topology induced by uniform structureTopologyInducedByAUniformStructure2002-06-10 19:49:182007-02-20 18:23:11Derivationuniform spaceuniform topology% this is the default PlanetMath preamble. as your knowledge
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% define commands hereLet $\mathcal{U}$ be a uniform structure on a set $X$. We define a subset $A$ to be open if and only if for each $x \in A$ there exists an entourage $U \in \mathcal{U}$ such that whenever $(x,y) \in U$, then $y \in A$.
Let us verify that this defines a topology on $X$.
Clearly, the subsets $\emptyset$ and $X$ are open. If $A$ and $B$ are two open sets, then for each $x \in A \cap B$, there exist an entourage $U$ such that, whenever $(x,y) \in U$, then $y \in A$, and an entourage $V$ such that, whenever $(x,y) \in V$, then $y \in B$. Consider the entourage $U \cap V$: whenever $(x,y) \in U\cap V$, then $y \in A \cap B$, hence $A \cap B$ is open.
Suppose $\mathcal{F}$ is an arbitrary family of open subsets. For each $x \in \bigcup \mathcal{F}$, there exists $A \in \mathcal{F}$ such that $x \in A$. Let $U$ be the entourage whose existence is granted by the definition of open set. We have that whenever $(x,y) \in U$, then $y \in A$; hence $y \in \bigcup \mathcal{F}$, which concludes the proof.