Remark 1

It can happen that a uniformity $\mathfrak{U}_1$ is strictly finer than a uniformity $\mathfrak{U}_2$ but that the two induced topologies are identical. The following example shows this:

Let $X$ be a non-empty set. For each finite partition $\varpi = (A_i)_{1 \leq i \leq n}$ of $X$, let $V_{\overline{\varpi}}$ denote

The sets $V_{\overline{\varpi}}$ then form a fundamental system of entourages of a uniformity $\mathfrak{U}$ on $X$. For if $\varpi$ is any finite partition of $X$ we have $\Delta \subset V_{\overline{\varpi}}$ and $V_{\overline{\varpi}} \circ V_{\overline{\varpi}} = \overline{V_{\overline{\varpi}}} = V_{\overline{\varpi}}$ (§ 1, no. 1, Example 2); and if $\varpi' = (B_j)$ and $\varpi'' = (C_k)$ are two finite partitions of $X$, then those of the sets $B_j \cap C_k$ which are not empty form a finite partition $\varpi$ of $X$, and we have $V_{\overline{\varpi}} \subset V_{\overline{\varpi'}} \cap V_{\overline{\varpi''}}$. $\mathfrak{U}$ is called the uniformity of finite partitions on $X$. The topology induced by $\mathfrak{U}$ is the discrete topology, since for each $x \in X$ the sets $\{x\}$ and $C\{x\}$ form a finite partition of $X$. Nevertheless, if $X$ is infinite, it is clear that $\mathfrak{U}$ is strictly coarser than the discrete uniformity.