a complete metric space. Each of the metrics $d_i$ induces a pseudometric $\varphi_i$ on $X$ by $\varphi_i(x,y)=d_i(x_i,y_i)$, and you can show that $\{\varphi_i:i\in I\}$ generates a complete uniformity on $X$ that is compatible with the topology.
A: If one has studied uniform spaces, this solution is understandable:
Let $(X,\mathcal T)$ be a Hausdorff topological space.
$1\to 2)$ : $(X,\mathcal T)$ can (hoemomorphically) be embedded in a product of metric spaces: $$\prod_{i\in I}(X_i,d_i)$$
$\prod_{i\in I}(X_i,\mathcal T_{d_i})$ is uniformizable say with uniformity $\mathcal D$. We have:
$$\mathcal T= \mathcal T_{\mathcal D_X}$$
where $\mathcal D_X$ is the subspace uniformity (compatible with $\mathcal T_{\mathcal D_X}$) . 
$(X,\mathcal D_X)$ is embbeded uniformly in $\prod_{i\in I}(X_i,\mathcal D_{d_i})$; where $D_{d_i}$ is the normal uniformity from metric $d_i$. I denote this embedding by:
$$(X,\mathcal D_X)\le \prod_{i\in I}(X_i,\mathcal D_{d_i})$$
We know each $(X_i,d_i)$ can be embedded isometrically (and so uniformly and homeorphically) in a complete metric space $(Y_i,\rho_i)$. So
$$(X,\mathcal D_X)\le \prod_{i\in I}(X_i,\mathcal D_{d_i})\le \prod_{i\in I}(Y_i,\mathcal D_{\rho_i})$$
So $(X,\mathcal D_X)$ can be embedded in a complete uniform space. As $X$ is closed, $(X,\mathcal D_X)$ is complete too. $\mathcal D_X$ is a complete uniformity on $X$ with topology $\mathcal T$.
$2\to 1):$ Let $(X,\mathcal D)$ be a complete uniform space with topology $\mathcal T$. As $(X,\mathcal D)$ is Hausdorff, a well-known theorem says it can be embedded uniformly (and so homeomorphically) in a product of metric spaces. $X$ is closed in such a product because it is complete and the product is Hausdorff.


I define a topological space $(X,\mathcal T)$ to be Dieudonne complete iff its fine uniformity is complete. Not hard to prove that this is equivalent to  proposition (2).

