A uniformly continuous function between totally bounded uniform spaces

Let $X$ and $Y$ is a uniform spaces. Let $f$ is a uniformly continuous surjective function $X\rightarrow Y$.

Conjecture: If $X$ is totally bounded then $Y$ is also totally bounded.

Solutions Collecting From Web of "A uniformly continuous function between totally bounded uniform spaces"

It seems that the conjecture is well known and can be proved straightforwardly. Let $f:(X,{\cal E})\to (Y,\cal F)$ be a surjective uniformly continuous map between uniform spaces and the space $(X,\cal E)$ is totally bounded. Let $F\in\cal F$ be an arbitrary entourage. Since the map $f$ is uniformly continuous, there exists an entourage $E\in\cal E$ such that $E\subset (f\times f)^{-1}(F)$. Since the space $(X,\cal E)$ is totally bounded, there exists a finite subset $A$ of $X$ such that $E[A]=X$. We claim that $F[f(A)]=Y$. Indeed, let $y\in Y$ be an arbitrary point. Since the map $f$ is surjective, there exists a point $x\in X$ such that $f(x)=y$. Since $E[A]=X$, there exists a point $a\in A$ such that $(a,x)\in E$. Since $E\subset (f\times f)^{-1}(F)$, we see that $(f(a),y)=(f(a),f(x))\in F$. Therefore $y\in F[f(A)]$.