Coarsest topology in Uniform Space

Question

The topology on $X$ induced by the coarsest uniformity $U$ for which the the mappings $ f_i ,i\in I $ are uniformly continuous is also the coarsest topology for which the $ f_i ,i\in I $ , are continuous?

Answer 1

It is well-known that the coarsest uniformity $\mathcal{U}_X$ induced by maps $f_i: X \to (Y_i, \mathcal{U}_i)$, $i \in I$ is generated by the subbase of all sets of the form

$$\mathcal{S}=\{(f_i \times f_i)^{-1}[U]: i \in I, U \in \mathcal{U}_i\}$$

which has as a base all finite intersections of such sets from $\mathcal{S}$.

Now let $\mathcal{T}_1$ be the topology on $X$ induces by this uniformity and let $\mathcal{T}_2$ be the coarsest topology that makes all $f_i$ continuous (w.r.t. to the topologies $\mathcal{T}_i$ on $Y_i$, induced by $\mathcal{U}_i$, of course).

Now, $\mathcal{T}_1$ is some topology on $X$ that makes all $f_i$ continuous (even uniformly continuous!) so by minimality of $\mathcal{T}_2$ we already have almost for free that:

$$\mathcal{T}_2 \subseteq \mathcal{T}_1$$

Now let $O$ be open in the topology induced by $\mathcal{U}_X$ and let $x \in O$. It follows there is an entourage $U \in \mathcal{U}_X$ such that $x \in U[x] \subseteq O$. So there are finitely many $i_1,\ldots i_n \in I$ so that $$U_1 = \bigcap_{k=1}^n (f_{i_k} \times f_{i_k})^{-1}[V_{i_k}] \subseteq U$$ for some $V_{i_k} \in \mathcal{U}_{i_k}, k=1,\ldots n$.

Now define $O_{i_k} \subseteq Y_{i_k}$ by $O_{i_k}=V_{i_k}[f_{i_k}(x)]$, which are neighbourhoods of $f_{i_k}(x)$ (not necessarily open) in $Y_{i_k}$ and now we claim that

$$x \in \bigcap_{k=1}^n f_{i_k}^{-1}[O_{i_k}] \subseteq U_1[x] \subseteq U[x] \subseteq O\tag{1}$$

All are obvious, and checking the first inclusion is just unpacking definitions: Let $x' \in \bigcap_{k=1}^n f_{i_k}^{-1}[O_{i_k}]$. Then for each $k\in \{1,\ldots,n\}$ we have that $x' \in f^{-1}_{i_k}[O_{i_k}]$, from which follows $f_{i_k}(x') \in O_{i_k} = V_{i_k}[f_{i_k}(x)]$ so $(f_{i_k}(x'), f_{i_k}(x)) \in V_{i_k}$, or otherwise put $(x,x') \in (f_{i_k} \times f_{i_k})^{-1}[V_{i_k}]$ and as this is true for all $k$, $(x,x') \in U_1$ by its definition, or $x' \in U_1[x]$.

But $(1)$ implies that $x$ is an interior point of $O$ w.r.t. $\mathcal{T}_2$, as the intersection of finitely many neighbourhoods of $x$ under the $f_{i_k}$ is a neighbourhood of $x$ in $\mathcal{T}_2$ (and each $f_{i_k}^{-1}[O_{i_k}]$ is a neighbourhood of $x$ by continuity of $f_{i_k}$ under $\mathcal{T}_2$). As $x \in O$ was arbitrary, $O \in \mathcal{T}_2$ too and we have shown

$$\mathcal{T}_1 \subseteq \mathcal{T}_2$$

so we have equality of these topologies. So one inclusion is by minimality and uniform continuity implying continuity, the other is because we have a concrete description of the coarsest uniformity. It's not hard, just a bit tedious.